Yeah, hi everyone. I can see that there are about five of you there Just a quick thing all those who are actually watching can please comment in the section so that I at least know and I'm aware Also, whatever topic you feel that you Want me to revise today, you know apart from the two things that we are already doing which is thermodynamics and chemical equilibrium You can please comment in the post, you know So there are two comments that are there on the YouTube one is a public comment And second is a live chat. You can use any of those that should not be a problem so just post, you know, who are a lot present if you can just post your names or You know, you can just post And name and if there is any topic or any question that you have any query you have you can also post that so that I can You know in the subsequent Session I can actually I mean as I go through the topics I can actually deal with them accordingly so I'm also going to post the same in the group Yeah, yes the class has started There are a few queries where people want to know that whether the class has started or not Okay Good. Okay. So Yes, let me check if you Have any queries as such Fantastic, so Sachi is here Mansi is here show me kiss your Arpita is there Ananya Sir, can you revise organic chemistry? Okay, good. I'll do that Sachi Okay, Ronak is there fantastic. So I think there are a few few guys and there is a request for organic chemistry And I think Aditya Aditya also wanted me to do but he is incidentally not there on the You know on the Call or this class today, so maybe you wanted me to do with thermodynamics, okay Aditya is also there fantastic Aditya you wanted thermochemistry is it Yeah, all anyone who wants any topic to be revised. We have two next two hours and you know the good part of an online session is that You know this will be recorded for one so you can always go back and revise it The second thing is that I have enough bandwidth to actually take in topics Which we all can do together, you know So there is no hurry as as happens in the physical class so we can deal with all these topics One by one and you know, you would also be comfortable having them together. Okay, so just please post on any comments Sachi is saying organic chemistry. So I'm gonna start with organic chemistry because that's what she wants Yeah, so give me one minute. I'm just sharing my screen with you guys. Yeah, there it is now Yeah, in a moment, I think you should be able to see my screen because I have just shared that Yes, so I think you guys should be able to see the screen now If you don't have any screen, there are two ways that you can quickly reach out to me because by the time you reach out to me Maybe I go to the next, you know, I have gone a bit ahead So feel free to immediately reach out one is of course, you can put up a WhatsApp message So I also have my you know, what's up with me or second You can just comment in the YouTube section from time and again every five ten minutes I'll be checking the YouTube section so that I am aware that You guys are there. Okay. I sure is also there. That's good That's nice. I think yeah, so there are more than enough people now Let's begin. So this is for organic chemistry because you guys had a query on that Okay, so One second Right. So I think most of the most of you do not have a problem in identifying the compound So the first thing in organic chemistry is basically classification. Okay, so what where? Where we are able to deal with what are the different types of compounds that we have, you know Where majorly there are saturated and unsaturated We also have benzene like compounds, but thankfully you don't have much of You know Equations on them. So basically we have saturated and we have unsaturated Saturated means something that cannot be add all those compounds where anything cannot be added any further The name itself says that it is saturated now saturated compounds definitely mean alkene Whereas unsaturated means your alkene and alkyne So an alkene and alkyne you basically have a bond which is at least one bond Which is a double or triple bond and therefore there are much more Additions that you can do mostly hydrogen is added but apart from hydrogen. You can also keep on adding Other compounds like chlorine HBR. So all of those compounds can be added So that's a saturated unsaturated the first thing that you need to know in organic chemistry is the classification Give me one second. I'm also Right now where where actually the things that you guys really get confused about I'm gonna touch Pay I'm gonna touch base on those first. So now I think by this time everyone is aware of let me quickly check if anyone has Just come in. Okay. These are the ones Okay, fine now there are 13 of you perfect So now there are two major points that people generally are confused about one is the nomenclature and Second thing that you guys are generally conflict confused about is the reactions What are the different reactions? How to really think about it? Etc. So I'm going to give you an overview of all of these And I'm also going to share some tricks and tips with you so that you know, you are able to really Give some very difficult to remember everything. That's a common Doubt that most of you have asked so how do we really remember all of these equations? and is there a way that I we can you know kind of have some Thumb rules through which we can actually be able to decipher what you want to Achieve in the reaction. So yes, there are thumb rules as well. So let's look at nomenclatures to begin with So these are the three things that we will see in organic chemistry. So let's let's keep this first maybe half an hour for organic chem and then we can go to thermo thermochemistry and Chemical equilibrium. So let's look at nomenclature. I'm quickly going to revise the rules I don't think so that anyone would ask you the rules of nomenclature per se as in saying that write down the rules But they will definitely ask you to name compounds or identify the Compound or the compound structure with the help of a given name The first rule is longest chain longest chain now I'm going to also tell you that You guys should be aware about two things now because earlier, you know We were not able to do because you were not exposed to functional groups and all so remember that functional group Functional group which is basically people are forgotten functional group is any group that defines the chemical properties. For example, OH Chlorine is also considered as a functional group. Although it is not considered in nomenclature I will tell this exception separately like OH then you have C double bond O Then you have CO OH to all these are functional groups and there are two more You know parts of the molecule which are also important Although they are not specifically called as compounds functional groups, which are the double bond and the triple bond Okay, so the double bond and triple bond I will write this if you don't understand this is the double bond and This is the carbon carbon triple bond. Okay, these two although are not directly technically functional groups They are important in nomenclature. So let's understand the first rule choose the longest chain. That's the first rule longest chain Now if you have a functional group except chlorine, so I'm going to mark this with a pink Except chlorine or for that matter halogens Except for chlorine and halogens if you have any functional group the longest chain should Invariably have the functional group with it. So must have functional groups Okay, so it has to be longest chain But only that chain which has functional group You will always realize that when used to have a functional group begin with the functional group from the from that end I'm going to give you an example. You have carbon carbon carbon carbon here Okay, there's this now incidentally everyone would think that okay if this was the compound given to me I would rather take this as my longest chain But you if there is a chlorine here if there is a chlorine here Sorry an alcohol or I told you that chlorine generally are not really considered as functional groups But now if there is an alcohol then in spite of this being the carbon carbon longest chain Which has six carbons this is treated as the longest chain. So this is correct and this is wrong I'm writing it here. Why this is treated as the longest chain is because with the functional group in Bibed in it. This is the longest chain that is possible. Okay, so I have two options here This is one two three four. I could have gone straight or I could have come down, which is one two three four five We have five carbons here. There is a chain which has six carbons Please note there is a chain which has six carbons that is a wrong selection Why because it does not contain the functional group So longest chain with functional group is the first rule if there is an alkene or alkyne in the compound Then the longest chain should have alkyne and alkyne as well. So I'm going to write this as functional group plus unsaturated bonds Which simply means that the double or triple bond unsaturated bonds There are a lot of possibilities that you guys can generally think about you might say sir If what happens if there is a functional group and double bond what happens if there are two or three double bonds in all of those There is a single rule that you should remember use maximum functional groups maximum functional groups plus bonds plus longest chain Okay, so wherever in whichever chain you will find maximum functional groups or maximum functional group plus bond So I'm going to make here a bracket sometimes you might find that there are three functional groups Sometimes you might find that there are two functional groups and one bond both of them are considered equal Having said this there is a priority list of functional bonds Which I have given to you the priority list if you still don't have I'm going to just flash this year in the Screen because it will be there in the YouTube and you can always anytime go back and check this in the check this list So if you want to have the priority list of nomenclature, I'm going to show you that Yeah, so you guys can also Google out there's there's no difficulty in it But just instead of writing it here, I'm going to write it you show it here So here it is clearly mentioned carboxylic acid if it there is a fight between carboxylic acid and alcohol choose Carboxylic acid if there is a fight between ester and nitrile choose ester so and so forth now a few of these Functional groups are not there for you. This is clearly there on the YouTube Just feel free to go to YouTube swipe here whenever you are in doubt pause Learn this and then you can come back. There is an acronym that will also be possible You can just Google out. There is a acronym which I'm not going to say here because For various reasons, but now you you can just have an acronym memorized and that acronym will give you a Priority of the order you like we have the acronym for chemical reactivity Metal reactivity. Similarly. There is also an acronym for this. Okay, let's go back to our rules Let me quickly check if everything is good on the YouTube Okay, nay trippin Neha and Kirthana has also joined good. Welcome guys the first Minutes of this video has been lost because I was silent So, you know, I will try and trim that after this, but you know, you can just follow it from wherever you have Okay, let me quickly check if I'm I if everything is fine Everyone are you able to see me? Can you can you quickly just confirm that if you're able to see my screen? Because I'm not sure if the screen is visible. I have not been able to see the screen anywhere here So can you please confirm if you are if you if you are able to see my screen on the YouTube? Yeah, I think you are able to see there's not no problem. Okay, good. So Yeah, so things are fine. Okay Okay, I think it's going good. I'm gonna pull this up. Okay, three puns saying, okay, it's visible. So that's good. Thanks, triple Excellent. So let's go to the next rule. So the first rule is clear longest chain with maximum functional group plus bonds And if you have a fight in functional groups also refer to the list that I have just mentioned That's that that's the first thing. Okay. Second. Now, let's go to the second rule The second rule says that the numbering the numbering should have the lowest sum Okay, the numbering should have lowest sum now this lowest sum is basically This also is diluted a bit whenever you have functional group So if you have a functional group simply start numbering from the carbon which has the functional group Without functional group if you have a carbon then you consider about Numbering lowest sum now what numbering am I talking about? I am talking about numbering of the branches I'm gonna quickly show you one more example. Let's say this is the one. This is the one. Okay, and you're you're you're okay There is one year a carbon two carbons here now. Now, this is a chain Now the longest chain is is typically this okay This is the longest chain the one that I'm drawing with the pink color. That's the longest chain now It's very easy to identify the longest chain In fact, if you really see this also is a longest chain and there is no problem with that Okay, whether I take this or I take this the compounds name will not be different This is a test for you to identify Symmetries listen to this carefully. This is a test for you to identify symmetries Whenever compounds by to through two or more different structures will have the same name It is okay to use that name but Even if you can find one structure which you have chosen in this longest chain Where you will get a different name you are in a problem and you have to check again For example, I have chosen this chain or I have chosen street chain Let me name them a and B and C and B whether I check whether I choose a B or whether I choose C B in both of these scenarios The name of my compound is not going to change because that is a symmetry absolute symmetry therefore, it is okay to choose Any in if the name changes then I have to think about all the other rules. That's point number one Let me just for your, you know Slight complexity, let me choose C B as a chain the moment that you see B as a chain Whatever I am drawing in blue are the branches are the branches. So look at this like a tree So this is the bark of the tree and these are its branches. So there is a branch here There's a branch here and there's a branch here, whereas this is the bark, you know The bark or the trunk, you know the trunk of the tree So the so the one that I've shared it with with a with a lot of you know blue blue thing Is that is that trunk and these two these three are its branches now? Whenever you have branches you number the branches. Okay, you number from both ends for example You can number from here. Let's call it as n1 or you can number here with let's call this as n2 Let me number it from B or we can call it as B as well B B and A so select B and C sorry in our case. So this is our B and this is our C Let's try and number them. This is one two three four five six seven If I number this way my branches are so I'm going to write here as BC the orientation of BC bar like a vector Like a vector. I'm saying BC bar my numbering of branches comes out to be three four and five if I take CB Sorry for a small Interruption in between Let me quickly check if everything is Okay, okay Yeah, so you guys I I lost lost a bit, you know in between I got lost a bit in between so There was a disconnection here because of some power issues. So I'm back. Yeah So I was saying that we have a two ways B and C So from one side we can go go as one two three four five six seven And I get BC bar as three four five whereas CB bar if I really do I get I get the connection the numbers as three Four and five as well. Okay, so if I go from CB bar here are the numbers. Let me draw it with a different color So from C, this is one. This is two. This is three four five six and seven So this is interesting. You see on both the sides. I'm actually getting as three four and five now in this scenario Had I had a case where I would have got two three four then Yeah, guys can someone quickly confirm if you are able to see me and If everything is fine for everyone if you guys are there anyone can just quickly put a yes like last time people did Okay Right Yes, okay perfect. I sure I saying yes, so it seems things are okay So great so in in CB bar, it's three four five now in this scenario the numbering from either side In the sense of lowest number does not matter because both of them are correct now How would I go about this here if there is a clash in number? The next rule will follow Which is the alphabetical order? Okay an alphabetical order. I know that this group check this I am writing this in another color again. This group is a methyl group Okay, this group is an ethyl group and This group is also an ethyl group and since there are two ethyls and ethyl isn't alphabetically Senior to methyl because of E instead of M. I would choose the numbering from this side Okay So CB bar is more correct in the sense the because the next rule will follow and the next rule here So a quick recap wherever we longest chain was the first rule numbering of lowest sum is the second rule Here are the lowest some incidentally where both the same So we end up going to the third rule and the third rule says that you always have to write the The substituents substituents are the branches. Please remember. I have shown you what the branches are you have to write the substituents in alphabetical order Girl order. Okay. Now, please note. Please note guys you pay attention to this This is very important. Both of them are ethyl. So this this compound is written as 34 di ethyl in this 34 di ethyl di D does not count ethyl counts E counts. This is what counts and then we write this as five methyl or methyl Okay, so it is the E and D E and M that counts So the group name counts the diet tri tetra does not count Also, please note that the numbers are returned and then all the groups are written together It is not written as stream ethyl and then again for ethyl. No, it is written as 34 di ethyl Why I'm writing di ethyl because there are two ethylene ethyl ethyl groups So there could be ethyl methyl propyl butyl and all of those right So this is a very important rule and then the third rule follows is Is a writer substitutes alphabetical order after the before the root okay before the root What is a root? The root is the longest chain that we are mentioned. So this is in this scenario This is three four dime ethyl five ethyl since there are seven it is an heptane Hey, now there is a very important point had this compound had an alcohol here I'm just saying showing a variation. So therefore I'm going to show with another color Now, let's say the compound had at this carbon at this car Alcohol I would have chosen this chain from your I would have started from your Irrespective of whatever else was there if I would have started from here I would have chosen I have two parts now one two three year and you're also one two three now in these two parts Where are the larger substituents? So I find on this part there is one subsequent on this part. Also, there is one subsequent. Okay now that is also same now Because there are one month substituents and the chain length both are the same I will again go to the alphabetical order and I will say okay here There is an ethyl. So the correct order would be from this carbon to ending at this carbon Okay, if you did not understand this, please go back to the YouTube channel Pause slowly listen hear me out and then try and make a sense if you have any doubts, please feel free to actually put in your You know doubts in the comment in the section and I'll be happy to Address them. So now we have seen the nomenclature You know in a short the the only difference that comes with functional groups is there is an additional rule in the functional group We always consider the root we always consider as the root as a derivative of the functional group And therefore if it is an alcohol, which is the OH group We would say that this is an let's say for example and heptanol Okay, the previous compound would have been an heptanol now in that also, please note It is not always that the OH will be at the end carbon. So you generally write hept That's whatever number it is two three four wherever it is write that number say two Nol hepta sorry Then to all okay, so it is just the oil that comes at the end Hep 10 to all okay There is an hyphen that you can use as a separator between the words and the numbers Please note hyphen is used only between words and numbers Generally between words and words or between numbers and between numbers and numbers definitely use a comma between words and words There is no need to use anything between numbers and words. You can use a hyphen So I have given you these rules very clearly and very distinctly although in the class We have mentioned the same but because of too many doubts that come in maybe you guys might not have really got a sense of It but now you're you can always rewind and check these rules and listen to them again and again. Okay, so that's a quick you know revision on Nomenclature you I think all you guys Rest of the things, you know carboxylic acids. We call this as Oic Oic acids then whenever we have a ketone we call that as You know for example in heptone or a pentone or so we ended with an own Similarly, whenever you're an aldehyde, we call it an al. So al. That's how you end it. For example, heptanol Heptanol, that's how you end it. So you you have all of these with you and now let's go to the next point If you have any doubts, please feel free to share, you know, sir saying that for example, you know You want a specific nomenclature of a of a compound? You can also send me a pic You can send me, you know comment on that any which ways which are whichever you're comfortable. Just get your doubts clear right now Okay, so now from your let's go to the To our reactions. Now, please note that for any functional group There are three things that have been mentioned in your textbook. First is the properties Properties second is actually reactions and third is generally uses, okay in the reactions There are two parts one is the preparatory reactions or preparations Second is simple reactions means after the molecule where it is where does it go? I'm going to touch upon a few few very basic and very important reactions The first reaction that I'm going to talk to you about and these are in no particular order. These are just Taken, you know out of the entire lot of reactions that you have but the reaction that you cannot miss out on if this reactions You miss out on then then you are in trouble. Okay, so I'm going to talk about all of those reactions That are extremely important for you The first reaction that is important for you is from alkene to alcohol any time you have a C double bond C Okay, you end up Doing a hydrolysis. Okay, so you end up doing a hydrolysis with H2SO4 and water. Okay H2O, okay, so this is a dilute H2SO4. Now people keep on asking me Dilute concentrated sir. This is a temperature that is a temperature now I think your teachers would like to have you the temperature that are there in your textbooks But if you really look at the J, you know, the larger exams, they are not very Particular about the temperatures except that you need to have the sense whether it is cold or hot You cannot say that okay something was at 400 degree Kelvin and then you directly say that okay You know dilute H2SO4 at room temperature that will be wrong But you can simply say this as Delta Delta means something that is hot or not. Okay, so So alkene with H2SO4 initially it actually ends up giving you carbon plus HSO4 Okay, and this further if you hydrolyze to the double bond your H HSO4 This is how the carbon breaks and this hydrogen goes here This this molecule gets here and when you actually go further and hydrolyze it you end up getting an alcohol Okay, so you end up getting a CH and a seed COH So this is one reaction that you should not really miss out on the second reaction that you should not really miss out on Is from alkyl halides. So when you have a carbon Which has a chlorine in it or any alkene in it for example, any alkyl halide in it For example halogen, you know like chlorine bromine iodine all of that in any of this situation whenever you use aqueous KOH Whenever you use aqueous KOH you will end up getting an alcohol So nothing it's just a substitution reaction that chlorine here goes out and the OH Minus from the act with KOH gets in so you simply end up getting an alcohol here I hope you understand the V that I'm writing. I'm not writing the entire compound I'm just showing you the action points where the real action of the Reaction happens. So this is a simple reaction of substitution Remember that with aqueous the important point here is aqueous KOH I'm repeating this again and again this aqueous part is important instead of aqueous if you use Alcoholic if you use alcoholic you'll end up getting an alkene So that I will write this here itself. So you end up using alcoholic here. Okay? You'll end up getting a alkene, okay Okay, so you end up. Yeah, that's that's one now Okay, so that that's that's one now Industrially, you know, alcohols are prepared in a lot of different ways The one of the most important ways that industry alcohols are prepared is alkenes That's by direct addition of water at low temperature and high pressure. So that's also one way So yeah, the reaction that I showed you here your extra so far is used as a You know just a What do you can say a catalyst? So basically you end up forming hydrogen sulfate and then water So that's that's one way that you can definitely use this as okay now the second the third reaction that I would like you guys to really know is dehydration of alcohols So you have this reaction? Sorry, you have alcohol here. So there is a hydrogen and there is an OH here and there is another compound here You use concentrated HSO. Please note in the last case. We had used dilute estrosophore on alkenes here We are using a concentrated estrosophore. Listen, estrosophore is a highly hygroscopic substance What do you mean by hygroscopic? It is an extreme thirst of water all the time In dilute estrosophore incidentally since it is dilute, it has a lot of water around it But in concentrated estrosophore, there is only estrosophore and it needs water and therefore it will get water extracted from wherever possible Therefore sometimes there are also small sachets with concentrated estrosophore crystals that are used to absorb water from Substances where we don't want any moisture So concentrated estrosophore is something that we really use here to end up getting an alkene So the water here gets out and the electrons from your get here to get a double bond So our alkene is formed with concentrated Estrosophore that is one of the most important reactions The fourth reaction that I you know you guys should be very aware about is the all the oxidation reactions Please note the oxidation reactions here. You use Km in a four You know alkaline came in a four then you use acidic K2 Cr2o7 or you can also use H2 H2 Cr2o7 all of those You know are okay Whenever you use this kind of a reaction, I'm only doing some specific reactions in the Interest of time. I'm not looking at all the reactions These are the reactions that you should not miss out on other reactions You can revise as you as you get time, but these reactions if you miss out on this It will be harmful not just for now, but even for your future studies, you know So both of these basically end up giving an oxygen. Okay, this oxygen. Why it is written in a Square bracket is because it's it's purely just an oxygen atom and nothing else So once you an oxygen atom The sequences that any alcohol will firstly go to become an aldehyde or a ketone and The aldehyde and ketone then further ends up being Carboxylic acid. Okay, I'm going to show you just one example Let's say we have propanol, which is CH3 CH OH CH3 When you have an aldehyde so since this is a two-degree alcohol the bond does break Okay, the bond does break here. So you end up getting two alcohols. So you will end up getting CH3 CH Oh CH double bond. Oh, so this is one one aldehyde and then you end up getting CH3. Sorry CH Double bond. Oh, so is it one to CH2 double bond. Oh, so this is formaldehyde and this is ethanol or Acetyldehyde and both of them will further end up giving you carboxylic acid The which is acetic acid and then you have formic acid. Okay now one very important thing one very important Please note because the alcohol was in the second carbon. I had to break the bond It's it's it's not simply that, you know, you end up getting A ketone, you know, so so a few of you might also think that sir, can it go through CH3 C double bond. Oh CH3. Yes, no problem. You can write this also. That is absolutely right. Absolutely. Okay But please note even if you write this in the end, you'll end up getting these two products only So the intermediary you can still write this as one of the products and Be satisfied for the moment. But in the end, you will end up getting carboxylic acids only, okay I hope everyone is there with me and uh, yeah, there are a few comments But I I would just again ask you if there is anything Please try and send out on whatsapp or your at this point. So I'm just coming to the end of The carbox the reactions in organic chemistry. This is one reaction that is very important The other reactions that are slightly important are tolens reagent. I'm just going to take names here. Uh, uh, the failings Uh, the solution. Okay, these two are some very commonly used commonly asked reactions than esterification Esterification these reactions are also very commonly used esterification is a very easy Reaction, you know, you can simply use a carboxylic acid rco OH plus You use alcohol, which is RO R-OH Please note carboxylic acid ends up giving OH. This is one another mistake that everyone does Carboxylic acid ends up giving OH and not H. Although it's an acid people think that H is given through here and ROH gives the OH. No, it's the carboxylic acid that gives OH So you end up getting R C double bond O O R dash. So this is the ester that you get. Uh, you can also have Etherification reaction ethers. So instead of this, I'm just going to show you comparison what happens when two alcohols are used So etherification in etherification, you basically have our OH plus H O R dash in this scenario you have One of them giving out OH which one gives please note the one that is Smaller the smaller one will give out OH so I'm writing here explicitly the smaller carbon groups So you will say sir what happens between if these carbon groups are same different what will happen The one that is smaller will give out OH The one that is larger will retain the oxygen and it will end up giving RO R dash So this oxygen comes from the larger one. So this is the larger I'll kill chain. I hope everyone is there with me and uh, you know, you're connecting With this, I think most of the major reactions are done. There are a few reactions Oh, yes one more one more that I recall which is also important is the reduction Reduction reactions in the reduction reactions. Uh, you can use li al h4 Uh li al h4, okay Or you can use uh, uh, h2 in presence of platinum and palladium as One second guys. There's a background noise. Okay, so you can use uh hydrogen in in presence of platinum uh platinum and palladium as uh the Uh, you know catalyst both of these do the same thing. They will take alcohols to alkanes Okay, simply reuse them Uh, they will take alkenes to alkanes Okay, uh, they will also take carboxylic acids for example Because carboxylic acids are on a lot on the acidic side on the oxidation side oxidized substance They'll bring carboxylic acids to alcohols Uh, so they can do that and further they can also be taken to alkanes Okay, depending on uh, what reagent you're using if you're using h2p tpd. It's It's possible to do this and it's possible till your Uh, uh, mostly alkanes will not happen for alkanes. You'll need much higher Reduction to be done Okay, so that that's that's a quick Recap on organic chemistry. I think this should be suffice for you to do. Okay If you are really looking forward to score very well, I think you should go back to your text and uh, look at the topics one more time um others Now i'm going to go to the next Topic of revision. Can you tell me anything that you specifically want me to do? I'm going to take up thermodynamics, but is there anything else that you want me to look at? I'm going to pause here for a second just to check the Uh, comments if there are any Okay, uh, so till you put up, uh, I'll give I will you know, just be here uh for the next topic and that will do as thermodynamics Okay, anything else that people want to really look at Okay, so I think I'll begin with thermodynamics. Uh Uh There's nothing specific that probably you have in mind so I can actually go through the general Aspects of thermodynamics So I'll give you an overview about thermodynamics in thermodynamics the topic firstly starts with understanding all the processes and parts of it For example, understand what is a system? Understand what is a surrounding system is what we study anything outside of that is a surrounding Boundary is something that Differentiates them then there were three types of systems One is a open system Second is a closed system And the third one is a isolated system. So there were three types of systems that we had studied Uh, then the processes that we studied for example, there were irreversible Uh, then there were reversible processes Uh, then we had Uh isothermal processes Uh, we had Uh iso Volumous, uh, or which we also call as isochronic process Isobaric process, okay? So we basically looked at an average of all what what all the different types of processes that were there This is just the basic setting of the chapter Okay, uh Okay Yeah, just to make sure that you guys are all there. I because you know, I don't have a direct connect Can you please again just put in your? Uh, confirmations that you are still here Uh, Sachi, Rona, Gaditya, Aishwarya Arpita Trippin, Kirtana, Mansi Uh, Ananya, Shomik. Yeah, sorry. I know I'm asking you again, but just put up a yes. No Amure, whatever you feel good like a quick check on whether you guys are still there You know because incident I want this to be Really connecting to you rather than just okay. That's good. I sure yeah Okay, Rona is there Where are others? Hmm Mansi is there. Arpita is there. Trippin is there Perfect. Okay. Good. Yeah. Thanks. Thanks guys. I think the rest are around. Okay. That that makes sense. Perfect So let's so this is the basic setting of thermodynamic chapter This chapter then revolves on two parts one understanding what energies different forms of energies are Which we also called as different forms of enthalpy enthalpy and energy are nothing but just you know kind of synonyms words equivalent to each other Said in different contexts nothing else And then we said, okay, what is the law of thermodynamics? Okay. What is the law of thermodynamics? Which is the first law that we have studied? Okay Now in the enthalpies of thermodynamics all that you have studied is the following points The first that you have studied is what is what are bond enthalpies? Then you have studied what are neutralization enthalpies Okay, then you have also studied. What is ionization enthalpy? Okay, ionization enthalpy Then you have also studied what is uh, uh, phase change enthalpy Phase change enthalpy. What is formation formation enthalpy? Okay formation enthalpy So there are multiple enthalpies that you have studied. Okay Please note the most important point in all of these definitions. These words themselves say a lot to you Okay So if someone asks you what is neutralization enthalpy You will simply say anything any energy that is released during neutralization And what do you mean by neutralization? Simply means that water formation from acids and bases or salt formation salt and water that is formed from acids and bases Someone says what is phase change enthalpy or state change enthalpy? Sir, whenever anything goes from solid to liquid or liquid to gas whatever energy is released or absorbed is Phase change enthalpy all that is good The problem happens in these two things and please please please note this very well The first thing that problem happens is Mentioning of one mole how much quantity because energy will be variant with the quantity right you tomorrow start The reaction with 10 kgs of the substance the energy released and absorbed will be different than what you do with one gram Right. So most of these are for one mole. So it is not even weight It is for one mole of a substance The second thing is Always mention always try and think as to what is the starting and ending point of the reaction? What is a starting and ending point of the reaction? I'm going to give you an example Starting and ending point. These are the two major things that you should really look at for example The starting point is in neutralization starting point is our acids and bases. Okay. Good ending point is one mole of water formed One mole of water formed not one mole of salt or one mole of acid and base reaction or one mole of acid one mole of base No Whatever the acid and base do I don't care after they have reacted one mole of water the moment it is formed Whatever energy has a timed out or has been given out. I will calculate that Face change. What is the starting point the initial phase and the final phase formation enthalpy? What is the starting point elements elements is the starting point in formation enthalpy? So you start from elements and you end up at compounds. So that is for which an enthalpy now Elements themselves cannot come from anything that is why they are called as elements So therefore their formation enthalpies is always zero Now elements start from this end and they end up being compounds. So that's Formation enthalpy now there are in some of these definitions some additional conditions For example in formation enthalpy the additional condition is that The reaction should happen with natural state of elements a natural state of compounds should be formed second Mostly it is it should happen at stp standard temperature pressure Standard temperature pressure is zero degree Celsius that is 273 degree Kelvin And one atmospheric pressure one atmosphere pressure. Okay, this is standard pressure temperature So this is one very important thing that you should remember Okay, so all of these enthalpies are there bond enthalpy neutralization ionization phase change formation now Whenever you are in a doubt. So this is the overview now. Where is the problem that happens? Or where is it that you guys really feel very hurt whenever there are signs that come up? Okay So first I told you about definitions where you should be careful The first thing is that you should be careful about amount second thing You should be careful about starting an ending point third thing you should be careful about some special conditions special Conditions for example stp or elements in their standard form These are some special conditions for some some of these enthalpies whenever we define them. Okay So now that that's one the next thing that you guys really Go wrong mostly mostly that you guys find a difficulty is with the signs the signs Now forget any and all definitions in that scenario always always think logically Whenever a bond is formed c bond c single bond. Oh, okay It is obvious that the bond was formed because they wanted to stabilize else Why would they stabilize or why would they go to form a bond since they stabilized implies energy must have been given out So I'm going to tell you the basic rule basic law. Remember this and all of your equations mostly you will be taken care of firstly Every reaction is happening for stability Every reaction is happening for stability. The second point is the stability means low energy Whose low energy low energy of the substances formed substances formed Okay Okay, that is why any reaction or something any any process is happening Now if the substance for some reason is not going to low energy Then you will have to give energy to sustain it Okay, so this is the third point give energy To sustain sustain means maintain that position. Why because that guy was not willing to you somehow push him up So obviously that guy is wanting to have you give him energy to sustain his position or state Okay, sustain his position or state, right? So these are the three rules that are always applicable So bond energies obviously since energy is given out bond enthalpies will be negative Formation energies obviously the compound is getting formed therefore compound ideally should be negative But let's say somehow you have given energy and made that compound formed some orbit compound like cl o2 Okay, but this compound would never want to be there You have somehow made it form in those scenarios your formation enthalpy will be positive because you have given them high energy To make them sustain there. Okay Then there are allotropic energies Which means change from one allotrope to the other so all of those enthalpies are are are the You know based on the science are Decided from some common sense now Change in enthalpy change any change is always defined as final minus initial So change in enthalpy will be final enthalpy minus initial enthalpy final minus initial position So check at initial. What was the energy of the substances check at final? What was the energy of the substances simply subtract final minus initial and you will be done Okay, so I am going to for example So so one thing that I want to do but I might not get enough bandwidth to do it now Is that I you know, you guys should be able to solve some problems There are a few students who are in touch with me and they keep on sending the problems multiple of them In fact, I would suggest that all of you actually do that I have enough problem if you feel that you do not have sufficient problems to solve Ping me and I will send you thermodynamics problems. I will send you chemical equilibrium problems This is again for those who are willing to get marks If you are if you're not willing in the first place, then you know I me me pushing you is not going to help much So we have a very free system here You can enjoy the class and retain whatever information you can But if you want to score better ping me and I will send you and I'll help you with the papers Okay, so uh with with additional sums. So that's that's uh Enthalpies now one more thing that you have to learn in thermodynamics is Hess's law Okay, Hess's law now, please note. Hess's law is nothing but accounting For example, something happened a plus b happened and it gave c Some other b plus b happened and it gave e But you were asked to actually talk about a plus 2b plus b Okay, giving c plus c So you will simply say boss when a plus b gave c. What was your cost? Did you take money from me or you gave money from me in turn instead of money here? You're asking energy So you will say, okay, this was h1 h1 was given And then b plus d what happened boss. So you will say it is h2 which was taken So therefore this is since it was this was given it was minus h1 Since the second thing was taken it is plus h2 So you would essentially simply say is that at uh guys The total energy here would be h2 minus h1 Okay, uh, so there's a request here. Uh, sir, can you solve an exemplary problem from thermodynamics? Okay, let's do a problem as well. So Uh, I'm not going to do a problem from your text, but I'm going to give you a problem that I already have so, uh Okay, let me just quickly give you a problem. Let me see how I can actually Show you the problem Uh Yes, so let's say that I'm going to read out the problem. Uh, and I'm also going to do, uh, You know, uh, I'm going to write the basic important points here. Uh, you guys can check What you guys can do after I've written the problem is write the answer in the comments. Okay So here's the first question for for you guys. Okay okay, so, uh So the heat of combustion here's a question the question is The heat of combustion. So therefore delta h combustion is given to us delta h combustion means something was burned of ethylene ethylene. They have not even the Uh, uh The formula they've given the name so you guys are okay to use the formula at 18 degrees Celsius At 18 degrees Celsius and at constant volume. Please note. It is constant volume Okay, at constant volume is minus 355 0.8 minus 355 0.8 kilo calorie kilo calorie When water is obtained in liquid state and here water is obtained in h2o is obtained as liquid state Okay, calculate the heat of combustion at constant pressure. They are asking calculate the same heat of combustion At constant pressure. Okay at constant pressure when p is equal to constant Okay, slightly tricky and not so easy a problem But doable. So they have said heat of combustion at constant volume is 335.8 kilo calorie per mole it should be per mole Uh, now they have asked that this calculate the same heat of combustion at constant pressure I'm going to give you a minute and then uh, I'm going to start discussing the same So hopefully you guys get it. Yes, so So I I will uh, I think now that you guys have got a Heads up and you might be thinking about it. I'm going to do this one problem directly for you Uh, you can always pause the video if you don't want to listen to the solution solve it at empty And then come back you can always do that as well But right now I'll just you know to for again in the interest of time I'm just gonna share the solution with you. So basically we have been given delta h c which is combustion at constant volume We know that delta h is equal to delta e Plus p p delta v Okay, plus p delta v Okay, now at constant volume The p delta v will be zero because the change in volume itself is zero because volume is constant and therefore Delta h v is equal to delta e And therefore delta e is equal to minus 335.8 now We know that p v is equal. This is the solution part. So I'm going to mark Separator between them aligning between them here also. We know that p v equal to n rt So therefore p delta v is equal to delta n rt The reason is the temperature is again maintained constant So the only thing that will change if pressure and temperature they've asked us when pressure is constant And they have already said that the temperature is constant So when pressure is constant then temperature is constant the only thing that can change is For volume is the delta n part now since we cannot directly find now Now see what we have got is this as part one. Okay now. We have been asked as delta h p Now delta h p is delta e plus p delta v where pressure is constant Delta v e we already know from this part. So it is minus 335.8 p delta v since I cannot there is no pressure given to us neither change in volume is given to us So I figure out that p delta v is equal to delta n rt and delta n is nothing but n products minus n reactants Now n products number of products for this equation. We will figure out is that they let let us write the equation So it's it's high lean It highly means c h triple double bond c h 2 C h 2 double bond c h 2 plus o 2 since it is combustion plus o 2 gives co 2 plus h 2 co 2 plus h 2 o now here we have asked this for liquid Okay, this is for liquid So therefore this will there will be this will not be considered in delta n because delta n is only considered for gases This is a very very important point. A lot of people make mistake in that co 2 definitely is in gases now firstly. I will balance this equation This is 2 c or 2 and whatever there are 2 and 2 4 therefore therefore 4 o 2 4 4 h 2 o now the oxygen that has been utilized are 2 to the 4 plus your 4 So that is 4 plus 4 8 and therefore that is 4 year Now therefore total number of products total number of products is only co 2 Please note h 2 o will not come into the scenario. Why because s 2 is in liquid form This is only gaseous equation. This equation is applicable only for gases And therefore I will have to neglect the 4 of h 2 o so the total number of product here is only 2 That is np. Okay. Now the total number of reactants is 1 plus Wait, oh, this should be 3. Sorry. This should be 3 this why this is 3 because 2 or 2 4 here and 4 year 1 second 2 c or 2 Oh, this should not be 4. This should be 2 year Yeah, I think thanks for pointing out guys. Yeah, so this should be 2 year So therefore this oxygen will be 3 So this is 2 2s of 4 hydrogens 4 hydrogens match 2 carbon 2 carbons matched oxygen 6 year 6 year match So this is the total number of products is 2 total number of reactants ends up to be being 4 Therefore delta n is minus 2 and since delta n is minus 2 My your p delta v will be minus 2. This is minus plus minus 2 into this delta n rt Instead of p delta v. I'm writing as delta n rt and that will be minus 2 into r R will be since it is kilo calories Therefore it will be 2 into 10 to the power minus 3 kilo calories And then I will write the temperature the temperature is 18 degrees Celsius. That is 291 Kelvin Okay, and so therefore if you really solve this this this will come out to be minus 3 35 point 8 This should be somewhere around 2 to the 4 4 into 300 somewhere around 600 minus 3 in plus 3 We will get cancelled out. So this will approximately come out to be about So that is 4 2s are 8 so somewhere around 10 I think it should be 10 or 1 1.9. I think approximately about 1.9 So this should be equal to about minus 3 36. 9 64 kilo calories Okay, so that's the solution for one quick problem Let's do a problem on Hess's law. That's something that I am expecting that will be there in your exam as well So we'll do a quick question on Hess's law. So Okay, so here's a question on Hess's law. So it says Before that, you know a quick revision on the thermodynamic laws because we have we have seen this problem But this problem was based on thermodynamic laws all we saw in the previous example was Hess's law and The different forms of energies and then the law the law basically says that law of thermodynamics is nothing but Law of thermodynamics Is nothing but conservation of energy conservation of energy Now what does this conservation of energy mean everyone knows conservation of energy? But what is so difficult here? Let's see basically if you have a container and you have a piston on the container and this container has a gas liquid or any kind of substance If I heat this container if I heat this container Q is the heat that I give Okay, if I give anything heat the energy of the internal substance will increase the energy is called as u It is called as internal energy and when the energy of the internal substance increases it has the Control to push this piston up if it pushes the piston up it will do some work What is conservation of energy conservation of energy said what we what did you give? What did you give? I said I gave heat q. Okay, they said that's fine So you write this as q then they say okay because of the q what happened to the energy of the gas I say it might have changed. So let's write it changed from ui to uf So uf minus ui is a change in internal energy. So I write as delta u Okay, but then all the q go to the internal energy. No, it did some work. Okay, the work was done by the gas So yes, I can write this as plus p delta v the work done But since we know that in chemistry we write work as Done by the gas as negative. We don't want any work to be taken out. So we write this q as Delta u so this work has to be negative But here I have a plus sign therefore I write this as minus w so w is basically nothing but minus p delta v Okay, so by with this equation, I simply get first law of thermodynamics, which is q equal to delta u minus w Now if you have thing a few of guys, you know yesterday had a Issue as to sir. What is q v q v means the heat given at constant volume if volume is constant Will I have a w no so q v is simply equal to delta u whatever heat you gave to the system It simply changed the internal energy of the system. It cannot do any work So q v is equal to delta u. What is q p sir q p if any heat that we give if it is changing only internal energy It implies that its volume will be constant But let's say if the volume is not constant if the pressure is constant and the guys are able to push the piston Then q p will be equal to okay internal energy will also change and work will also happen So q p is equal to delta u minus w now a lot of people also have questions on saying sir Sometimes q is equal to delta u plus w. Sometimes it is delta u minus w. I will tell you Forget about all of these equations You will never go wrong if you are doing by the fundamentals and what is the fundamental Anything that is done on the system is good and therefore positive Anything that is done by the system is bad and therefore negative So therefore what you guys have to really look out as whenever you're writing the equation Forget about the equations and science and all of that Do your calculations well account for your energy is pretty well If something is getting into the system call it as positive if something is coming out of the system call it as negative In the end, whatever you gave in should come out. That is conservation of energy as simple as that So, you know, don't get trapped into the Terminologies and words and science and all of that or it already is good to know them If you in the worst case scenario a smart guy would not really be concerned about it He would simply be concerned about his accounting of the energy if his accounting of the energy is good The rest will be taken care of Okay, so I think you guys are also With me now what I'm going to do. So this is the one that we have already done So the revision of the technical things of thermodynamics is Complete there is nothing more in thermodynamics apart from this so How do we calculate bond energies allotropic and all of these enthalpies? I've already told you like a major overview of everything now. Let's go back to knowing You know doing a problem. Okay, so Here's the problem. So the the question says that Use the following data and so it says use the use this data Use the data Uh at uh, you know, all values are in kilo calorie kilo calorie per mole At 25 degree Celsius that is room temperature. Okay So all of these values is at this and we are asked to calculate We are asked to calculate the bond energy the bond energy of cc bond carbon carbon bond and ch bond Okay, this has been asked what is the data given to us the data given to us is delta h combustion Delta h combustion of ethane of ethane Is equal to minus 372 point zero Delta h combustion of of propane Propane is equal to minus 530 point zero Delta h Uh conversion of carbon solid to carbon gaseous Okay is given as 172 kilo calorie per mole Bond energy of h h bond is given to us bond energy Of h h bond is given to us and that is equal to 100 and 104 And delta hf Delta hf of H2o liquid Is given to us as minus 68 Point zero and delta h formation that is okay and delta h that not forget that not now Not simply means that at standard conditions. Okay, and delta h not of co2 gaseous Is given as minus 94 kelvin This is the data given to us and uh, uh, you know the combustion energies are given and Uh, the rest energies and even we have to find the cc and ch bond energy Please note they have given combustion energies of ethane and propane both that is for a reason because between ethane and propane you have Two, uh, you know one carbon extra and two hydrogens extra So you can use that and do this reaction So i'm again going to pause here for a few seconds for you guys to really be correct if you want to pause you can pause After about 30 seconds. I would be starting the solution Okay, so now a lot of these are given to you, uh, you know, you can just note down these Point, you know, or you can always pause the video and take down these values Meanwhile, I go ahead a bit. So here's here's the Solution for it. So when we have c3 h8, which is our propane and when this propane is actually heated You know, basically the the delta h formation for propane is Something that if we know, uh, then propane is actually made from Three carbons three carbons, which is in solid state Please note those carbons has to be in natural form and for hydrogen which in the elemental gaseous form So if we know this delta h Uh, then this is like we will know be able to know the carbon carbon energies and hydrogen hydrogen energies. Okay, so, uh Let's call this as delta h1 also for ethane, uh, let's call this as from two c Uh, plus three h2 two c plus, uh, you know, three h2 giving Uh, c2 h6, let's call this as delta h2. Okay Now, uh, for carbon, let's call the cc carbon energy as x kilocalories and the hs carbon energy as y kilocalories, okay Uh, now what we're going to do is uh from the equation one, we know that since it is three carbons and four hydrogens uh No, not h. Sorry. This should be ch the ch carbon energy because that's what we need to find. We want to find that as uh, uh, you know As y okay So from the from this equation delta h1 from delta h1 is given as is it should be equal to 2x Plus 8 y why because in in c3 h8 this gives c3 h8 So in fact, let me write not for all the arrows here. Let me write this as dot. I'll see you guys will again get confused So i'm saying that for c3 h8 and for these guys, okay Now c3 h8 has two cc bonds. It has two cc bonds So therefore and the energy is released when carbon carbon bond is found energy is released Therefore i'm putting a minus sign. Please note minus sign because energy is released and two carbon carbon bonds are formed so that will be 2x And eight carbon hydrogen bonds are formed because this is c3 one two three One hydrogen two hydrogen three hydrogen one four hydrogen five hydrogen six hydrogen Seven hydrogen and eight hydrogen there are eight hydrogens formed and therefore this will be and this energy is released Therefore 2x plus 8 y will be the released energy This will release energy But while this energy is being released the carbon is you know, there is there is energy that I have to give also And therefore I will put this as plus what energy I have to give I have to convert carbon to carbon gaseous That is point number one and I have to convert hydrogen to hydrogen atom Plus hydrogen atom. I have to break these two So i'm making a star here and both those both of those stars i'm putting it here so that you will recall remember again Once you understand you can always rewind and get back to what i'm saying So now how many carbon I have to convert to carbon gaseous three So I will multiply this by the three carbon carbon carbon gaseous form which is how much Yeah, so carbon solid to carbon gaseous is 172 So I will multiply this three carbon atoms per because it's per mole. Please note. It is per mole There's a three moles that are getting to gaseous form. This is three into 172 plus Now this is the energy that I have to give so everything I'm putting in plus And then I have to break four hydrogen bonds. So I'm going to write this as four into 104 this is four into 104 that is what I will write so basically what I end up getting and similarly I will write delta h2 As as minus x minus x y because in in in ethane, I have cc1 bond and there are six hydrogens So, uh, the the the energy that will be released is x cc. We have considered here as x So delta h2 will be one x is released because of carbon carbon bond Plus how many high carbon hydrogen bonds are there six? So six y is a release because of carbon hydrogen bonds And then what is the energy that I'm giving in that I'm going to put as plus I have to now convert again these these carbons here these carbons now these carbons I have to convert So let me write that with the hash probably now So here I'm writing hash this carbon and here again a double hash So now these carbon and hydrogen I have to convert So that they react to the gaseous form So this is the solid carbon that goes to the gaseous form. So that will be two into 172 plus three moles of hydrogen are going to break down. So that is three into 104 So these are the two reactions that I've got now, but I also know that they have also given me that the combustion energy of You know ethane is given as minus three 72 and propane And bond energy of hydrogen is given. So this is just the bond energy of hydrogens that we have used here now, uh, we have also been given as You know combustion of uh, see for the formation energy of etch co2. So carbon carbon really going to see or to Uh, so this is carbon C Now this goes to Plus o2 being going to co2 And this has been given to us as minus 94 kilocalorie Okay, and we have also been given as hydrogen because these are formation and enthalpies of hydrogen co2 and H2o Oh, so hydrogen with half o2 is being given as h2o. So we have been given this also which we Uh, are going to write it as minus 68 kilocalories Minus 68 kilocalories. Okay. So we get co2 as minus 94 and we get Uh, h2o as minus 68. This is given to us. Okay, this is given to us now When we write the equations for c2 at 6 c2 at 6 plus Plus, uh, so, so please note. These are the formation enthalpies delta h1 and delta h2 are the formation enthalpies Let me use a different color. This is formed. This is formed from its elements This is the elemental form elemental form elemental form elemental form So this h1 are basically hf's. These are formation enthalpies So whenever I write c2 at 6 and combust it here, I'm basically writing the formation enthalpies of this and The other guys. So c2 at 6 plus 7 by 2 o2 gives basically twice of co2 plus H2o where it will be three thrice of h2o Now this is given to us as minus 372 kilocalorie. Okay. This is given to us minus 372 kilocalorie Now we already know that, uh, uh, you know, if we Really this uh, okay once again Yeah, so now we know this co2 and h2o's energy. These are the formation energies So this is 2 into minus 94 And this is 3 into Plus 3 into minus 68. So this is the energy that is given out to us. So therefore I'm writing this as minus minus This is the energy that I have to give and this will be delta h1 Delta h1 that I have to give in so all this delta h1 is the energy that we have got That I have to give so therefore I will write it as positive And this will be o2 will be zero So delta h1 is simply going to be equal to all of this Similarly, this is one equation that I will get and let me write it with a capital i Now similarly, I'm going to write a second equation where I have c3 H8 c3 h8 plus oxygen giving co2 plus h2o Now oxygen if I have to balance this equation this has to be 3 co2 This has to be 4 h2o and oxygen will be 5 Please note in all of these scenarios. I have kept the This as one mole. This is one mole. Why this has kept as one mole. I although this is I've written as 7 by 2 I have not written this as 2 in c2s 6 and 7 or 2 is because This energy is given me per mole is given to us per mole So therefore per mole of whom it is combustion energy of ethane So therefore it will be per mole of ethane Therefore I've kept this as one mole and then I've written everything Similarly, I'm keeping propane as one mole and then I'm doing everything else Now this energy is nothing but delta h2 because this is the formation enthalpy of this And it's positive because I have to give this energy to break down these reactants This energy will be zero and going further. This energy will be given out to me Therefore this will be 3 into the co2 energy, which is minus 94 Plus this energy will be 4 into minus 84. Sorry minus 68 Right, so we and let's call this as equation 2 now From these equation 1 and 2 we will get 2 x and y equations So here is a 1 x and y equation So we solve all of this and we get 1 x and y equation And we get second x and y equation a little bit delta h2 delta h1 we liquid to this and delta h2 we liquid to this I'm just going to directly going to tell you the answers You know if you when you solve this you will get x as equal to 82 you can you know whenever you find time you can rewind this video and you can again go through it x is equal to 82 kilocalories and y is equal to 99 kilocalories Okay, so that's the answer So I think this is a perfect and classic and think very complex and all the components of Hess's law is considered here Please note that we the principles. I'm going to revise the principle that we used We use the enthalpy of formation from their elemental states So please check these were solids and we have used enthalpy from solid whereas h2 has to be gaseous So this also has to be gaseous and this has to be again gaseous gaseous So we have kept this Now the second equation that we used is since solid cannot directly react We have converted them to gaseous we have broken down h2 into h and h This we did because this is where the reaction happens If I simply put a carbon gaseous and a hydrogen besides it there is no energy exchange This is where the reaction happens, but this will need energy and therefore those energies have to be accounted This energy once we have got this delta h1 and delta h2 They have nothing but formation energies of propane and ethane these formation energies We have further used in their combustion reactions So we have formed c2 h6 and then we have combusted it To give co2 and h2o co2 formation energy we have used h2o formation energy We have used their formation energies and these formation energies We have simply subtracted to get this delta h here And this delta h is already given to us as 372 Which we have used in our further calculation delta h1 We know in terms of x and y and similarly delta h2 We know in terms of x and y and we will be able to solve both of them I recommend that all of you guys actually go through this example once again after this class Spend like half an hour 50, you know Probably 15 20 minutes if required and please see what all the steps that we have done try to attempt it yourself Try to do it yourself do the calculations come back If you are still stuck you can always ping me one on one I'm again checking if you have any doubts. I recommend that if you have doubts you please send them now I've said this n number of times. It's up to you. If you have anything, please let me know Now, uh, so that's that's about thermodynamics. I've also given you a Two problems on thermodynamics. There won't be any any problem if you understand what we have done in these two problems I guarantee you even the bond enthalpies formation enthalpy Combustion enthalpy all of that have been taken care in these two examples. So Check out these two examples and see if you understand them. You should be okay with it Now after this I'm going to go to your chemical equilibrium One one another topic that most of you are guys are struggling or probably might have a issue on so in in uh Chemical equilibrium what we will study. So here's let's understand the overview of chemical theorem like we have done for all the other two topics as well Right. So in chemical equilibrium, uh Okay, one second Yeah, so what is the overview of the topic the overview of the topic is it starts with understanding what is a, uh, you know, what is basically, uh, uh Reversible reaction. Okay. So a reversible reaction is when you have forward and backward Equations reactions both happening simultaneously, then we understand. Okay. If there is a reversible reaction, it does not mean that There will be, you know, the reaction will simply keep on changing its concentrations at all point of time The forward and backward reaction eventually at some point of time Come to an agreement where both the concentrations of let's say a a plus bb giving cc Plus db, you know, both of these actually come to a stencil So that is when we say that equilibrium is achieved Which means that the forward reaction and the backward reaction have now got Balanced to have have have balanced each other. This is where we come to say, okay If there are these rates, then what is the rate about and this rate is basically defined as Amount change in amount so delta amount. I would say delta amount Let me write it again. I am Delta amount divided by, uh, delta time Okay, so, uh Now amount can be written in any quantity can be written in any way for example kg moles per liter density volume even color even, uh, you know the the Texture size or any all of these can also be used to define rate But what we basically use here is is a common language where you and me both should be able to Talk in the same language and therefore we end up doing the rate of reaction in terms of moles per liter per time so right so amount we write in terms of moles per liter and We take this with change in time So this is a standard convention in which rate of reaction is written as once the rate of reaction is understood Then the next thing comes as to okay. What does this rate really depend on and there comes your law of mass action Where law of mass action says the rate of any reaction is basically dependent on the concentration That is proportional to the concentration of The substance and why it is proportional to the concentration of the substance is because the reaction technically happens in the one second Yeah, so the so why is it proportional to the concentration is because now concentration means how many substances are present per unit volume And the more the number of substances present per unit volume The more collisions will happen between this substance and all the other substances The more collisions happen the more reaction will proceed the more there will be a You know There will be a transformation of these substances to give products and vice versa So it's a very simple theory. This theory is called as collision theory just for someone You know people who want to read more, but I know you don't have time till tomorrow, but at some other point if you have any Any discussion you can simply think of this as a collision theory Uh, then okay, so this is law of mass action where we say rate is proportional to the concentration of the solution So we say rate of forward reaction rf is proportional to a and b both So therefore it should be a to the power a into b to the power b where a and b are concentrations of Of of substance a and substance b now in the since there is a proportionality if you have to make it an equation We'll simply put it as constant k that is k into a to the power a and b to the power b simple as that Okay, good. So this looks so far. So good once this is done now just for your understanding Law of mass action, uh, you know, if you want to really have a Definition for it if you can say that at constant temperature The rate of a chemical reaction is proportional to the product of the molar concentration of reacting species With each other. Okay, uh, and this each term raised to the power equal to the Numerical coefficient. So, you know, what we call as numeric. This is a and b are the numerical coefficients So each of them, uh, raised the power of numerical coefficient of that species In that chemical equation. Okay, so that's uh, That's the description of law of mass action I have said this on the video so you can again as I said always pause and come back and revise it if required Uh, now so that's law of mass action Now essentially when law the reaction begins, so let's say a plus b is giving c So a plus b will have a very high concentration. So let's say this is the concentration And this is the time that we are using So a plus b will have a very high concentration is concentration will drop and eventually it will become constant uh, similarly So instead of dropping like this problem, you know, uh, actually it will drop in a the the curve will be very Quite different. So the concentration actually drops like this. Yeah, it will drop like this and it will become constant Uh, the concentration of products. So this is the reactants. Let me call this as the reactants here But in the on the other side the products will start from zero and then it will become constant So let let let's call this as products And then essentially they become please note again a lot of people who do the mistakes is that they make both of these concentrations same It is not necessary that the concentrations of products and reactants will be same not possibly not necessary not Uh, uh, you know, not a part of any any sort of equilibrium The only thing that actually becomes equal is the rate is the rate So if I if I have to draw here, I think it will be too much If I have to draw the rate of reaction graph Then if this is the rate and this is time Then the rate of forward reaction Would actually decrease And the rate of backward reaction will also increase and they both will become the same after a certain point of time Where here rf equal to rb So this is the rate of forward reaction and this is the rate of backward reaction As as the forward reaction decreases, you know decreases the backward reaction increases and then they become The same so this is this is the first major major major part of Chemical equilibrium after this rf equal to rb We write their expression saying that kf into You know rf is with a to the power a and b to the power b Is equal to kb into c to the power c and d to the power d Now we write kf by kb because kf and kb both are constant So this entire division should be a constant and we call that as kc And therefore because of this kf by kb kc turns out to be C to the power c and d to the power d instead of a on the top and b on uh and c on the bottom Okay, so this is why the expression comes out to the way that it is it is on their screen right now So this is kc, which is concentration of products divided by concentration of reactants Now the rules for writing this kc is uh, they are always written in moles per liter. That's that's very important The second thing is uh, it should always be homogeneous or of these has to be homogeneous Now, please note this rule is nowhere said, but i'm giving it to you just as a thumb rule again If you have gaseous products, if you have liquid products, if you have solid products and Sorry, I have to put in an aqueous here as well Gaseous Okay liquid Then solid An aqueous, okay Sorry, it will be aqueous and solid here. It would be aqueous and solid Or it was right. I'm sorry. It was my bad. Okay. So it was solid and aqueous. Yeah So now what does this mean? What does this mean? This means that if you have a gaseous product in the reaction a a plus b b giving cc plus Dd then if any of the product is gaseous forget about writing all the other products You don't need to write all other products. Your k will simply be that product Now, let's say you don't have gaseous You have liquid liquid and all the other products then only consider only consider the aqueous product Okay, only consider the aqueous product. So what i'm basically trying to say here is that gaseous product is considered better uh than uh Okay, so it i'm gonna i'm going to change this again for you Gaseous is considered better than aqueous considered better than liquid considered better than solid So unless all products are in solid unless all products are in solid You never write any of all of the any any solid product And so if you have gaseous products write all gaseous products and forget about any all other products If you have aqueous products write all aqueous products and forget about liquid and solid products If you have a liquid product write all liquid and forget about any solid product So always take these side always consider this side as priority or preference is what what i'm saying here I'll give you an example say a bb cc and ed here you have one product which is aqueous And all the other products are liquid liquid liquid and liquid Okay, all three are liquid and this is aqueous your kc will nothing but we simply d to the power d only aqueous You don't even have to write any other liquids in the reaction Similarly, if you have even one gaseous product you only write the gaseous product now Please understand why does this why why are we doing this? You're doing this because if you have a solid product its concentration barely changes It's so small because it is not occupying any volume. It is so concentrated So its concentration does not change So what is the point in keeping it in the in this expression because here it will simply remain A huge number without any variation to give an example This can let's say a a will be 10,000 and b will be 0.1 So this 10,000 will simply be lying here and any change in 0.1 going to 0.11 or 0.09 Will not make much of a difference to 10,000 But it will make a huge difference if we take 10,000 to the kc side and made it or make it as one And then we change 0.1 to 0.99 and all of that then the numbers will change very rapidly of k So that is that is one important aspect So anything that has a huge concentration with respect to some anything else that we will consider as constant Okay, so that's the simple rule or logic that we are going to use here This is for heterogeneous products for homogeneous you don't have to worry you can take everything Okay, homogeneous means all our liquid all our solid all our aqueous or all your gases. Okay So that's that's one Now the second thing that we are we had studied was kp Where in kp we said we will write it in terms of pressures This is pressure of a to the power a pressure of b to the power b Pressure of c to the power c and pressure of d to the power d So these are all partial pressures And the earlier one were concentrations. I'm just going to give you one more quick Good rule is p since pv equal to nrt and we are talking about You know gaseous substances homogeneous gaseous substances We know that p is equal to n by vrt But this is nothing but moles per liter. This is nothing but moles per liter And therefore this is nothing but concentration of a substance c into r and t Concentration of a substance c into r and t. Okay, so since it is the concentration I can always write pressure in terms of concentrations and rt So if I write these pressures in terms of concentrations, this will be simply a to the power a r t to the power a Into b to the power concentration of b to the power b into rt to the power b and divided by c to the power c into rt to the power c and d to the power d into rt to the power d Now guys, please note here. I have taken c downwards and a upwards Does not mean that so this this is always products. I have just taken a general case You guys will again get confused sir in the last question you you took c upwards and now you're taking a upwards No, I'm clearly mentioning it here on the video also is that the upper numerator has always to be the products and lower numerator denominator has always to be the reaction So in this this case is our where cc plus dd is actually giving us a a plus bb So we have changed the chemical reaction. I'm just showing an expression So now I have written all of these pressures. I have all of these pressures I have written in terms of concentrations Now therefore I end up getting kp is equal to a to the power a into b to the power b Upon c to the power c into d to the power d Into rt to the power a plus b minus c plus d okay Now this is equal to nothing but kc and a plus b minus c plus d is basically delta n because this is rt to the power np n products minus n reactants This can also be sometimes we call as delta n. So this is delta n which is this So this is a this is a relationship between kp and kc And that also sometimes is good and useful for you to have So that is the second concept in chemical equilibrium The third and most important concept the last concept of this is No, there are two one more concept. Yeah, the last concept. This is now about reaction quotient So we are saying that look, uh, you know at equilibrium all the concentrations are fixed So at equilibrium kc is equal to have a fixed concentration of cc dd and a and bb Okay, but I am saying that okay. If that is the concentration then Okay, this is the concentration Then what happens if I change the concentration of let's say the reaction is not reached equilibrium How do I know that when will it reach equilibrium? Where is it going? What is happening to it? Etc. So what I do is I simply take a ratio and call it as q this the ratio may not be at equilibrium This is not at equilibrium necessarily. This is not at equilibrium necessarily. It might or might not be so therefore It is not at equilibrium necessarily Okay, necessarily, okay I've done some mistake here over whatever. Okay, not an equilibrium. So I simply say that q is equal to this ratio simple ratio as c to the power c d to the power d and a to the power a and b to the power b Let's say I end up getting q to be greater than the k value which is given to me It implies that my numerator is greater and my denominator is lesser Therefore the reaction will bring down the numerator and increase the denominator Which means that the this will go from products to this is the products, right? This is products to reactants and therefore the reaction will go Backwards this will that this is a q is greater than the implies that the reactions will go backwards if q is less than k implies that my Products is lesser than my reactants and therefore the reaction will go forward And if q is equal to k implies that the reaction is already at equilibrium and there will be no change No change. So this is one reaction quotient Problem that will come to you where you might be asked for a multiple times as to What are the different ways of Really doing, you know, what are the ways of The progress of the reaction based on the qc and k values Okay Now the last topic which is important There is so there is one topic where we have different factors that will actually affect the chemical reactions But more than that the most important factor is lech atelier's principle So so there's only one Yeah, so lech atelier's large atelier's principle or lech atelier's principle Simply say that anything that you do to the reaction Now when you say that you are doing to the reaction means you are doing to the surroundings For example, you increase the pressure of the surrounding you increase the temperature of the surrounding you put some substance in the surrounding Now this surrounding will start affecting the reaction Now the reaction earlier was in an equilibrium with the surrounding you came in and you did some disturbance you added pressure or you added some extra substance or you simply Put in heat whatever you did the reaction will now Try to achieve another equilibrium with the surrounding. I'm going to tell you a story about it It's like saying you were living very peacefully in the community in the society in the building that was that was happening Suddenly someone came and started parking in your in your parking slot Obviously your routine is disturbed and you are angry about it. So what will you do? Then you go to the society and you say, okay, look, this is my parking spot The society guy says look they they don't have a parking and there is a problem with the society Let's sort it out So what do you do then you wait for a society to sort it out then you both come to an agreement If there is no other space then you will say look keep it for some days But after some days allow me also to park if there is a solution you will say, okay Take take that space there or put it outside or whatever it is So you will come to an agreement and understanding with the society over a few days. That is exactly what is happening here The reaction was in equilibrium with the surrounding Someone came and increased his temperature put in some substance increased his pressure or whatever it is The reaction now wants to have the equilibrium again So what does the reaction do the reaction nullifies or the reaction reverses whatever you did To the system. Okay. So for example, you increase the temperature of the surrounding The reaction will behave in such a way that it will absorb the heat absorb heat So you increase the temperature the reaction will absorb the heat so that the temperature of the surrounding comes down You increase the pressure the reaction will actually start compressing itself The reaction will work in such a fashion that it will compress itself So that the external pressure is brought in and this increasing pressure is again brought down to the normal value Sometimes it so happens that the entire pressure cannot be consumed. So you will come to some equilibrium point, which is okay Which is okay, right? So let's say you increase the pressure by 10 So reaction will not be able to accommodate all the 10 Atmospheres it will accommodate five and it will keep five in the surrounding and we'll say, okay Now it is an agreement between you and me that the 55 pressure we maintain This agreement is decided by the k value The k value so whenever any any change has happened you basically try and find the q After the q you again change the you know weight till the q goes to k because the q will tell you whether reaction is going forward or backward and it will again go to either You know a reaction Going forward or going backwards depending on what is the case Now what i'm going to do is i'm just going to quickly do a problem for you and Then you know, maybe you you'll have an understanding of it in a much deeper final for a deeper manner, okay, so let me give you a problem on chemical equilibrium A quick question is everyone there with me? Okay. I think every everyone is there Yeah, people you have questions. Please post don't feel feel free if you don't want to put it on the comment You can always text me personally My whatsapp is just besides me. So feel free absolutely to be You know On the on the group, etc Or not on the group even if you want to give it to me send out personally that's also okay, so Okay, so I see that I don't have any other questions or queries or problems Okay, okay fine. So then I'll proceed towards the problem so So here's a question, okay, it says that For an exothermic formation of sulfur dioxide sulfur trioxide So here's the question. So the question says that twice of s o 2 plus o 3 Both are gaseous in form gaseous gaseous in form Okay, and it is it is giving out It is given out giving out twice of s o 3 gaseous. Okay Uh, essentially they are saying is that k p since they see not all of them are gaseous So therefore I have no other no other choice but to use k p because that is the most Uh, uh, you know, uh, comfortable and convenient uh formula to use so 40.5 per atmosphere And this is k p for this reaction is given at 9900 kelvin and Uh Delta h of the reaction is given to us Delta h of the reaction is given and this delta h of the reaction is equal to minus 198 kilo joule per mole. Okay, this is the this is the data given to us The first thing that they have asked us is write the equilibrium constant expression for equilibrium constant k p for this reaction Okay, the k p is question mark they have asked and then the second thing that they have asked us is At room temperature that is at 300 kelvin approximately at 300 kelvin Uh, what will be k will k p be larger? They have asked k p will be Uh, larger or larger or will it be smaller or will it remain same? Okay Or equal equal to k p at 900 kelvin. Okay, so they have asked this as the second question The third question that they have asked is How will the equilibrium be affected if the volume of the vessel containing the three gases is reduced? So volume is reduced keeping the pressure temperature constant volume is reduced with t equal to constant Uh, what happens? Okay, so they have asked what happens with that and the fourth question that they have asked is What is the effect of adding one mole? So we are adding plus so we are adding plus one mole of Helium gas Achieve gas why we are adding helium because helium is unreactive to a flask which contains to a flask Which contains uh so two or two and so three at equilibrium to a flask which contains above reaction At equilibrium. Okay, so this is what they have asked now Again, I'll pause here for 30 seconds so that you get time to pause this Whenever you are listening once again, uh, you know after this session and you can I mean well if you have questions you can post them. I'm just looking at if you have any other questions Okay, so there are 13 of you heard a lot of you are not there. I don't see the adityas and anirudh and All of the other people even tanisha is not here. Uh Ravi kiran Okay, then momika is not there. Ruhi is not there. Dhatri is not there. All of them are I don't know where these guys are Hopefully they watch this video one again once again and probably get Enough of the understanding for tomorrow's exam. Maybe that helps them Okay, coming back. So you could pause here and now I'm I'm going to give you the solution in just in case if you uh Again wants to recheck so check now from the equations that it is firstly and exothermic reaction So that is very important to know the second thing is this is at a very high temperature Exothermic reaction at a very high temperature Kp is 40.5. Okay Now so first thing that they have asked is what is the expression for kp The kp expression will be nothing but partial pressure of so3 Back to the squared it will be squared by because it is 2 so3 divided by partial pressure of o3 Plus partial pressure of so2 Square uh into into sorry, this is not plus this is into into partial pressure of so3 square Now there was a one one very funny question that I had Okay, is that someone actually questioned me saying that sir Uh, the delta h h of formation of elements we take as zero So why are we taking that in partial pressure here of elements now? Please understand these are completely unrelated topics The uh, delta h formation is enthalpies and energies of the elements This is chemical equilibrium. That is thermodynamics where we are talking about energies here We are talking about rate constants and rate constants elements will be taken care of They have to be put here into Into the rate expressions. Okay, so, you know, please don't relate one something from someplace to something else completely unrelated Okay, so this is kp where we have this is the expression kp for that the first question is solved Then the second question at 300 kelvin kp will be larger smaller or equal to kp now, please note At 300 kelvin, we have decreased the temperature. This is an exothermic reaction Since we have decreased the temperature the reaction will run forward This will run forward Like anything why it will run forward because it wants to give out heat to compensate the decrease in temperature So it will want to give out a lot of heat Okay, so since it wants to give out heat s o 3's pressure will increase Uh, and s 2 s o 2 and o 2's pressure will decrease because the reaction is going forward and therefore kp Year will be larger. So the answer here is very very large In fact, it will be very large than what it was at 900 kelvin So at 300 kelvin kp will be much larger than at 9100 kelvin So that is that is the second scenario. Okay, I'll I'll repeat again because this is an exothermic reaction and You know, we are decreasing the temperature So decreasing the temperature favors exothermic reaction Because the reaction will want to give out more heat at a lower temperature to balance out the change in the effect The third thing is what happens if the volume is decreased now? Please note the volume of the products is two moles. Okay two moles two moles is at products And the total volume here is three moles. Okay three moles at reactants So if reactants are more Then volume is more if products are more than volume is less Now we are decreasing the volume means the reaction will want to increase the volume So the reaction will go from products to reactants. So if volume is decreased The reaction will will go backward. The answer here is backward. Okay So the reaction will heavily start moving in the opposite direction And what will happen is that the s o2 and o3's volumes will increase a lot than s o3's volume S o3's volume will diminish so the proportion of s o2 and o2 in the reaction mix will be more The last question that they are asking what happens if one mole of h e gas is added Now, please note one mole of addition of h e gas is doing nothing but except for increasing the pressure So this has nothing to do with the reaction It is simply going to because the more one more substance has added into the solution the total pressure The total pressure of the system will increase Why because in the same volume now one more mole of hydrogen has Helium helium gas has to be added So the total pressure has now increased as the now please check here. There is two moles. So therefore there is lesser pressure Here there is three moles. Therefore there is more pressure and we have increased the pressure since we have increased the pressure The two more pressure will go to lesser pressure to accommodate the increase in the pressure. That is leach at least principle So when we increase the total pressure the reaction will go forward the reaction will go forward. Okay, so that's That's the quick quick answer to to the question. So all the four Answers have been given to you and I think that brings us to the last topic in chemical equilibrium If you have understood to this lecture, I think we have really done about whatever is going to come to your paper tomorrow 60 to 70 percent will be from this whatever we have discussed apart from you know I mean in these topics not from the other topics in these topics I think now the only remaining part Remains is how much you assimilate and really are able to reproduce tomorrow. Okay So now again, I'm going to pause. I think what my agenda was there for today's class I'm more or less at the end of it. I would like you guys to again question anything if you have I'll wait for a few seconds if you have anything to talk about else. I'll I'll you know try and You know, we will try and connect on something else. I'm as I said, I'm again available on the Phone if you want, please let me know and I'm here So I'm just pausing here because I think there's much that we could do and We we could connect Yeah, okay, so it seems that Yeah, so I it seems that all of you are content Yeah, we are also coming to the end of the time So can you guys quickly tell me whether you guys were so instead of just saying, you know, we connected and all Please please try and Share what what what did you feel? How is the exam? Are you prepared not prepared? Can we just put in a few comments and let me know how you guys were? How do you guys feel about tomorrow's exam? Yeah Are you ready? Do you feel confident? Do you feel okay? Do you feel stressed out all well? Huh? Yes, I'm I'm looking forward to your connects to your Messages Okay, so there are I think only five of you you're around now If there's anything from you five guys, just let me know else.  I should you are trippin kirtana Sachi Aditya Shamik Yeah If any of you are here, just please let me know Or else Yes, please Okay, so it seems that all seem to be good. I don't see any responses coming on It seems that you all are okay and doing good then So almost done, but still a bit nervous Okay, good. I think being nervous is also a good healthy sign. Just don't stress yourself But I am sure that you will do good. I think I sure you have been studying as well. So I expect you to do well, you know, so don't worry about it But if you have understood what we've done today and if you feel confident about it I think you will do pretty well in the exam Not an issue Okay, trippin arpita vansi. I think you all will anyways do good kirtana also will do good Uh, yes, I think you have been connecting. Aditya, if you have any issues Ping me on whatsapp as you've been doing, you know, I will also try and help there Ronak, if you have anything you can always let me know Okay So I would not take much of your time. I think you guys have been fantastic already with all the work all the very best to you My best wishes just feel free to connect. Also, please do keep do send me a message How did your people go tomorrow and I really think that That it would be only all the good news that would come in and not really Anything else. So So let's pause here for a bit. Uh, and you know, you guys are free to study, you know, use your time well Okay, all the best take care. Bye. Bye. Thank you The video will be here on the youtube in another five minutes. You can always refer to that Yeah, take care guys. Bye. Bye