 Okay, so we had discussed enthalpy of formation now the second term we have you write down the enthalpy of company that is delta C H enthalpy of conversion. Okay, one more thing you can write down in enthalpy of formation enthalpy of formation could be negative as well as positive it can be anything enthalpy of formation could be negative or positive anything. Okay, write down that enthalpy of combustion is always negative here. This is always negative. Okay, statement write down, it is an enthalpy change occurs. It is the enthalpy change occurs when one more again you see one more one more of the substance of substance is burned in excess air excess of air. The combustion process. Okay, so if you look at this example. Carbon fight. I'm taking plus O2 gas. This forms CO2 gas right and one more we have boron solid is the reference state of boron boron solid plus three by four O2 gas. It gives half of B2 O3 solid. Okay, so obviously the enthalpy of combustion is negative always negative. So here in this case, whatever energy releases suppose delta H for this reaction is suppose X. You know, it releases. So X, I am assuming X, June, or let it be a magnitude I'm not taking I'm just saying here will have enthalpy of combustion in this. But since the elements are there in a standard state, right, one more law of substance is forming for this reaction we can write the enthalpy of combustion is equals to the formation of CO2 isn't it. Now suppose in this process, X, June, of energy releases, X, June, of energy releases. So enthalpy of combustion here is, we can say X, June, of energy, or if you want you can put the negative sign minus X, June, energy. If you talk about the enthalpy of formation here, right, since half mole formation we require X, June, so one more formation because enthalpy of formation is defined for one more. Half moles give X, so moles gives you two X, June. So for this equation, what we can write the enthalpy of combustion of Oran is equals to half of the enthalpy of formation of B2 O3. So it's not like always enthalpy of combustion and enthalpy of no formation will be equal, it depends upon what equation we have, what relation we have. So I will just keep that in mind combustion is always an atomic process. Next third term write down, enthalpy of hydrogenation, it is delta of hydrogenation of H. See hydrogenation is addition of hydrogen, the term also if you can understand hydrogen, hydrogenation is the addition of hydrogen. Definition write down, it is the enthalpy change, enthalpy change when one more, when one more of an unsaturated, one more of an unsaturated compound, saturated compound is converted into, converted into one more of a saturated compound by addition of hydrogen, done? Okay. What is hydrogenation you see? Suppose you have, first of all, what is unsaturated compound? Could you tell me? What do you understand by unsaturated compound? And what do you understand by saturated compound? Unsaturated compounds are those compounds, which is, which has double or triple bond, right? Which is double or triple bond. So in this we have alkene or alkyne, the example, double or triple bond. Saturated compounds are alkyne, right? Unsaturated compounds are those compounds which shows addition reaction. In saturated compound, addition reaction is not possible, substitution is possible. Okay. So what happens here you see, you have this, suppose this molecule you have an alkene, C double bond C, right? When it reacts with hydrogen, H, H bond we have, then this sigma bond, this pi bond breaks and it converts into this compound, which is an alkene here. And some energy comes out in this process. This energy, if one mole of substance is forming from one mole of substance, then this energy is called the enthalpy of hydrogenation. Number of moles must be one for both. Correct, enthalpy of hydrogenation. Energy releases in this process, right now. Why energy releases that we'll discuss in organic chemistry when we discuss the mechanism of addition of this particular reaction, right? So heat or energy releases in this process always. Energy releases in this process. Exothermic, correct? Look at this example. Suppose we have CH2, double bond CH2, when it allowed to react with hydrogen, then there will be some catalyst also we use for purpose. And the product would be CH3, CH3. Ethene converts into Hane. If you have benzene ring, three moles of H2 we are taking for one double bond, one H2 to three double bond, three H2. It converts into cyclohexane. Energy evolves this enthalpy of hydrogenation. Now you see one question here we discuss. Questions carried the enthalpy change when acetylene is hydrogenated to ethane. Hydrogenated to Hane. Enthalpy change in this process. See, and there are some data given. Data is what? We have enthalpy of combustion of carbon. Enthalpy of carbon is given in 90 kilojoule per mole. Okay. We have enthalpy of combustion of hydrogen. This is 290 kilojoule mole. Enthalpy of combustion of C2 at 6. It is minus 1560 kilojoule per mole. Enthalpy of combustion of sorry. Enthalpy of formation of C2 H2. It is 230 kilojoule per mole. Copy this data first. Okay. See how do we approach this kind of question. Question is, calculate the enthalpy change when acetylene is hydrogenated to ethane. Acetylene, the reaction is this. H C triple bond. Acetylene is hydrogenated to ethane. Two molecules of H2O2. So it is CH3. CH3. This is the reaction we want. Right. Okay. Now, this data, what would be the reaction for this? It is a combustion reaction. So carbon plus O2, solid we are taking O2 gas gets CO2 gas. We have the energy involved. It is given. Conversion of hydrogen. So hydrogen plus gas it is plus half of O2 gas. It gives H2O. Enthalpy is given minus 290. What is this reaction? Enthalpy of combustion of C2. So C2H2. Right. Solid plus O2. Gives CO2 gas plus H2O liquid. What is this number of moles of O2? Because we need to balance this reaction. How do we balance this? This balance over here. We have two CO2 here. Four plus one five. We'll have five. Now, what is the delta H formation of C2H2? 2H6. This reaction. This should be what? This should be product minus reactant. Right. So what do we write? Okay. Okay. One thing you see. Okay. We can do one more thing. Fine. This we need to form. Correct. Because when I write on this, you see it's here. When I write down this of C2H2. Right. This equals to what? We form this constituent element. We need to form this carbon and how do we have here? And we need to form C2. Right. We'll do from these three equations. We'll try to form this equation right down here. To form C. Solid plus. Gives C2H2 to get. Form this equation from this three. Can we do that? Obviously it is balanced. 2C will have over here. How can we do that? Tell me. How do we get this equation from this three? We need to add one plus two. We need to multiply this one with two. Multiply this by two. Add the second equation and subtract the third one. Isn't it? So can I write this equation one into two? Plus equation two. Minus equation three. Can we do that? Tell me. I don't have space much. Is it fine? Tell me guys. Where is H2? Which one Anusha? Tell me. Which equation you are talking about? Yeah. Do you understand this really? Are you? Please respond. Did you understand this relation? So first equation is enthalpy of combustion of carbon. So what I'll write here? You see, I'm just scrolling this a bit up. Okay. The first equation is enthalpy of combustion of carbon. So we'll write here. Yes. Yes. Right. In order to get the desired equation. So in. We light the first one two into. Then help. Plus enthalpy of combustion. Hydrogen. And minus. What will write. In therapy. Combustion of. Right. Right. In therapy of combustion of. Acetylene C2H2. But one thing here we have. Since we have. Invert the equation. Okay. We have. Now what would be this? This value is. Two in. Three. Nine. Plus. Mine. Two. Nine. Minus enthalpy of combustion of C2H2 is. It is not given in the question. In therapy of combustion of C2H2. Is not given in questions. I'll write down this as it is. Delta C H. C2H2. But this form. We have this side here to 30. So could you find out enthalpy of combustion of C2H2 here. Enthalpy of combustion of. C2H2. Equals two. When you solve this will get minus one three. Double zero. We need enthalpy of combustion of C2H2. See what I have. And I see this. When you write this properly. And I don't have much space. That's why maybe you're getting views. See what I have done. We need to find out the. Enthalpy of this equation. So if you know the enthalpy of combustion of. C2H2. In the commission of H2. Enthalpy of combustion of C2S6. We can do this. That would be. This minus this. So. Everything is given. Enthalpy of combustion of given in the question. Is this. But C2H2 not mentioned in the question. Enthalpy of combustion of C2H2. Not mentioned in the question. Right. So we'll try to find out the enthalpy of combustion. Of C2H2. That finding it out from this equation. Because formation is given. How this C2H2 forms. The C2H2 forms by the condition of carbon and hydrogen. So with the help of this. We'll find out the enthalpy of combustion of C2H2. Which is this. Which is this. Understood this. Now for the desired equation. Which is this here you see. This is the desired equation. I'll just change the color so that. Get confused. Kilojoule. And the desired equation is this one. In this the enthalpy. Of this reaction would be. Enthalpy of reaction would be. Enthalpy of combustion of. C2S6. The product minus. We have. Enthalpy of combustion of. Enthalpy of combustion of. Of combustion of C2H2. Plus. Enthalpy of. Combustion of hydrogen. Into two. This is what we need to write. Because we have two H2 here. No. Two we need to multiply. We just need to substitute the value into this. In order to get the answer. You see here. Enthalpy of combustion of C2S6. Given in the question. Minus one five six zero. Minus. Enthalpy of. We have calculated. Minus one three zero zero. Minus. Two times. Of two ninety. For hydrogen. When you solve this you'll get. Approximately. Minus three twenty. Kilojoule. This is the answer. We have. Enthalpy of this reaction. Just you need to. You know. Add and subtract. In order to get the desired answer and desired. You know. Equation just you need to. Add and subtract. But while adding or subtracting. You must take care of that. You need to multiply with the number of moles. If it is given. Which equation or you are talking about. The equation is not given. Just a second. You see. With the question. Question is what you see. Calculate the enthalpy change when. Acetylene is hydrogenated to it. Acetylene is this. Hydrogenated to it. So for this reaction. You need to find out the enthalpy change. You see for this reaction. You need to find out the enthalpy change. That's the question. How did I write down these equations? I think this you are talking about. This one right this equation. Yes. Yeah. From this data you see. The enthalpy of combustion of carbon means what? Carbon plus O2. You see O2. Hydrogen H2 plus O2. If this data is given. Then obviously corresponding to these data's the equation would be this. Yes. Similarly for. Combustion C2 S6. The equation is this. Now with the help of these data. And delta F of C2 H2. We'll first find out. The enthalpy of combustion of C2 H2. Which is this I have done over here. Which is this. In terms of C2 H2 forms by this equation. In this we have enthalpy of combustion here. This enthalpy of combustion you will get this. Now this will use to find out the final answer. In the help of combustion of C2 H6. Minus enthalpy of combustion of C2 H2. Which is this we have calculated. Minus enthalpy of combustion of H2 into two. That is what the equation we have written over here. And this. Basically you need to just write down product minus reactant. That is it. Nothing. Correct. Understood. I would request all of you to solve this question also on your own once after the class. Okay. Then probably you'll understand what we have done. Okay. Yes, guys. Did you understand this? Yes. Once you solve on your own. The solution you have already I will share the notes also. You will understand that what I have done. Okay. Because the space is not that much. If you have changed the board, then again, I have to again come back and look at the values. Okay. That's why I'm not going to the next slide, but I will share the notes. You can, you know, whenever you stuck, you can go through this solution. You will understand. Just one thing you need to keep in mind. That product minus reactant you need to do. With the given data, we'll write down the reaction and we'll add or subtract in such a way. So that we'll get the desired equation. Yeah. Right now, next enthalpy of neutralization. If you go a bit. We can finish chapter today itself. Right on enthalpy of a new addition. Neutralization process. What do you understand? Acid base reaction, right? When one mole of an asset gets neutralized by one mole of a base or one equivalent. Also you can say the amount of energy absorbs or releases. That is the enthalpy of neutralization. Very basic definition. We have all the definition. If it is neutralization, we are taking one more. If it is neutralization, then we have to consider asset, asset base. If it is combustion, then reaction with O2. If it is what, if it is formation, then reaction with the constituent elements. One more, right? So I know the definition. Then help you off. It is defined as. It is defined as the energy absorbs or releases. Energy absorbs or releases. When one equivalent of H plus. In dilute solution. When one equivalent of H plus in dilute solution. Combines with. One equivalent of OH minus. One equivalent of H plus in dilute solution. Combines with. It is the enthalpy of neutralized. It is defined as. It is defined as the amount of energy. Evolve or absorbs. When one equivalent of H plus. When one equivalent of H plus. In dilute solution. Combines with. One equivalent of OH minus. To form water molecule. Right. So whenever you have strong acid or strong base. Generally we take monoprotic acid. Strong acid. When one equivalent of H plus. When one equivalent of H plus. In dilute solution. Combines with. One equivalent of OH minus. And the strong base if you. Reacts. It forms salt and water. Salt and water. Right. Generally we have one equivalent of it will take. One equivalent of it will take. Monoprotic acid. Like we can have at CL. We can have HNO3. Only one H plus donor. And monoprotic base also. Like we have any OH. KOH etc. Then obviously the energy release. Like mostly the energy releases in this. Like mostly the energy releases. Delta N. Of H. In therapy of neutralization. This value is find out to be. Whenever you have a strong base. A strong acid reaction. It is coming out to be minus 57.1. Kilojoule. This value you must remember. Okay. Strong acid is strong base. Minus 57.1 Kilojoule. Or in kilo calorie. It is minus 13.7 Kilo calorie. Right. So this you must remember. In healthy of neutralization for the. Monoprotic acid or base. Strong acid or base. It is always minus 57.1 Kilojoule. Okay. Understood. Similarly, there are a few more. You know, enthalpy term we have. I'm not going to dictate you this. I'm just telling you listen. This is not that important. Like suppose we have enthalpy of solution. Inthalpy of solution is what? When a solute is dissolved in solvent. The amount of energy releases or absorbs. Is called enthalpy of solution. Are you getting it? During the formation of solution. Right. One more of a substance that is solute. Dissolve in excess of solvent. The amount of energy releases or absorbs is enthalpy of. Solution. Enthalpy of hydration is what? Hydration means one more of a substance. You dissolve in water. The amount of energy releases or absorbs is enthalpy of hydration. Are you getting this point? How do we have these definitions? Yes, all of you. Please respond. Tell me. The definition you are getting. Do I need to dictate all these definitions? How do we. Get a solution. Solution we get when a solute is dissolved in solvent. Correct. So how do we say the definition? One more of solute. Dissolve in excess amount of solvent. The amount of energy. Absorbs or absorbs. How do we get solute? How do we get solute? How do we get solute? How do we get solute? How do we get solute? The amount of energy absorbs or releases is the enthalpy of solution. Clear? So this kind of definitions are there you can go through and said. I have done the important one I have done. Okay, you won't have. Any other requirement won't be there, but if you want you can go through once, okay. Because I want to take up some topics, which is there to understand, right. We have 15, 20 minutes we have. So I think we can finish this today. right now next bond enthalpy bond enthalpy okay this is generally applicable for gases applicable for gases okay write down write down it is defined as I'll write down wait it is the energy required to break one mole of a particular bond one mole of a particular bond in gaseous compound in gaseous compound to form product in in gaseous state that's why I said it is applicable for gases only see try to understand it is important and I don't know but most most value I have seen most residents are confused with this okay you see it is also defined as the average energy which is required to break one particular bond see for example we have methane right we have methane over here the bond enthalpy would be what bond enthalpy would be the average of the energy required to break this this this and this bond all the four bonds are you getting my point right so it says what the average enthalpy require to dissociate the said bond present in different gaseous compound is called bond dissociation enthalpy I will take one example and then I'll come back to this CH4 again okay with this example you will understand the difference in bond energy and bond enthalpy there's difference in bond energy and bond enthalpy suppose we have water okay we have water H2O gas gases state if you break the bond of oxygen and hydrogen you will get hydrogen gas right an OH gas just to break the bond we have this suppose enthalpy required to break this bond is it is given 501.87 kilojoule per mole to break this one oxygen and hydrogen bond further if you want to break this oxygen hydrogen bond here this bond if you want to break then what happens you see you'll get oxygen gaseous atom hydrogen gaseous atom right but if you think that this enthalpy should be same like we had earlier no it is not this enthalpy is different this enthalpy is 423.38 kilojoule per mole what it so if you if I ask you to calculate the bond enthalpy of H2O molecule then the bond enthalpy of H2O molecule would be bond enthalpy of H2O molecule would be it is one and two so it is the average of this two means it is delta H1 plus delta H2 divided by two which is 501.87 plus 423.38 divided by two if you take the average here you get 462.62 kilojoule per mole so bond enthalpy don't get it get confused over here bond enthalpy and bond energy are two different things bond enthalpy is the average energy which is required to break the bond all will be in gaseous state yes understood so in CH4 also you can think that this is suppose we have delta H1 this is delta H2 this is delta H3 and this is delta H 4 right so what is bond enthalpy here bond enthalpy for CH4 would be the average of all these right so delta H1 plus delta H2 plus delta H3 plus delta H4 divided by four this is the bond enthalpy tell me understood now yeah look at this question on this the question is one mole of of phosphorus gas atom is converted into into p4 gas molecule it's converted into p4 gas molecule we need to find out we need to find out the bond enthalpy in this delta H so you see what happens here p4 sorry phosphorus gas converts into p4 gas balance this equation you'll get 1 by 4 over here so delta H for this conversion would be what the bond energy of reactant minus the bond energy of product I forgot to write down the data given over here the bond energy is given phosphorus phosphorus bond energy is given here and it will be given in that question it is 500 kilo joule per mole right so we don't have any you know phosphorus phosphorus bond here right so it is 0 minus here it would be 1 by 4 into into the bond energy of this what is the bond energy p4 molecule you see it is a p4 molecule this you should know the structure this is the p4 molecule 1 2 3 4 5 6 bond we have so 6 into 500 which is minus 750 kilo joule right so basically if you see this here if I write down the enthalpy of this would be what there are six bond right so 6 into bond energy of phosphorus phosphorus and this should be the average energy that should be the average energy right since we have 1 by 4 over here so 1 by 4 millimeter right and here we always write reactant minus product not product minus reactant you must keep this in mind bond enthalpy if you need to find out so reactant minus product right if I write down here like in the case of methane here also we have this energy you see the bond enthalpy this delta H bond enthalpy sorry the bond enthalpy of this would be four times of the bond energy but this bond energy is the average bond energy of CH bond average CH bond energy right because average is nothing but this so this average bond energy you can substitute at this term four and this four will get cancelled this will become equal so whenever we have this kind of molecule like p4 we have we have s8 we have we have CH4 then we have this expression that number of bond into the average energy that we have this energy which is given this average only okay one more thing which has been asked in j this you have to memorize okay it's a three-dimensional things probably you won't be able to visualize or understand this what type of molecule as in any molecule it is applicable right but we always take here the average bond energy any molecule it is applicable if not then each bond energy will be given if average is not applicable each bond energy will be will be given because without that you cannot find out this like here average is not written but we are considering this as average okay one last thing here the question is this was asked in j as well I would say this you memorize okay the term is is there you should memorize this the question is n atoms are present every corner of truncated octahedron truncated octahedron find delta H