 hope this is visible to all. Yes sir. Good. So energy change in chemical reaction, right? Basically energy is what? Energy is just the capacity to do work or just a capacity to transfer heat, okay? Now another term comes that was heat. Okay, so we were talking about energy that is a capacity to do work or transfer heat. Okay, now heat. What is a term heat? We have never seen that, right? Heat is a form of energy that flows between two objects. See here, you can see some smoke is moving from its higher concentration to lower concentration, right? Because of heat. So heat is a form of energy that flows between two objects because of difference in temperature, right? That means heat moves from higher temperature to lower temperature. Basically we have different different types of energies. So first light energy, then electrical energy, kinetic and potential energy. You all must be aware with kinetic and potential energy very well, right? Now let's talk about thermodynamics. What is thermodynamics? Thermo means thermal or heat. Dynamic means movement of heat. So basically in this branch of science, in this branch of chemistry, we learn movement of heat, how heat changes its phases, right? Or how heat moves from higher temperature to lower temperature. Thermodynamics is the branch of science that deals with relation between heat and other form of energy, right? Like mechanical energy, electrical energy, chemical energy and it also deals with the relationship in between all the forms of energies. As we have discussed in the class like entropy, enthalpy, internal energy, what are the relationship in between internal energy and enthalpy, right? So all those things are part of the thermodynamics. Now in thermodynamic, we have a system that we have to study, right? So basically what is a system? System that means a quantity of matter or region of a space chosen for a study. Or what should I say that the part of observation, right? This system is surrounded by a boundary, correct? Which separates this from surrounding, right? So boundary, it can be real or imaginary, okay? Please keep this thing in your mind that boundary of a system can be real or imaginary. This separates the system from its surroundings. Now what is surrounding? We all know very well that what is surrounding, right? But just a simple thing, surrounding is not the part of our observation, okay? This is just a physical space outside of the system boundary, right? Now in thermodynamics, we have three kind of systems. Close system, open system and isolated system. We'll discuss about this. First of all, closed system. It's in closed system, I just can say very easy example, very normal example. We all see in our home, if suppose we boil something, right? And if we cover it with a lead, so this become a closed system. Guys, please excuse me just for a second. Yes, so we were talking about closed system. So the closed system is the arrangement where heat or energy is possible. But exchange of mass is not possible. Heat can be exchanged but mass cannot be exchanged, right? It's a simple example. If suppose we are boiling water in a vessel and if I have covered the vessel with the lead, then what will happen? If I touch the vessel after some time, I can see that, I can feel that it is hot, it became hot, right? So that means heat is transferring, energy is transferring from the system from the vessel but mass is not transferring. This closed system is covered by two kind of boundaries. The one which is fixed boundary, this one, and another one which is moving boundary, okay? So sometimes your boundary can be fixed, all the four direction boundary can be fixed or sometimes you can just choose according to your requirement, according to your observation needs. Next we have open system. Open system is the system where energy and mass both can be exchanged by surrounding, okay? Next we have isolated system. Here we'll see difference in between open system, closed system and isolated system. So in heat as we have seen that energy and mass both can be exchanged. In closed system only heat or energy can be exchanged but in isolated system heat and energy both cannot be exchanged from the surrounding, right? The best example for isolated system is what? Thermos, what we use in our home, right? Okay, now in thermodynamics we have different different kind of processes and the path of those processes. So basically how many processes or which kind of processes we have? The first is isobaric process, isobaric. Baric that means pressure, right? The process which is going on or which is proceeding at constant pressure. Isothermal, that means thermal, again related to temperature. So the process which is going on at constant temperature, which is proceeding at constant temperature. Isocoric that means constant volume, then isentropic means at constant entropy. I think you might have learned about all of this thing. Isentropic must be new for you, right? So this is related to entropy of the process at constant entropy. Now what is the path of the system? Path is the steps or series of states. A system passes through during the process, right? Steps of the system or the series of states which system passes during a process. Okay, now we'll see examples of these processes. The first if I say isothermal process. What was entropy like in the before slide? Entropy. Entropy that means randomness or disarrangement or order of randomness, order of disarrangement. You know in the universe every planet is going far from each other, right? The distance is not constant, that distance is increasing, continuous increasing, right? So what is happening? The entropy or the disarrangement of the planets is increasing. Entropy is increasing. I'll just say one small example. You know that you guys know that I have a small kid, right? So before having a kid like a suppose he is going to school, my home is well arranged. Everything is at the proper place, right? But when he comes back, what happens? He throws everything everywhere, right? He stores his clothes. So what happened? That disarrangement of the things is basically entropy, right? So come back to the chemistry topic. Now if we say what is entropy in chemistry? So suppose you have some gas molecule in this vessel CO2 at the normal temperature 25 degrees Celsius, right? Now if suppose I provide some heat. So if I provide heat, what happens? Kinetic energy of gas molecule increases, right? They start moving randomly in random direction with high speed, right? So what happens? Expansion and molecules, right? Disarrangement in molecules. So that is known as entropy. Got it? Yes ma'am. Okay, good. So the next we'll see some examples of isothermal process, isobaric process and all the processes what we have learned. Isothermal process freezing of water at minus 10 degrees Celsius, right? Isobaric process heating of water in open air. In open air. Why open air? Because pressure is constant. The atmospheric pressure is 1 atm. So if that pressure will be constant, right? Then isochoric process. Now isochoric process that means we have to keep volume constant. So the example can be heating of gas in a sealed metal container, right? If suppose you have this metal container and if you are heating this metal container, whatever gas is present into that, volume cannot be changed, right? Because it is sealed. Okay, next we'll see reversible processes. Reversible processes also we have learned. Reversible process that means a reactant can convert into product and again product can convert back into reactants, right? See I'm talking about reversible. I'm not talking about equilibrium, okay? So in the system if the system is close to equilibrium at all times, alteration of the conditions can restore the universe, right? Like if I say universe, that means what? System and surrounding to the original state. Next we have cyclic process. Cyclic process you might have learned some cyclic compounds in organic chemistry, right? Something like this or something like this, right? What does it mean? That means once you start from one side and then you end it by the same side, right? By the same molecule and you are coming up at the same molecule. So cyclic process that means the final and initial state are the same. Final and initial state are the same. What does it mean? Suppose you are using solid and gaseous reactant. It converts into let's say gaseous reactant but the product of these gaseous reactant is again solid and gaseous product, got it? So this is basically cyclic process. Again the state is same. Okay the next is what? Adiabatic process. You have learnt about this adiabatic process in the glass, right? In this process DQ is zero during the process. DQ that means what? Heat, okay? No heat is added to the system or no heat is removed from the system. Ritu please mute your mic. Okay so adiabatic process that means again it's no heat can be added to the system or no heat can be removed from the system, okay? Next we have internal energy. Our main topic starts from now. Internal energy is generally represented by U or E, okay? Now what does it mean? What is internal energy? Internal energy is just the amount of energy associated with the system due to its existence, okay? If any system wants to exit in this world whatever energy that system is required, whatever energy that system is carrying that is known as that is internal energy. This is a state function, right? Internal energy is state function. What is a state function now? A state function is basically the property which depends upon the state, right? Extensive property. State function and extensive property. State function that means the value depends upon the state, extensive property, it depends upon the amount of the substance. See logically also I think you can see this it's applicable, right? Internal energy is depending upon the state definitely water if water is present in liquid form its internal energy is different. If water is present in solid form its internal energy is different because the state is different, right? And again it depends upon the amount also, right? Internal energy is the sum of all the energy present in that substance, all the energy. I have written only 5, 6 here but there are so many energies present in the substance, right? E t stands for transition energy then E r this is for rotational energy, E v this is vibrational energy, right? And E i, i is interaction, okay? E l is nuclear energy of the molecule. E is electronic energy or electrical energy, okay? Apart from that we have potential energy, kinetic energy, right? So basically U is what the sum of all the energies present in the substance. Mom, I and E is for what? I, I is interaction energy of molecules, interaction because the substance whatever we are talking about let's say CO2. So CO2 in that molecules of CO2 must be interacting, right? With each other or with some other molecules so it must have some interaction energy. I and E you were asking E is electrical energy, okay? Time let's say your substance is not electrically charged let's say your substance is neutral, okay? So electrical energy will be zero, right? But sometime your substance is having charged particle like OH negative. It has electrical energy, right? It has one charge, it has one negative charge, right? So that is how all these values are important to calculate internal energy, huh? Tell me. It's written here. I can't, it's written. Yes, yes, tell me. I can't see your screen. I don't know why. I've logged in and logged out, I've logged out and logged in like quite a few times and I've switched off my computer but I still can't see it. Oh, okay. I cannot stop sharing the screen because if I stop then again it will take two, three minutes for me to upload this PPT here. So can you just contact to Ankur sir or to Shar sir for this problem? Okay. And just try once again, logged out again and then try to log in. Okay. Arthesh, you have messaged me. What an example, example for what? I didn't get your point. You can unmute your mic and then you can discuss with me. Okay. So let's, I was talking about internal energy and internal energy as I have told you that it is the sum of potential energy and kinetic energy both, right? So suppose at a time if I'm saying that my car is moving. So what happened? If it is in moving condition then car must have some kinetic energy, right? Kinetic energy is not equal to zero. But at that time potential energy will be zero, right? So it doesn't matter that any energy is zero but the energy value should be added, should be there, right? Okay. The next part of this internal energy is what? Higher the temperature, higher the internal energy. Okay. That's important. If temperature is higher then internal energy will be higher. Okay. Change in temperature. We need to monitor change in internal energy. Okay. Don't get confused by this E. This E is the another representation of internal energy only. Guys, am I going very fast? No, mom. Okay. Great. So now the first law of thermodynamics. It's very simple. Mom, can you move to the last previous slide once? Higher the temperature, higher the internal energy. If anybody is missing out anything, then please don't worry. I'll share this PPT. Okay. And these things are basic. So don't worry about it. Thank you, mom. Done. Okay. Now the first law of thermodynamics. First law of thermodynamics that is what? Energy cannot be created cannot be destroyed. Energy cannot be created, not destroyed, right? Total energy of the universe is constant because obviously that energy cannot be created cannot be destroyed. So the sum of total energy will be constant ever, right? What happened? Energy can convert from one form to another. Okay. So energy can be converted from one form to another. Let's see this equation. This proves the first law of thermodynamics. See how delta U is equal to QP plus W. This QP stands for heat of the system. So delta U is equal to QP plus W. Internal energy is the sum of heat plus work. Okay. Internal energy is the sum of heat plus work. Suppose I give any kind of energy, I give just energy to any system. Definitely, what will happen? Expansion in my molecules. Let's say I'm here I have water or let's say any gas. I'm heating the system. What will happen? Expansion in gas molecules. Definitely it will happen, right? Expansion in gas molecules and moreover, this vessel will be hot. What happens? You are not able to. Okay. Let's continue. This vessel will be hot, right? Then what happened? The whole system, this whole system, okay, is having internal energy U that is equals to work by this gas. Yeah, I have an Android phone. Okay. Okay. Let's continue. This is equal to the heat plus work done by the system or done on the system, right? So the change in internal energy, why the change in internal energy? The energy before heating this vessel and energy after heating this vessel. Okay. Now let's see change in internal energy at constant pressure. Definitely we have to do it at constant pressure, right? But temperature cannot be constant. If you put temperature also constant, then how will you provide heat? Am I right? So at constant pressure, delta U that is change in internal energy is equal to Q, it should be QP, Q plus W, right? Q stands for the heat of the system, W for work done by the system or done on the system, correct? We have discussed this work done by the system or done on the system in the class in detail, right? If you have any confusion, any doubt, ping me in this chat box or message me personally, I'll explain here. Now because W is equal to P delta V. Why it's P delta V? Because pressure is constant. If the pressure is constant and temperature is changing, then there will be variation in volume of the gas, right? So at constant pressure, change in volume. W is equal to P delta V. Now if I rearrange this equation, then I can calculate the amount of heat of the system. Q is equal to delta U plus P delta V, right? Delta U is equal to QP minus P delta V I have used here and here I have used plus W. Then what does it mean? That W is having value minus P delta V. I'll show you when W has minus P delta V, okay? See here, work done on the system, then W is positive, okay? If work done on the system, then W is positive. If work done by the system, in this case, W value will be negative. Remember? Yes. So let's see some differences in between heat and work. Like what is heat exactly and what is work? I'm sorry to intervene, Sraddha. Can you hear me? Hello? Yes, yes. Yes, I can hear you, sir. Yeah. The thing is, I think, you know, have you shared the entire screen or one particular app? Sir, I have shared only this presentation. That's it. Actually, what happens? What is happening is few of these students are not able to see the PPT. They can only see you writing over there. So, yeah. So if there is nothing, you know, if you can manage with this thing, you know, you can share the entire screen, maybe, or re-share your application. So what application you are using, by the way? I'm not using any application, actually. In this Moxtra only, it is share file, right? No, it's not being. Okay. So you have shared the file and then you are annotating over it? Yes, yes. Okay, okay. Fair enough. So now what should I do? Actually, what you can do is you can share the entire screen. Okay. If I share entire screen, then how will I, how can I use that annotation? No, no. Okay. So you are using a PPT and then... Okay. Okay, fair enough. So, you know, do one thing, process once again, you know, because others can see. So maybe you can make once again and then maybe that will happen. Anyone has any difficulty in viewing the screen? Anyone? Guys? No, sir. Can you see the PPT and the annotation over it, right? Yes, sir. Never mind. So, Shradha, I request you to just share once again and then I think... Sure. I'll do that. Okay. Thank you, man. Okay, guys. I'm uploading PPT again. I think we should wait for a few minutes. I'm so sorry for the inconvenience. No, no, no problem. It's getting uploaded. So, is it visible to you right now? See this PPT? Yes, ma'am. Yes, ma'am. Okay, great. Thanks for the carry on. Thank you. Sure. Thank you, sir. Okay. So, can you just remind me what we're doing? Difference between heat and work. Thanks, Vaishnavi. Yeah. So, difference between... Not exactly. I should not see that difference actually. What is heat and work basically, right? So, heat is just the simple thing that transfer of energy from a different temperature zone. Simple thing. The different temperature zone, okay? And here, energy expanded to move an object again. Important, right? What is work? Work, the work means energy we are using to expand or just to change the position of any object. Okay? This is in the term of physics. Okay? Let's come back to the chemistry part. What is work? Work means change in volume of any gaseous substance or any liquid substance. Okay? Change in volume. Let's say if I have some gas here and I provided some kind of energy, heat energy to this, what will happen? My gas molecules will start to expand. So, I can see that that there is some work done by the system, right? If I have covered this with a non-moving piston, okay? Then what will happen? This gas molecule cannot expand anymore, right? Right? Then there is no change in volume. So, no work done, right? Work done can be calculated by change in volume. See the formula says itself, W is equal to P delta V. If there is no change in volume, that means no work done by the system or no work done on the system, okay? You all might have learned one formula that W is equal to F into D. D is displacement, F is equal to force. So, if displacement is zero, if object is not moving, that means your work is what? Zero. It becomes zero. Similar thing is here. If your molecules, atoms are not moving, they are in constant volume position, that means work is zero, okay? Now, we have signed conventions for heat and work. Let's say for heat, if we have endothermic system, that means energy absorbed by the system. So, because energy is absorbing, energy is adding to the system, this Q will be positive, right? The Q must have positive value. And in case of exothermic system, because energy is releasing out from the system, Q will have negative value, right? Similar thing we have in work. If work done on the system, that means work is done on the system, like we are doing something for the system. So, for us, work will be positive, right? And if work done by the system, that means for us, the work should be negative, okay? If you have any confusion in to understand work done by the system and on the system, you can discuss with me. Okay. Next topic, enthalpy. Basically. I have a doubt in the previous slide. Tell me. This Q over there, the single Q written over there, what does it symbolize? Actually, this is the symbol of heat, QP. Symbol of heat, that's it. If I see U, delta U is equal to QP plus W. So, you just can simply say the symbol of heat, heat plus work, heat at constant pressure, that is why QP, okay? Okay. Okay. The next is what? Enthalpy. Enthalpy is very simple thing. Just the transfer of heat, exchange of heat or amount of absorbed heat or released heat. That's it, okay? So, enthalpy is what? Heat transfer. It is denoted by H. H is equal to U plus PV, okay? But we cannot calculate enthalpy of the system. We only can calculate change in enthalpy of the system, okay? So, change in enthalpy delta H is equal to change in internal energy because the system is converting into the another form, like your reactants are converting in the system into another form. So, change in internal energy plus delta PV. Now again, I'm saying you that P should be constant. So, if P is constant, then delta H is equal to delta U plus P delta V, right? This is our main formula. Delta H is equal to delta U plus P delta V, correct? Now we know from the previous slide that Q is equal to delta U plus P delta V. So, can I write delta H is equal to Q delta P, right? So, enthalpy is what? It's just the heat transfer, got it? Okay. Now, is used to quantify the heat flow into or out of the system, correct? So, change in enthalpy is equal to enthalpy of product minus enthalpy of reactant, right? We'll see some numericals based on this. Change in enthalpy is what? Heat given of or absorbed during the reaction at constant pressure. Don't forget to write this in exam, okay? This is the key one. Change in enthalpy is heat given of or absorbed during a reaction at constant pressure, okay? Let's see how. Here we have two reactions, okay? In one production of water or formation of water. In another one, dissociation of mercury oxide, right? So, formation of water is exothermic reaction, okay? Formation of water is exothermic reaction. Why? Why it is exothermic reaction? Because formation of water is about formation of new bonds. Ducation of HGO is endothermic reaction, okay? It needs some amount of energy to break these bonds, okay? So, in exothermic reaction, we all know that delta H for exothermic reaction is what? Negative, right? Delta H for all exothermic reaction is negative and delta H for all endothermic reaction is what? Positive, right? Let's see how and why. See here. In this slide, don't worry, I will go through the slides what I have skipped, right? Here we have two kind of reactions, exothermic and endothermic. In exothermic reaction, the energy, this is what? This is the energy or enthalpy, anything is fine and this is the reaction path or reaction progress, okay? I can say reaction progress or reaction path. Now, if I say that my reactants are here and my product is here, here is my product and here is my reactant, definitely that reactants are at higher energy level. So, when reactant convert into product, it has to lose some energy, right? So, because of this losing of energy, this graph is going in downward slope and what is the reason? The reason is HR is greater than HP, correct? So, what is our formula? Our formula is delta H is equal to HP minus HR, correct? Now, if I say your R is greater than P, what will happen? My R is greater than P, then my result will be negative, right? Similar thing we have for endothermic also, delta H is equal to HP minus HR and here because product energy is more, so can I say HP is greater than HR? Definitely, when HP is greater than HR then delta H will be positive, got it? So, that is why the signs for delta H is negative for exothermic and positive for endothermic. Now, how will we represent this energy in thermodynamical equation? As the water, formation of water is exothermic reaction, so in product side we have to add energy, right? If we have exothermic reaction, energy must be added towards the product side. If we have any endothermic reaction, then energy must be added to reactant side, okay? In endothermic reaction, energy must be added to reactant side and if you write in a way of delta H, then delta H is equal to negative amount of energy. Here in endothermic reaction, delta H is equal to positive energy, okay, leave it. Now, the same thing what we have discussed, enthalpy is what the measure of heat content of a substance at constant pressure, we cannot measure the actual enthalpy, so we always measure enthalpy change or change in enthalpy. All these things we have discussed just now, right? Okay, let's do this question. Which of the following reactions are exothermic or endothermic? The first reaction is carbon react with hydrogen to form C4H10 butane and the delta H value is negative. Chal, guys, I am waiting for your answer. The first one is what? Exo or endo? Please message. Very good. I got answer from most of you, right? Okay, so the first one is what? Exothermic. Okay, the next one is, tell me next one is fast, good. Next one is endo. Very good. Third one is, third one is exo. Guys, please check your answer. Some of you have sent me the answer as endothermic. No, the third one is exothermic. See, energy is positive here in the product side. If we add energy in the product side, it is exothermic. Why? Because the representation of delta H will be minus 45 kilo joule for this. And again, delta H value, if you have a negative sign, then it should be exo. Got it? Last one, very good. The last one is endo. Good. Very good. Chal, the next, okay, these graph we have discussed, right? So, let's do one question based on this graph. Consider the following reaction. This is this. Give the graphical representation for energy change in the bow reaction. So, basically, it's a formation of order and the important thing is what? Change in enthalpy given here. It's negative. First of all, what do you need to understand? If delta H value is negative, then HP will be, okay, what I was saying is, yeah, if delta H value is negative, then HP value is less than HR, right? That means HP is less than HR. So, let's draw a graph. This is energy or enthalpy. This is path of reaction or progress of reaction. So, in this case, let's say, here I have reacted, here I have product. H2 plus half O2 and here I have H2. This is the enthalpy for H2. This is the enthalpy for hydrogen and oxygen. Now, because as we know that the enthalpy of reactant is greater, so it has to lose energy to come to the product side. Looks better. Okay. So, what do you have to write? That HP is less than HR. So, delta H value is negative. Done. Let me check if I'm forgetting something. No, that's enough. Good. Yes, actually, this arrow indicates a release of energy. Okay, release of heat. Release of heat. Okay. The next, delta HR. Delta HR means what? Change in enthalpy of reaction. This R stands for reaction. So, delta HR is equal to sigma delta, okay. I think this delta sign change has been changed when I was changing the letters. Okay. So, this is your basically delta H. Just write delta H, okay. Only delta H. Don't even write C. Sigma delta H of reactant minus product. Is it in the opposite way? Okay. I'm correcting this formula here. Delta HR is equal to delta H of product minus delta H of reactant. Sorry, guys. This was about the delta Hc. When we do calculation in the reaction of combustion, enthalpy of combustion reaction, then delta Hc of reactant minus delta Hc of product, it happens. But I think this is not in our portion. So, just leave this equation. We will learn this equation in 11th grade. Okay. For you guys, just learn this one. What is happening? Who is joining and leaving again? I don't know. Okay. Leave it. So, delta HR is equal to delta Hc product minus delta Hc reactant. Okay. Another formula for change in enthalpy of reaction, what we have, energy required to break bonds minus energy required to form formation of bonds. Okay. Hope you are done with these formulas. Let's do one question based on these formulas. Calculate the enthalpy change for the reaction. The reaction is what H2 plus Cl2 gives to HCl. Guys, do it and send me the answers. I'll wait for 30 seconds and then I'll try to do it here. 246. Isn't it? Take your calculation. I don't think so. It's 246. But good. At least you tried. Okay. I'm doing it here. So, first of all, what is the reaction? H2 plus Cl2 gives to HCl. Now, H2 and V, what are the things we have? HH bond enthalpy, ClCl bond enthalpy, HCl bond enthalpy. Right. So, let's see how this reaction happens. First of all, this H2, that means hydrogen and hydrogen. This react with this chlorine-chlorine. So, what is happening? First of all, hydrogen, hydrogen bond will break and chlorine-chlorine bond will break. Right. So, what is happening? Cleavage of these bonds. Cleavage of these bonds. After this, they form new bond. They form HCl, HCl. Okay. So, here, formation. Correct. Now, what is our formula? Formula is energy required to break the bond minus energy required to form new bonds. Energy required to break this bond is HH is 437, 37 plus because here we have chlorine also. So, chlorine is 242 minus HCl. HCl is 433, but don't forget we have two molecules of HCl. So, enthalpy must be multiplied by 2. Okay. 437 minus L. Okay. Let's see how many of you have got the answer. Hmm. Minus 187. Good, Smithi. Before that, who was there? Shadha, Adarsh, it's minus, beta. Tanei, good. Bhoomi ka, it's minus. Okay. Minus 187. Good. Let's go. Next. Determine the enthalpy change for the following reaction. CH4, Cl2, CH3, HCl, blah, blah, blah. Okay. You try. You try it by yourself and I'll also do it here. Hope you guys are done. I didn't get any answer till now. Okay. So, CH4 plus Cl2, that means chlorine and then chlorine convert into CH3Cl, CH3. So, CH3, H, H, H here and H and then Cl. Okay. So, basically here we have four CH bond. Four CH bond. This is the CH bond. Its value is 412, 4 into 412 plus Cl2, it's 242. So, we just have one chlorine-chlorine bond. So, 242. Then we have three CH bonds. So, 3 into 412 plus one Cl bond. Because of this one Cl bond, we have to use this Cl bond enthalpy, 338 plus one HCl bond. So, 431. 890 minus 2005. That should be equal to 2 minus 5. Very good. Awesome, all correct answer. Okay. The next question is, calculate the enthalpy change for reaction and it's not about the bond enthalpies and all. It's a simple formula, delta H is equal to H of the product minus H of the reactant. So, carbon dioxide and hydrogen react to form carbon monoxide and water. Even enthalpies are carbon dioxide, carbon monoxide and H2O. They have given. Now, we have to calculate enthalpy of the reaction. Simply, we have to calculate delta HR value. HR is equal to H of the product, sigma delta H of the product minus sigma delta H of the reactant. Okay. Again, excuse me. I'll be here in a minute. Sorry, guys. I need to take a break for, I need to take a break for two minutes. I'll be back in two minutes. Okay. Thank you. Yes. So, I'm back here. Okay. So, your answer is 38.4. Do this. Please take your answer again. Yes. It's actually minus 38.4. Good. Does anybody want me to do this here? Does anybody want me to do this here? I have got answers from Shraddhash, Smiti, Anand, Adarsh, Harshini, Ritu, Raghav, Medhanj, Bhumika. Should I do it here? Nobody's responding. Okay. I'm doing it here. It's not minus. It's positive. It's not minus. Yes. Yes. It's positive only. Okay. See, I'm doing it here. Given entropy, CO2, carbon monoxide, and this is given. Okay. How did you get that? Smiti, Shraddha, no, ma'am. Okay. Okay. Fine. I understood your point. Yes. Harshini, actually, enthalpy of hydrogen is zero. Why it is zero? Because anything which is having in pure elementary form, that enthalpy is always zero. Okay. Because currently, it is not reacting with anything. So, the enthalpy of pure element is always zero. Let's say if I say O2, carbon, nitrogen, always if you find this kind of elements, any kind of elements, the enthalpy for that element is particularly zero. Okay. So, for this, minus 241.8 plus minus 113.3 minus minus 393.5 plus zero. So, this comes minus 353.1 plus 393.5. So, it should be equal to 40.0. Guys, your answer is 38.4 is wrong. Sorry. It's 40.4. Isn't that minus 355.1? Yes. That's 353.1. Ma'am, no, ma'am. It's 355 when you calculate. Okay. Let me just do the calculation again. 241 plus 113.3. Okay. Yes. Yes. Correct. My bad. It's 355.1. Good. Okay. So, next we have standard enthalpy of formation or sorry, standard enthalpy change. Why it is a standard actually? As we know that standard, that means the concentration of your compound or product, whatever product it is, it should be 1 mole per liter or 1 mole per DM cube. Okay. The most important thing, if we write standard enthalpy, it is noted by not, this not stands for the 1 mole concentration, 1 mole per DM cube at, at 180M pressure and 25 degrees Celsius temperature. Okay. This is known as NTP, normal temperature and pressure, 180M and 25 degrees Celsius. Right. So, at this condition, whatever the change in enthalpy comes for any element, for formation of any element that is known as standard enthalpy. Okay. Now, enthalpy change of formation, enthalpy change of formation or I think, I can say standard enthalpy change of formation. Okay. So, first we discuss and we will discuss enthalpy change of formation when 1 mole of a compound is formed. Okay. So, that is usually known as enthalpy of formation. But if I say 1 mole, that means it must be for standard enthalpy of formation. Okay. Standard enthalpy of formation. But if the compound is like, maybe compound is forming 2 moles, 3 moles, whatever, you can just simply say delta HF. No need to write this, not thing for that. Okay. For example, carbon react with oxygen to form carbon dioxide. So, for this reaction, the standard enthalpy of formation, because carbon dioxide is 1 mole, the standard enthalpy of formation is minus 393.5 kilojoule. Okay. For which reaction? For formation of carbon dioxide. Similarly, for water also, we have a constant value minus 285.5, right? Because it's 1 mole of water. See here, if I change this amount of water, if I use 2 moles of water, then I have to multiply its change in enthalpy also. Okay. I will show you in one question. Now, again, we have another example, formation of alcohol. So, again, whatever the enthalpy will come, that will be enthalpy of formation of alcohol. See here, now, only 1 mole of product on RHS of the equation, elements in their standard state have 0 enthalpy. Guys, please pay attention here. The first point is, in the RHS, we must have only 1 mole of product in formation enthalpy, in standard formation enthalpy. Second point, the second important thing, elements in their standard state have 0 enthalpy. And why I have written graphite here? Because the third point, carbon is usually taken as graphite allotrope. Obviously, we don't react with diamond, right? So, carbon is usually taken as in the graphite form. And another form is the coke also available for that we have seen in metallurgy, right? Usually, carbon we use in a graphite form. So, this allotrope we have to use in this reaction. Next, we have enthalpy of reaction from enthalpy of formation. Okay. Simple thing. Now, here we have to calculate HR with the help of delta HF. Okay. We have this equation delta H is equal to sigma delta HF of product minus sigma delta HF of reactant. We know this very well, right? So, let's say this example. Please don't see the answer. Calculate the standard enthalpy of change following reaction, given that the standard enthalpy of formation of water nitrogen dioxide and nitric acid is 286 plus 33 and negative 173 kilojoule per mole. The value of oxygen is 0. Why? Because it is an element. So, now the sum of the product of all the sum of the products of enthalpy. What I am saying? Sum of the enthalpies of the product. So, the first is because we have 4 HNO3. So, HNO3 enthalpy should be multiplied by 4 minus 2 into hydrogen water enthalpy 2 into minus 286, right? Because we have 2 molecules or 2 moles of water, sorry, then nitrogen dioxide because we have 4 moles of nitrogen dioxide. So, we have to take 4 into 33. Okay. And 0 because oxygen is an element here. So, that this as in 0. After doing all this calculation, the answer is minus 252 kilojoule. Okay. Now, I am moving to the another question. This is very, very, very simple. Enthalpy change for following reaction is hydrogen plus half molecule of oxygen is reacting to make 1 mole of, sorry, 1 mole of hydrogen react with half mole of oxygen to react with 1 mole of water to produce 1 mole of water. So, for this delta HF is minus 286 kilojoule. Now, we have to calculate for this reaction. Tell me. Okay. So, I have got the answer minus 572. Why minus 572? Because minus 286 we have for 1 mole of H2O. As in the question, we have 2 moles of H2O. So, we have to multiply this enthalpy by 2. Okay. So, this is minus 572. So, I can say that delta HF or delta HR is equal to minus 572 kilojoule. Now, listen to me. You all have done this mistake in the symptom test in the class, right? Here I cannot write kilojoule per mole. I cannot write this. This is not applicable because this is not per mole. Okay. So, kilojoule per mole is only applicable for standard enthalpy or formation enthalpy. Okay. Okay. The next question. Yes. Calculate the standard enthalpy of formation of C2H4 from following reaction. Okay. Guys, do it. Now, the first reaction is incomplete, right? No, no, no. Actually, I think there's an arrow mark there. Yeah. Arrow. Yes. Arrow was missed. C2H4 combines with oxygen to form CO2 and water. So, this is basically combustion reaction. We have given the reaction enthalpy. Okay. See this properly. This is minus 1, 3, 2, 3 kilojoule. So, again, what they are asking is standard enthalpy or formation of C2H4. So, we need to calculate it for C2H4 so that delta HF reactant we have to calculate, correct? So, delta H0 is given. I'm writing it here minus 1, 3, 2, 3. Product, two moles of carbon dioxide and two moles of water. So, 2 into minus 393.5 plus H4 plus 0 because oxygen enthalpy is 0. So, this will be minus 1, 3, 2, 3. This is so bad. 1, 3, 2, 3. 2 into 293.7 will be, will become minus 287 plus 498 should be. So, now, it will be minus 1 to 85 minus delta HF of C2H4 is equal to minus 1, 3, 2, 3. Now, delta HF of C2H4 is equal to 1, 3, 2, 3. 1, 285. That is equal to 38 kilojoule per mole. Why it's plus? Why it isn't plus? Guys, please check your answer once again. This is in minus. This is in minus. If this goes there, that minus, minus, we will cancel out. Why it isn't plus? Yeah. Shruti, unmute your mic. Yes, it's positive. Okay. Everybody, everybody answered me negative 38. Please, yeah, don't do this kind of mistake. It's positive 38 kilojoule per mole. Okay. Now, why kilojoule per mole? Because we are calculating enthalpy of formation of C2H4 one mole of C2H4. Okay. So, enthalpy of formation of one mole of C2H4 is equal to 38 kilojoule per mole. Okay, Adash, I might have not seen your answer. Good if you have sent it to me in positive. Okay. Now, I'm moving to the next question. Okay. We'll see now enthalpy change of combustion. We all know very well that what is combustion reaction? Combustion, that means burning of anything, burning of any element, burning of any compound, right? In presence of oxygen. So, when one mole of any substance undergoes combustion, complete combustion reaction, then we say delta Hc. Okay. One mole of substance. Okay. Now, why we are not using that not sign? Because delta H0c, C here, that means all the reactants and products must be in their standard state. Okay. Must be in their standard state. So, because here it is 302, here it is half O2, it is not in their standard state. So, we cannot use delta H0c. We can just simply say delta Hc. For formation also, we can simply say delta Hf. Okay. What is the important part about this combustion reaction? All combustion reactions are always exothermic. That's very, very, very important. All combustion reactions are exothermic. All neutralization reactions are also exothermic. Right. What is the important part here? What is the tricky part here? When we write all combustion equations, you must have observed one thing. Whatever the number of carbon you have towards reactant side, same number of carbon or carbon dioxide gas will be produced towards product side. Correct. See the first example, one carbon will produce one carbon dioxide. See the last example, 2C2H5, sorry, C2H5OH, that means here we have two carbons. It will produce two moles of carbon dioxide. Okay. Similar thing is for hydrogen as well. Whatever the number of hydrogen we have, it will produce exactly half number of water molecule. See in the second example, H2 we have two hydrogens and so it will produce one mole of water. If in third example, ethanol, here we have six hydrogens. Six hydrogens. So it is producing three moles of water molecules. Right. So now come to the important points. Always one mole of what you are burning on LHS of the equation, air to balancing the equation. Okay. Now what we need to remember, we will get one carbon dioxide molecule or every carbon atom in the original form and one water molecule for every two hydrogen atom. The same thing what I have explained, right? Now if we are done with the amount of carbon dioxide and water, we need to check amount of oxygen. Okay. So now I think it should be very simple for you. It should be very easier for you to go from here. Go. Let's do this example. Heat produced by burning of one mole of ethanol in oxygen, gas to produce carbon dioxide and water is 1367 kilo joule per mole. Now all of you in the last test of the symptom test, you have got confused in between sign of this 1367 because they are already saying that heat produced, heat produced then this is very obvious that one it is negative 1367. Okay. It doesn't mean that there is no sign. So you will consider this as a positive. They are not talking about the enthalpy change. They just change the terminology and then they are saying about heat produced. That is equal to enthalpy change, right? So it is negative. It's not positive. Okay. Now the first, right? The thermodynamic thermo dynamical equation. And then second, we need to calculate the mass of ethanol to produce this much of energy 4 into 10 to cover 3 kilo joule energy. Okay. So the first what can be thermo dynamical equation? Burning of ethanol to C2H5OH because it is in liquid form. So liquid plus O2 as I have told you the same number of carbon atom it will produce carbon, same number of carbon dioxide. So two moles of carbon dioxide plus because we have six hydrogen atom in this. So six, sorry, three H2O. Okay. Now we need to balance amount of oxygen. So seven here, seven here, three. This is the equation. And this is liquid. Don't forget to write these states. Otherwise, you know that what will happen? Okay. Next, the next question. What mass of ethanol is to be burned to get 4 into 10 to power 3 kilo joule energy? One mole ethanol is equal to 46 gram. 46 is molar mass of ethanol. So 46 gram of ethanol is producing 1367 kilo joule energy. X gram will produce 4 into 10 to power 3 kilo joule energy. So X is equal to 4000. This is 4 into 10 to the power 3 into 46 divided by 1367. This will become 134.60. Let me check your answers. 134.6. Good. Just one minute. What I am saying is if you write your answer in this way, what the answer we have got? 134.6, right? 134.6. So there is a two way, there are two different ways to write the answer. 134.6 because they have asked in the question how much ethanol we need to produce this. So this is 134.6 gram. Okay. Now I am talking about the another question which is actually the same similar kind of question but they have used different terms. If suppose they are saying that substance X is burning after combustion it produces 150 kilo joule energy. Okay. X will produce how much of energy? Okay. Or they might ask what will be the value for delta H for burning of two X substance? Okay. So the calculation is very simple that you will multiply 150 by 2. Am I right? 150 by 2. But the writing way of the answers is different. How? Here in the first one they are asking, two X will produce how much energy? So you can say 300 kilo joule energy. Simply 300 kilo joule energy it will produce. Okay. But if they say what will be delta H value for two X then you have to write minus 300 kilo joule. Please try to understand this difference. Okay. You have to write minus 300 kilo joule in case of delta H value and if they want your answer in words then you can end it up with the positive 300 kilo joule energy is produced. Okay. Let's move on to the next question. Enthalpy of neutralization. So in enthalpy of neutralization we all have learned that when any strong acid and a strong base neutralizes the enthalpy of neutralization is constant that is minus 57.1 kilo joule per mole. Right? In the previous slide I have told you that value for enthalpy of neutralization is always exothermic. Correct? This 57.1 is what the standard value of neutralization. So definitely it must be for one mole only. Right? Okay. Let's write one formula here. Enthalpy of neutralization is equal to enthalpy of neutralization of strong acid acid and strong base plus enthalpy of ionization of ionization. Right? If I write in the symbolic form then can I write this delta Hn is equal to 57 minus 57.1 plus delta H of ionization. But true. Please write this formula properly. Okay. Now I'm moving to the next slide. One minute. Can you move to the previous slide? Okay. Okay. Okay. Done. Now let's do some numericals. Strong acid and strong base react with each other to liberate this, this, this. We know calculate energy released when 0.1 mole of HL is added to 0.2 moles of NaOH. So many of you were confused just before the exam. Right? It's so simple. Yeah. Don't get confused. 0.1. I'm doing the first one. 0.1 mole of HCl is reacting with 0.2 moles of NaOH. Right? Now because 0.1 mole of HCl we have. So it will release 0.1 mole of H plus ion. It will neutralize only 0.1 mole of OH ion. Am I right? So because the neutralization is now happening in between 0.1, 0.1. Right? It doesn't make any sense for us if we have more OH ion because it will not be neutralized by H plus ion. We have limited number of H plus ions. So 57 minus 57.1 into 0.1. Okay? Now how this thing comes? The answer should be I think 5.7. Is it? Yes, minus 5.7 something I guess. Okay. So why this formula we have? Like why we are multiplying this 57.1 into 0.1? Because we know that we have a formula. The enthalpy of neutralization of strong acid and strong base is 57.1. So for one mole of a strong acid and a strong base for one mole of H plus ion and OH ion neutralization as you can say. Okay? The energy required for this is minus 57.1. So for 0.1 mole, how much energy required? That X? X kilojoule. Correct? Now we need to find out the value for X. So X is equal to 57.1 into 0.1 divided by 1. This one comes from this one mole. So now we have reached to this equation. Right? Any doubt? Anybody? No question? No doubt? Good? Okay. So the next is what? 0.2 mole of H2SO4 is added to 0.5 moles of KOH. Oh, shit. Wait, wait, wait, wait, wait. Maybe it is not working here. Oh. So the next we have 0.2 mole of 0.2 mole of H2SO4 will be added to 0.5 mole of KOH. Sorry, I forgot to leave one more slide for this. I am doing it here. Okay? In blue color then. 0.2 moles of H2SO4, that means 0.2 moles of H plus ion. Correct? No, not correct. 0.2 moles of H2SO4 will release 0.4 moles of H plus ion. Why? Because one mole H2SO4 will release 2 moles of H plus ion. Correct? So it will neutralize 0.4 moles of OH negative ion. See here we have 0.5 mole of KOH. That means we have 0.5 moles of OH negative. But 0.1 is remaining. Now it is not useful for us. So final titration is happening in between. Final neutralization is happening in between 0.4 moles of H plus ion and 0.4 moles of OH negative ion. So the enthalpy of neutralization should be multiplied by 0.4. Negative 0.457 into 0.4. Please check my answer. Is it correct? Minus 22.84. Is it correct? Good. Good Vaishnavi. Good Shruti. Okay, let's move on to the next question. Enthalpy of neutralization of NaOH by acetic acid is 55.03. Calculate the enthalpy of ionization of acetic acid. Good. This is a different kind of question. Now because we have discussed that enthalpy of neutralization of strong acid. Strong acid and base. And base was minus 57.1. But here we have strong base and weak acid. Okay, that is why this enthalpy is also lesser. Now we will use our formula. What we have learned in this slide. I don't know where is it? Yeah, here. Enthalpy of neutralization is equal to 57.1 plus delta H of ionization. So we have this delta HN value. Delta HN enthalpy of neutralization is minus 55.03 kilojoule and delta H of ionization we have to calculate delta H. I should write here. Ionization we have to calculate. And a standard enthalpy though definitely we know that. So 55 minus 55 minus 55.1 is equal to minus 57.103 sorry plus X. I am just considering this as an X. So X is equal to positive 2.7. Am I right? Adarsh, isn't it positive? Smithy, good. Adarsh, I am asking something to use and it positive. Yes, Raghav, good. Okay. So actually the answer is positive because it's minus 55. This 57 will come to left hand side. So it will be positive. Okay. Next we will see bond dissociation enthalpy. Okay. Bond dissociation enthalpy that means the energy required to break one mole of gaseous bonds. One mole of gaseous bonds. Okay. Please don't forget to write this word in your definition gaseous bond. Okay. To form gaseous atom. And for value for this is endothermic because I have told you dissociation in dissociation we need energy to break. Okay. So always decomposition reactions are maximum at maximum time. Decomposition reactions are endothermic. That is why we say that thermal decomposition if you remember. Right. For example, chlorine gas Cl2 it is converting into chlorine atoms. See how CLCl that is why I have written Cl plus Cl to Cl. Okay. OH is converting into O and H. Right. See one thing here. Making bond is exothermic as opposite is breaking a bond. Okay. So making bond is exothermic as opposite of it is breaking a bond. Right. For diatomic gases the bond enthalpy is 2 into enthalpy of atomization. Hope you all remember that what is atomization. This point is very important. Smaller bond enthalpy is equal to weaker bond and easier to break. Right. Smaller bond. Yes. We don't remember. Atomization. Oh simple thing beta. See here I'll just write one example which is given here itself. Cl2. Cl2 is what? Is it an atom? No. It's molecule. Right. So it will break the molecular bond will break into chlorine plus chlorine. Okay. This was gas. This is also gas. This is also gas. So now this molecule has converted into two different atoms of chlorine only. But two different atoms. So this is known as atomization or enthalpy of atomization. That means in formation of atom how much energy is consumed. Anthelpy of atomization. So enthalpy of atomization is always positive because it's endothermic. Positive. Why? Because it's endothermic. Right. Anthelpy of atomization. Hope you understood now. Okay. So here are some mean values given for bond enthalpies. You don't need to remember these enthalpies because this will be provided in the question itself. Okay. But if you have any confusion, if you want to go through this, this I will send you this ppd. You can go through these values. Okay. The last part of our chapter is Hess's law. Right. Hess's law that means the enthalpy change is independent of the path taken. What does it mean? It's very simple thing when reactant A converts into B product B. So the enthalpy reaction, reaction of the, sorry, the enthalpy of the reaction is delta Hr. Right. This is what, this is the sum of all these enthalpy, all these steps enthalpy. Okay. So whether you go through this path, this three different steps, or you just go directly from A to B, your enthalpy will be same. That is why enthalpy change is independent to the path taken. This is Hess's law. Okay. And what is the representation of this? This diagram A to B. Okay. It is not necessary that you have to take H1, H2 and H3. You can take H1, H2 or H1, H2, H3, H4. Doesn't matter. Okay. The delta Hr should be equal to delta H1 plus delta H2 plus delta H3, or how many steps you have considered because delta Hr is equal to delta H1 plus delta H2 plus delta H3. But, but let's say that I don't have value for delta H2. Instead of this, I have value for delta Hr, H1 and H3. So I can rearrange this equation. Right. I can rearrange this equation to calculate the value for delta H2. Okay. Don't get confused with these things. Okay. Hess's law is what? The change in enthalpy that occurs when reactants are converted to product in same weather the reaction occurs in one step or series of steps. Okay. I know that so many of you might get confused with this definition. So if you, if you don't want to go through this, you can just refer this the simplest one. Okay. Okay. Now the important thing for Hess's law. Remember the rules for manipulating equations. Remember what kind of rules? If suppose you have to multiply your equation by two or you have to divide it by two. So you have to remember that thing. Okay. Whether do you need multiplication or not or you need to reverse it. Right. Okay. Next is what? Add delta H value for each step together after proper manipulation to obtain overall enthalpy for desire reaction. Why this is so to avoid some minor mistakes. Right. We need to add delta H value with each step when we are manipulating the steps. Right. Okay. Next one simple question based on this. Calculate the standard enthalpy of formation of methane. Excuse me. The standard enthalpies of combustion of carbon, hydrogen and methane are given. What we aim, we need to calculate, calculate the standard enthalpy of formation of methane. Okay. And the combustion enthalpies are given. First of all, I am writing the given things. Combustion of carbon C plus O2 gives CO2 and delta H for this is minus 394. Okay. Then H2 plus O2 gives H2O delta H is minus 286. It is not balanced. So okay. We will balance it afterwards. Half H2O2. Then now combustion of methane. So CH4 plus O2 gives carbon dioxide plus water. Okay. What we want? We aim to formation of methane. Right. That means we wanted C2 plus 2H2 forms CH4. This we want. Correct. Now let's see all these three equations and try to convert, try to convert our equations into this form. What I need to do? I have to reverse this equation. Delta H value for this is minus 890, I guess. Yes. Okay. Now first I need to reverse this equation. Okay. I am reversing it. So 2H2O, 2H2O plus CO2 gives 3O2 plus CH4. This is the first thing I did. Now second thing, what I want? Because I have 2H2O. Sorry, I forgot to write it here. 2H2O. So I need to multiply my second equation by 2. The delta H value will become positive now. Now 890. Okay. Now second equation, I am multiplying hydrogen 1 by 2. So 2H2 plus O2 gives 2H2O. Delta H also I have to multiply. 2 into 286 will become minus 572 kilojoule. Okay. The last equation is carbon plus O2 gives CO2. I don't need to change. I don't need to do any change in this equation. Minus 394. So after, oh God, I don't have space, minus 394. So I will add all these equations and let's see what happens. This will be cancelled out. This also, this also 3 oxygen here, 2 oxygen I have here, 1 oxygen I have here. So carbon plus 2H2 forms CH4. See the exactly same equation we have got, right? This one. Now after doing all this, all this calculation, put Dave Shruti minus 76. Great answer. I have got only 3 answers, minus 76 kilojoule. Okay. It's 2O2 in methane. So no, no, 2O2 in methane. Yes, yes, yes. Sorry. It's 2O2. Correct, correct. It's 2O2. So I should cut that. Yes, my mistake. Good. Okay. So I saw your answer. So now we are done with Hess's law as well. Right? Correct in O2 in H2. I have corrected that. Good. So now what do you have to remember after doing all of these things? Explain now what should be, what should you able to do? Explain the difference in between endothermic and exothermic that I know you guys can do, right? Understand the reason for using standard enthalpy change, right? Now recall the definitions of enthalpy of formation and combustion. Please do that properly. Write equations representing enthalpy of combustion and formations. Okay. Now you need to know how to apply Hess's law, right? Recall the definition of bond dissociation enthalpy. Calculate the enthalpy of neutralization of weak acid and weak base. Okay. Then calculate the standard enthalpy change using bond enthalpy values that we also, we have done. Calculate the standard enthalpy change using enthalpies of formation and combustion. Think you can do all of this. Initially I was thinking to do your textbook question. Then I was like, we all, we have done all the questions from the textbook, right? So I did some different, different questions and I have designed a worksheet, similar question to a textbook, some twisted questions. Okay. So I will share that worksheet with you. Please go through that and solve as soon as possible and send me your answers. And guys unmute your mic and tell me if you want me to do textbook questions here. So I will, I can do it from the next class. No ma'am. No ma'am. Okay. Good. That is good then. Then I'll do some different, different questions which will be little more tricky and from the other sources. Okay. Chalo please do worksheet because worksheet may, there are lots of questions. Okay. Bye. Good night. See you.