 percentage of CNT such that in the slide it can be seen that some reactions only produce about 17 pure torsion of fog over the reaction, generally they look at how much is the reactants getting decreased and how much are the products getting fogged and and as this process happens with time whether it is half slope or reactions is much more detailed but we are just trying to understand the cons of the evaporation. Let's look at some equivalents for example you know the water is going to change to gaseous water changes with time then we say that the rate of reaction is very low but basically clear water and gaseous water are in less gaseous producing liquid this is this is one type of equilibrium the other type of equilibrium for gas which generally associates into twice of an order so when end forming an order and settles down we call it as how this concept really looks like for example you know let's say we had a cookie store and in a cookie store we found that going forward the production or addition is the rate of addition of cookies to the store so as the chain keeps on eating cookies we have to add more cookies right then someone else comes up and he has some more of them support and so on and so forth we keep on adding for our store so that the this is called as equilibrium so if we had decided that you're going to have a preside upon it it is getting consumed so will we have leads us to understanding of how separation of is same that the rate of sale of cookies but there's small expression you know when we have so how does this brown come out so n2o4 breaks down into molecules which is what gives it the brown color so let's look at the beginning of this some molecules of n2o4 and you can you can find that realize that the screen is there so the n2o4 is less and the brown one are the starts getting established you will find that the white molecules are formed their concentration under which no matter how much time you wait the color does not really you know change so that's the position where we say that they have done in such a fashion that you know we don't have any distinct happenings understand what what do we mean by law of mass action okay so what law of mass action basically means is that the rate of reaction is proportional to the product of the active masses of these reactants so what is an active mass generally active mass is the substance that actually is participating in the reaction but in our situation active mass most of the times we will take as the molar concentration you know raise to power of that number of species let's look at some examples so for example in the in the example a giving p plus q it is proportional to k1 into a to the power of 1 okay because we only have one as the coefficient of a but on the other side we give the rate that is k2 into a into b okay so depending on the different reactions you will find that the rate actually depends on you know you know it was the product in 2a plus b you will find that the you know the the component of a is basically raised to the power of 2 because we have 2t going forward right so 2a plus b is what gives us the rate of reaction so coming back to understanding what is active masses active masses is nothing but your molar number of species in our situation the molar concentration is actually you know so when we write square bracket of a it essentially means the molar concentration of a going forward okay so that is the way that point which is the rate of reaction where n204 is equal to equilibrium happen okay so the rate of forward reaction becomes equal to the rate of reverse reaction which when equilibrium is achieved okay as the formation so how leads to formation of the molecules of n204 okay n204 formation is what shows that it is pretty dynamic which means that the forward reaction is actually equal to the backward reaction both of them are happening simultaneously forward reaction going going up can be written by a small half arrow so this shows us the backward reaction right now the idea of writing half arrows is to show that both of the reactions are happening simultaneously so half arrows actually mean that the reactions can be going in each of these directions but at the same time the second reaction also has to be you know accommodated in the process now come back to equilibrium concept one more time is that you know a decreases to a constant while the concentration of b which is the product increases from zero to a constant give a constant concentrations which will be concentration of n204 is actually decreasing with time concentration has come up from a large one to a lower one but as as you really find n204 also keeps on increasing and at both block of slightly grayish color that you can see here is there the concentration of both n204 are stable okay so from this point onward with n204 you'll have the concentrations of n204 much much larger than the concentrations of n202 so this is the point where n204 is starts becoming constant after a certain point of time but you'll realize that the concentration of n204 also has increased so you'll realize that their concentration decreases while the other one's concentration is definitely going to increase and depending on whether they are further than equally beyond equilibrium then the backward reaction will dominate if they are before equilibrium then the function is going to dominate equilibrium have any concentration of of so if you really see this the equilibrium constants achieved are 0.204 equilibrium concentration achieved of n204 is 0.0898 now if you really look at n o2 by n204 less n o2 square by n204 turns out to be 4.63 so this basically is showing us how the equilibrium actually the value of n o2 square divided by n204 experimentally all of them are more or less the same what basically is changing is the concentrations of n o2 and n204 and the equilibrium constant concentrations again are different because you started with different substances and keeping the you know the entire divided by n204 actually makes the number to be constant right so this is constant now how do you really write an equilibrium constant no matter where do you start from the final expression in equilibrium constant is seems to be the same so for example you know for any general reaction a a plus v b giving p p plus q q where a small a small b small p and small q are the coefficients and capital a capital b capital p and capital q are the concentrations of these are the concentrations the concentrations at equilibrium okay these are not the concentration where you started or where at at any point in the in the entire system these are concentrations at at equilibrium so once the concentrations are you know which are at equilibrium you'll realize that the expression p to the power p and q to the power q divided by a to the power a divided by b to the power b this remains constant you know and this is constant for a certain temperature so it remains constant for the temperature and drink no irrespective of where do you start in this reaction now let's look at n204 giving twice of NO2 okay now for this expression our equilibrium constant is NO2 square divided by n204 concentration okay now can the value of this it turns out to be experimentally equal to 4.63 into 10 to the power minus 3 a plus bb cc plus dd you know this expression is what we call as law of mass action as well as we have seen in the earlier situations we are talking only about the rate of reactions here we are actually talking about the equilibrium where rate of forward reaction rate of backward reaction both are being considered okay now in the great constant and one then we'll realize that the reaction is almost proceeded to the product and therefore the equilibrium will be will lie more towards the right hand side when we say right hand side the flux is going to be much much larger than the concentration of the reactants whereas if you you know it has pivoted the products but if the rate constant is much less than one then we'll realize that this equilibrium is going to lie much to the left and therefore it would have favored the reactants more than the products okay so these are the different ways that you can you know really work out with this right there are sometimes homogeneous equilibrium now equilibrium are different types okay the large two types are homogeneous and heterogeneous in homogeneous equilibrium we basically find that the the phase of the reactants and the products both are absolutely the same okay so you have n204 gaseous giving twice of n204 gaseous okay going to be the same you know in the gaseous form now if we write an expression for them the constant definitely be written as n02 squared divided by n204 but there is a different type of expression that we can also write in terms of their pressures where we can simply write the pressure part pressure of n02 squared divided by the pressure of n204 if you really see the you know and this is also one of the questions that that we had in our exam coming up so if we if you really see that that the pressure kp kp basically means is that the equilibrium constant in terms of in terms of it may not be equal to kc of course because in terms of pressure so their values have to be different but both of them can actually help you understand equilibrium in much better fashion right now you will also realize that they can both be related to each other in terms of kp being equal to kc into rt to delta n how basically happens is you know if i if i may have to you know that pv is equal to nrt and in this you know if we divide n by v this is nothing but it turns out to be our concentration so p into n by v into rt is nothing but crt right so when we absorb concentrations of p you are there maybe 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 rt okay so this is p by rt and we can write cb also as pb by rt and write cp as pp by rt and so on so forth right similarly the last one is pq by rt now you will realize that you end up doing an expression which is p a into pb divided by p p in rt into power delta n this delta n is basically the difference between the moles of the product will be at the top and reactants will be at the bottom so it is the products minus the reactants that will survive of rt right and this can basically be taken minus reactants and this is basically taken in our you know slide here where we write this as rt is equal to you know rt into the power delta n where delta n is moles of gaseous products and minus moles of gaseous reactants okay v minus a plus b now in homogeneous reactions cs3coh plus h2o we will find that you know h2o is a huge so this was our original reaction let's call it as casey dash this reaction is not very visible because h2o is a huge constant you know the concentration of h2o is much much larger than any other concentration so you'll find that the casey turns out to be cs3coo minus into h plus divided by cc0oh so here is a quick rule that we should actually remember of course the general practice that we do not take up any units in the equilibrium constant the units are summarily excluded although equilibrium constants do have units for example in this scenario the unit of equilibrium constant is moles per liter why because the concentrations would be in moles per liter into moles per liter divided by moles per liter we actually taken out so therefore casey will end up having only moles per liter in this scenario now the equilibrium expression of how do we write equilibrium expression for some reactions for example here in n2 gaseous plus 3h2 gaseous twice of nh3 the equilibrium constant can be written as ns3 square divided by n2 into you know h2 square okay no problem you know to understand this in much more depth the equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 at 74 degrees Celsius are CO giving 0.01 to moles chlorine is 0.05 for moles and CO and 1.04 moles calculate the equilibrium constants casey and kp right now we have been given all the concentrations of COCl2 the first thing that we need to write it is the chemical equation and it is very important to write the chemical equation in a balanced format because only in the balanced format will you realize that what are the coefficients that are necessary to you know have this reaction happening now you'll find that casey can be written as COCl2 divided by CO into Cl2 now if we add the reactions you know concentrations at equilibrium which is nothing but 220 right now in the same thing if we write kp you know then we'll have to take up all the gaseous products and we'll have to write delta n as well if you can see delta in here is nothing but 1-2 because we have only one mole of gaseous products whereas we have two moles of gaseous reactants so you know is the delta n that we have and therefore we'll end up getting delta n as minus 1 so you'll realize that the relationship between kp and kc of course we'll write t value but the relationship between kp and kc is rt to the power minus 2 turns out to be 7.7 you see kp is 220 but kp turns out to be 7.7 you know at the end of the you know calculations let's look at one more example the same you know NO2 example but this time NO2 is actually decomposing to give gaseous oxygen now what is the equilibrium pressure of oxygen is what is being asked if this and pressure of NO is this right then you'll realize that the kp that we can write is pressure of NO2 square into pressure of O2 divided by pressure of NO2 square right look at this this expression you'll find that the pressure of oxygen is nothing but kp into pressure of NO2 divided by the pressure of O2 if we if we know oxygen pressure of oxygen 158 pressure of definitely given to a relatively pressure of you know NO is even substituting the values we end up getting pressure of O2 as 340 atmosphere so expression 2 to be really deal with you know equation in different phases so you'll see that let's let's take one example CO3 giving CO plus CO2 in this expression you will find that not in the gaseous form in the solid form whereas one is in the gaseous form so you realize that the one that is in the gaseous form you know acts creating a different concentration you know these are pretty low whereas the concentrations of solids and liquids are pretty high so if we write the expression of equilibrium here we will get CO2 into CO plus CO3 but CO3 and CO are very high in concentration so generally we would neglect their change with time and therefore we'll take them as constant and therefore we own concentration of CO2 which is Kc dash into CO3 into CO now given like p the kp value will simply be the pressure of CO2 that is there in this you know equation still included in the expression for the equilibrium constant okay now this can also be looked at in in the in the graphical format if you really see CO3 giving CO plus CO2 you realize that CO3 and CO are being solids in the entire equilibrium would not really matter right so the pressures of only CO3 and CO are what is going to matter and you'll find that these pressures you know is it will change with time so what we can simply it is pCO2 being constant very quickly achieved because concentrations of CO3 and CO are pretty even think of the concentrations of CO3 or CO expression so the thing that we do is we write concentrations of reacting species in the condensed phase you know almost do not appear in the concentrations of the substances expressions per se and and therefore you'll find that expression is not something that we really when we quote value for equilibrium constant you know you we must definitely specify the balance equation and the temperature because 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 the last is that if the reaction is expressed as the sum of two or more reactions then the equilibrium constant for the entire that is basically a product of the equilibrium constant concentrations called as the IC method in this we firstly it's a trick of the species one species has changed what will be the different changes in other species and after that at equilibrium when we find the concentrations we find out how much of it was really changing in the entire time for example if you have Br2 giving twice of Br you know we would have the equilibrium constant for this reaction let's say is 1.1 into 10 to the power minus 3 and the initial constant taken is 0.06 that we had taken initially now let's see how do we really solve for this question okay so you'll find that let's actually the change in the concentration of Br2 this Br2 will come up to become twice of Br the concentration was 0.63 and 0.02 so this is called as the I in the IC method the change let's say one Br2 decomposes then twice of Br is formed right so whatever is the decomposition that happened you know twice of that will be Br because per mole of Br2 decomposition twice of Br is getting formed now in that scenario 0.63 minus x while the Br form will be 0.012 plus twice of x now I think is where we write the initial the change and the equilibrium one below the other and you'll find that you know KC would work out to be you know 0.012 which is theorem concentration square you're about 0.063 minus x that is 1.1 into 10 to the power minus 3 okay this is actually a quadratic expression but that's okay you know you can always use the root of quadratic expression as minus v plus or minus root of v square minus 4 c upon twice a now with that you'll end up getting x is equal x two solutions for x which is minus and plus now 0.0105 and this thing now obviously x is minus it implies that forward but the reaction actually went backward and therefore at equilibrium you'll find that the bromine is actually you know minus 0.09 molars is what and so what means is that that cannot be really negative you know so the only factor that we can have is 0.00844 mm is what the solution could be right so you'll have to take the second value of x and you will realize that you know since the concentration of bromine was getting to minus we cannot take 0.010 the value of x but we'll have to take 0.00178 as the value of x and find the concentration of bromine once we have found the concentration of bromine you can be similarly found out by putting the value of x and in fact we know that you know the reaction went backward to actually form more of br2 than twice of br so that is the i c ratio similarly you know for one more such expression again twice of hi giving h2 plus i2 the expression for you know k equivalent for this reaction would be taking i2 product divided by the hs square where we will have minus 2x giving plus x and plus x because initially we only started with hi this is 2.42 minus twice of x and xx at equilibrium so therefore our key equilibrium will be pressures of x2 divided by 0.242 minus twice of x the whole square to give you an expression of 12 divided by you know 0.242 minus twice of x whole square now actually equal to the equilibrium constant that we have which is 1.26 into 10 to the power minus 3 so you'll realize that here we were asked for the equilibrium concentration of hydrogen which is this x so we just need to solve for x and we will get the answer for it also a bit of rearrangement actually really does help and you will find that you know some mathematical tricks can be used here for example we can you know we can solve this expression by actually neglecting a few values taking out all the 10 to the power minus 3 so you multiply the entire expression by 1000 you'll realize that you will end up getting 0.995x square plus 1.22 into 10 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 root of quadratic equations to solve this problem and we end up getting x in terms of 0.00802 so of course the physical system the concentration of hydrogen cannot be negative our answer here only is 0.00802 as concentration of hydrogen at equilibrium similarly you know we can also solve it for the end of the problem here we can calculate kp in kp we will simply write the ics in terms of shares and we can i would like you guys to stop pause the video here and actually work out this problem and see if you are able to solve this with kp for this reaction okay now moving on from here let's say k the k equilibrium is really small for a particular reaction basically it means that it's 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 you know always approximate you know 0.20 equal to 0.20 only because that's what is really going to give you the concentration of reactants in kp equilibrium initial concentrations is around 3 orders of magnitude or more then you can always go for this approximation and you realize that this approximation always works out to give you the solutions for this you know problem you don't need to do much calculations but easy calculations can be early work okay since you'll have to use the quality to really solve the problem now again looking at you know i2 and 2i here we'll you know if you really look at the problem we have to calculate the concentrations at equilibrium at 600 kelvin and 2000 kelvin where the kc for both of these expressions have been given so the kc for 600 kelvin is given as 2.903 power minus 10 whereas the kc for 2000 kelvin is given as 0.029 okay 0.209 so having said this is the kc values are different question you'll find that we will have to write the ic's for the different equilibrium the ic's will be the same but we'll relate them to the solutions of different equilibrium for example in the first scenario we will relate the i2 to i2 to 2.904 power minus 10 whereas in the second one in the first one you'll realize that it is almost if you know three orders of magnitude between the initial concentration and x and therefore so all we can do is we can write this as 2x square is equal to 0.20 minus x can be written the same as 0.20 and you know 2x square divided by 0.20 can be equated and the equation can be solved for x but for the second equation which is 209 we cannot do this approximation because it will be very close to 0.20 and therefore our answer will go wrong so the approximation can only be used if there is a difference of three orders of magnitude going forward so in this scenario we go through the quadratic method and we get our roots for our solution now let's understand what happens when you know you have you know concentration of two reactions working together for example a plus and c plus d giving a e plus f in this scenario let's say the equilibrium concentrations are kc dash and kc double dash right so our concentration for kc double dash which is cd and ap or double dash divided that is equal to ef divided by cd you realize in this scenario you know you know ad are common to each other so if we add this reaction the net reaction that we get is only a plus b giving c e plus f and e into f divided by a into b but if you really you know also look at the multiplication of both of these kc double dash if you use the product the c and d at numerator and denominator actually cancels out and therefore the net reaction can be of disease of their individual reaction so the equilibrium constant net is nothing but the product of individual reactions so if the reaction is expressed as a sum of two or more reactions the equilibrium constant of the overall reaction is actually given by a sequence of individual reactions so expression equations this is the that we have been looking in looking from beginning so n204 giving twice a panel two will be divided by 204 which is 4.68 into 10 to the power minus 3 in the reverse format our equilibrium constant changes which is n204 divided by n2 square now which will be basically one by k and before 216 so please note that change in the values from 4.63 into 10 to the power minus 3 it has gone almost to 2.16 so when the reaction is written in the opposite direction the equilibrium constant becomes the rest is equal of the equilibrium constant of the original one the reaction course related by substituting the initial concentrations of the reactants and products say you know we do not know whether the equilibrium is being achieved or not what we will generally do is you will you will find the reaction quotient at any point of time so what is 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 have so for example you know when we have qc that is greater than kc the ultimately we proceed to the left from right this is because you know the products are more in number than the reactants and hence the you know expression has to go towards the towards back downward similarly if you have now qc that is equal to kc the system then is at equilibrium and you will realize the good so the reaction is not going to move towards any direction as such the last is when qc is actually less than kc we know that the products are less in amount than the reaction will grow so forward which is towards left from left towards right this is the different ways to actually look at qc and kc so when qc was basically qc has to achieve kc and therefore it will now when qc and kc are the same there is no net change and when qc is more than the kc the reaction you know the reaction has to proceed more towards reactants than products to get the value of kc that is necessary the next you know point of this topic is to understand layshacklier's principle where we will understand whether there is an external you know to the equilibrium so let's say equilibrium and we change some parameters of the system then you know whatever we change we is going to you know disturb the equilibrium and what the equilibrium then is to nullify that change that we brought in you know at this point of time the here by external space we mean that some parameters of the equilibrium and it is what the equilibrium is what layshacklier so how do we do this is we change for example you know let's say we change in the concentration you know so let's look at this example you know the end to end to the Hebert process now in this scenario if we add more of ammonia obviously the reaction will have more of the concentration now and they try to go towards the different side producing more of so here's the in the time that everything was the same the night concentration hydrogen and ammonia were equal suddenly added ammonia so the there is a spike in the in line here line you realize that the concentration of suddenly increase now they will try to you know balance the concentration of the concentrations of hydrogen and nitrogen so so this is the way is that you know you can actually have the hydrogen is now improved from from from the earlier ones to much more right similarly the nitrogen concentration also has increased okay now let's look at the pressure you know for for a general reaction like a a plus bb giving cc plus dd in this general expression if you start adding uh you know uh in in the uh the reaction is to move towards the different sides so increase in concentration of products it's going to shift the equilibrium towards left similarly increase in if you remove the products the equilibrium is going to shift towards the right hand side the decrease in the production of products you will be able to try and shift towards the right hand side right similarly if you add reactants the equilibrium right hand side but if you add products or you remove the reactants they will shift towards the left so this is one way where you have you know the concentrations of a b c and d looked at uh in different dimensions either you change the reactants or the products as an effect of both of them happening okay uh let's look at the aspect okay especially in case of situations where you have gauges reactants and products so whenever you increase the pressure with the fewest small gas will it will go to right so in this is a decrease of pressure less moles and therefore the equilibrium will decrease the pressure it will shift more towards a and t because more number means more the pressure so if you're decreasing the prep to increase the pressure and therefore you 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 the volume you'll find that it will go to the sides which has fewest number of moles which is basically c so this is what lech atelier lech atelier's principle says is that the the at equilibrium if you do bring in any changes you will find that you know the reaction moves in that direction so as to nullify the change let's have our system you know n204 and n02 here is you know the n204 system is where if you increase the temperature you take this you know the d of k degrees why because if you are increasing the temperature you know n204 giving you know is is actually a reaction which is favored by temperature if you increase the temperature more of n02 is formed and therefore it is more brown in nature as you can see in this hotter bulb but if pressure more of n204 is formed so this reaction is favored by the increase in temperature and therefore the k becomes much more as because the exothermic process okay so similarly in an similarly if you process you know if you increase the temperature your k is going to decrease decreasing k means that you know you will have less of products and more of reactants okay so temperature actually matters and equally for endothermic to thermic processes similarly for decrease in temperature it will be the reverse because the temperature for exothermic processes you know your k is going to increase because it will want to have more of the products why because it's an exactly process so you decrease the temperature it will try to warm the system up and process of forming system which is going to create more heat but heat is created only while you form products and therefore more of products will be formed okay now this is a typical energy diagram for a catalyst now adding a this actually does no difference you know it only increases the speed of forward and backward reaction but having said the reaction simply keeps on moving in in both the directions in with larger speed having said this you know there is a little change in terms of the concentrations or the equilibrium does not shift only what happens is the activation energy changes so it does not change the k it does not shift the position of equilibrium only the activation energy changes and the equilibrium is achieved faster so you can see that the catalyzed reaction lower activation energy whereas on catalyzed as a higher activation energy so catalyst lower is both ea for forward and reverse reactions and equilibrium has other difference in the entire system okay now we have we have seen the habit process quite quite some time right so we have we know that nitrogen and hydrogen basically gives ammonia that is delta it's basically endothermic process of the form of ammonia so basically sorry it's an exothermic process so you have ammonia as ammonia is formed heat is given out almost 92.6 kilojoules per mole to realize that as you increase the pressure you'll find that the mole percent of ammonia actually increases why because as you increase the pressure the system will want to go towards that side which will have less number of molecules which is nothing but ammonia so increasing pressure actually helps similarly if you also see if you increase in the temperature the green shows N2N H2 whereas NH3 red so if you keep on increasing the temperature you'll find that the moles percentage of NH3 decreases but this is this process is exothermic so clearly temperature exothermic would resist that you know entropy you know they would want to have higher temperature because they can absorb more of heat but the exothermic you know that they will they will not be favored by this reaction so basically if you really see you know this is one of the reasons why we actually keep on taking as much as possible while you know this process keeps on happening because as we take on ammonia you'll find that N2N H2 will keep on reacting more because we are taking out substances from NH3 equilibrium so let's just happen so if you if you change the concentration yes the equilibrium shifts but the equilibrium constant does not change if you 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 there is no change in equilibrium constant but if you find that the equilibrium does shift but there is also a change in equilibrium constant so equilibrium constant does depend on the temperature but it does not depend on the catalyst as a final one so if you change if you put the So this brings us to the end of this presentation and we will be sharing some more problems with you pretty soon and I hope this will help in understanding the equilibrium chapter. I would have any questions you please post it out on the thank you so much. I'll get back to you.