Thermochemistry | One Shot | NEET 2024 | Wassim Bhat

So whatever topics I shall be discussing today expect two questions directly on or indirectly from the today's session for sure guys for sure. So people the first topic which we shall be covering in the today's session what is that going to be that is going to be enthalpy of the reaction. Enthalpy of the reaction. What is meant by the term enthalpy of the reaction? Let's try to understand My dear students. Enthalpy of the reaction. It is defined as It is defined as the amount of heat. The amount of heat absorbed or released. The amount of

heat absorbed or released during a chemical reaction. During a chemical reaction. The amount of heat absorbed or released during a chemical reaction comma carried out carried out at constant pressure carried out at constant pressure. Carried out at constant pressure. What it means exactly? Let's have a look. For example, for example, I'm taking a general reaction over here in which reactants are getting converted into products. This is a general reaction in which reactants are getting converted into products. Okay, this one general reaction. My dear students, I'm assuming that this particular reaction it is carried out at

constant pressure. I'm assuming that this particular reaction in which reactants are getting converted into products. I'm assuming that this particular reaction is carried out at constant pressure. Now my dear students, now my dear students when this particular reaction will be carried out at constant pressure definitely there will be some amount of heat absorbed or released during the reaction. Absolutely My dear students that amount of heat that amount of heat which is either absorbed or released during this particular reaction that amount of heat which is either absorbed or released during this particular reaction which I'm representing

with QP amount of heat absorbed or released during this particular reaction which is carried out at constant pressure. the amount of heat absorbed or released during this particular reaction which is Carried out at constant pressure that is something which I call as enthalpy of the reaction. So first of all first of all what is enthalpy of the reaction? How do you define the term enthalpy of the reaction? Let me quickly tell you it is the amount of heat which is absorbed or released during a chemical reaction which is carried out at constant pressure. Right? I

believe this is clear to every view. Yes, I'll be covering all the topics. All the topics in three to four hours. All the topics. Yes. Okay. Even JS parents can watch it. Yes. I hope I'm clear. Please and please do not spam now. Okay. So my dear students from now onwards how exactly you are going to define the enthalpy of reaction? Either you call it enthalpy of the reaction or you can call it as enthalpy change in the reaction. The choice is all yours. Either you call it as enthalpy of the reaction or you can

call It as enthalpy change in the reaction. What is meant by it? The amount of heat which is absorbed or released during a chemical reaction which is carried out at constant pressure. Point number one. Point number two. My dear students, imagine the same reaction. Imagine the same reaction was being carried out at constant volume. Imagine the same reaction was being carried out at constant volume. At constant volume, whatever amount of heat will be absorbed or released during the reaction. At constant volume whatever amount of heat will be absorbed or released during the reaction that is

something which you call as internal energy of the reaction or you can call it as internal energy change in the reaction. So there are two terms which I have defined. One is enthalpy of the reaction. One is internal energy of the reaction or you can call it as enthalpy Change in the reaction internal energy change in the reaction. Let me define both the things again. Enthalpy of the reaction or the enthalpy change in the reaction. It is basically the amount of heat which is absorbed or released during a reaction which is carried out at constant

pressure. Internal energy change in the reaction is the amount of heat which is absorbed or released during a chemical reaction which is carried out at constant volume. Right? Which is carried out at constant volume. Yeah. Now my dear students in thermodynamics I believe all of you would have come across a relation. And what is the relation? You should be knowing about this relation. Delta H is equal delta H is equal delta U plus delta NGRT. I believe you all would have come across this particular reaction. My dear students since right now I'm Talking about the

reactions. So I'll write enthalpy change of the reaction internal energy change of the reaction. This is one more result which I shall be using. This is one more result which will be relating enthalpy and internal energy of a particular reaction. This is the result which shall be relating this is the result which shall be relating enthalpy change and internal energy change of the reaction. Right? There will be some reactions whose deltaU will Be given to us and delta H we will be supposed to calculate. There will be some reactions whose delta H will be given

to us. Delta U we will be supposed to calculate. For all those questions I shall be using this particular result. Right? A particular set of questions asked. Particular set of questions is asked in which delta U will be given. Delta H you are supposed to calculate. Delta H will be given. Delta U are supposed to calculate. So for those sort Of questions I'll be using this particular result over here. Okay. Now people few more things few more things. Few more things my dear students based on the sign of delta H. Based on the sign of

delta H. We classify reactions into two types. Based on the sign of delta H. We classify reactions into two types. Remember it. Those reactions whose delta H is positive. Those reactions whose delta H is positive. What is delta H? Enthalpy change. You can call it like this. Enthalpy of products minus enthalpy of reactants as well. Right? Delta H enthalpy change. You can say enthalpy of products minus enthalpy of reactants. Now my dear students that particular reaction whose delta H is positive whose delta H is positive. Delta H positive means HP minus HR this term is

positive. When can be this particular term positive? This particular term can be only positive if Enthalpy of products is greater than that of enthalpy of reactants. And let me tell you those sort of reactions which have got delta H positive. What do you call those reactions as? You call them as endothermic reaction. You call them as the endothermic reactions. Point number one. Point number two, those particular reactions which will be having delta H negative. Delta H negative that means enthalpy of products minus enthalpy of reactants. This term Will come out to be negative. When can

be this particular term negative? When can be HP minus HR negative? This can be only negative. If enthalpy of products will be less than that of enthalpy of reactant. And those reactions for which delta H comes out be negative. You call those reactions over here as the exofothermic reactions. So my dear students on the basis of the sign of delta H you are going to classify the reactions into two types. One is going To be endothermic one is going to be exothermic. For endothermic reactions delta H is positive. For endo for exothermic reactions delta H

is negative. Let me know once in the chats if all these things are clear. Let me know once in the chat if all these things are absolutely clear to you. Quickly in the chats with some thumbs ups. Everyone, quickly my dear students, quickly my dear students. Everyone in the chats. Everyone in the chats quickly. Everyone in the chats. Some people are saying video is not clear. You need to change your settings. I think you're watching the video at uh 140p, right? Just go to the settings and increase the quality of the video. Yeah, perfect people.

So one thing which I want every one of you to know from now onwards if I ask you any time from now On what is enthalpy of reaction? What you should say? Enthalpy of reaction is nothing but amount of heat absorbed or released during a chemical reaction carried out at constant pressure. Sometimes if I ask you what is internal energy change of the reaction, you will say it is nothing but amount of heat absorbed or released during a reaction which is carried out at constant volume. If I ask you what is the result between enthalpy

of the reaction and internal Energy of the reaction, you will say delta H is equal to delta U plus delta NGRT. Right? Okay. I believe this is clear. On the basis of sign of delta H, we classify reactions into two types. Endothermic exothermic endothermic reactions are the ones whose delta H is positive. Exothermic reactions are the ones whose delta H is negative. As simple as that. Okay. Now people let's move ahead. Let's Move ahead. My dear students there is something which is what we call as standard enthalpy of the reaction. Okay, right now I defined

Right now I defined enthalpy of the reaction. Now I'm going to define one more term that is standard enthalpy of the reaction. Standard enthalpy of the reaction. What is meant by the standard enthalpy of the reaction? Let me make it clear to you. My dear students, standard enthalpy of the reaction is nothing but it is the enthalpy change. It is the enthalpy change. It is the enthalpy change. What is meant by enthalpy change? Enthalpy change means the amount of heat absorbed or released. Enthalpy change means the amount of heat absorbed or released. Standard enthalpy of

the reaction is the enthalpy change which is measured Which is measured under standard conditions. Which is measured under standard conditions. Now the point is what are standard condition? Standard conditions are pretty much simple here guys. when pressure is kept as 18m and temperature is kept constant which is generally taken as 25° centigrade. Now what it means exactly? What it means exactly? My dear students imagine you have got the reaction in which your reactants are Getting converted into products. I believe this particular reaction is carried out at 180m pressure. I believe this particular reaction is carried

out at 180 pressure. I believe this particular reaction is carried out at constant temperature as well. I believe this particular reaction is carried out at constant temperature as well. Assume that the temperature at which reaction is taking place is 25° centigrade. So can I say this particular reaction is Carried out under standard conditions? Absolutely. This particular reaction is carried out under standard conditions. At this point of time, whatever amount of heat will be absorbed or released during this particular reaction, that is what you call as standard enthalpy of the reaction. That is what you call

as standard enthalpy of the reaction. That is what you call as standard enthalpy of the reaction. Am I clear? Am I clear people? Am I clearly? Am I clear with this? Am I clear with this quickly in the chats? Yes, we are going to solve questions as well. Am I clear with this? So I took a reaction which is carried out under standard conditions. These are the standard conditions. Under standard conditions, whatever amount of heat will be absorbed or released during the reaction, that is something which you Call as standard enthalpy of the reaction. Now

people, if I ask you, if I ask you, is there any difference? Is there any difference between these two terms? What should be your answer? Is there any difference between these two terms? Yes, there is a difference. This is something which you call as this is something which you call as enthalpy change. What is meant by enthalpy change? The amount of heat absorbed or released. This is something which is what you call a standard enthalpy change. This is something which you call a standard standard enthalpy change or standard enthalpy of the reaction. Okay. Perfect. Whenever

enthalpy of the reaction, whenever enthalpy change of the reaction is measured under standard conditions, you call that particular enthalpy change which is measured under standard conditions as the standard Enthalpy of the reaction. I believe this is clear to everyone. Now people, if this is clear to everyone, then let me tell you something very very very important. Let me tell you something very very very important. You need to remember that and understand that one by one. Try to understand people. If I talk about standard enthalpy of the reaction which is the amount of heat absorbed or

released during a chemical Reaction which is carried out under standard conditions. Sometimes you call the standard enthalpy of the reaction as sometimes you call it as standard enthalpy of formation which I'll let you know in some time sometimes you call it as the standard enthalpy of combustion which I'll let you know in some time sometimes you call it as standard enthalpy of neutralization right okay you would have heard about these terms I Believe sometimes my dear students you call it as enthal enthalpy uh standard enthalpy of whatever sublimation, right? Sometimes you call it as standard

enthalpy of fusion, right? Sometimes you call it as the latis enthalpy. Sometimes you call it as the latis enthalpy. Sometimes you call it as the enthalpy of atomization. Right? Sometimes you call it as For example enthalpy of hydrogenation. Perfect. All these different types of enthalpies of the reaction we have to discuss one by one. Right? So first of all enthalpy of the reaction under standard conditions. Simple amount of heat absorbed or released. Amount of heat absorbed or released during a chemical reaction which is carried out under standard conditions. Okay. Now this enthalpy of reaction sometimes

you Can call it as enthalpy of formation, enthalpy of combustion, enthalpy of sublimation, enthalpy of fusion, enthalpy of vaporization, enthalpy of atomization, right? Enthalpy of hydrogenation etc etc. Now all these things one by one we need to discuss and from these questions are basically asked in your ne examination. Right? Correct. Right people so all these things All these things aid this is not standard condition which we do in mole concept it's not STP this is the standard conditions where temperature is kept constant generally 25°C okay remember it moving ahead pressure you can keep one atm

or one bar which is approximately the same okay now one by one we shall be discussing about all these enthalpies of the reaction try to understand people what exactly I'm going to talk about okay try to Understand what exactly I'm going to talk about the first the first enthalpy of the reaction which is what you call as which is what you call as enthalpy of formation enthalpy of formation enthalpy of formation I'm representing it as delta H not F and I'm defining it under standard condition. I'm defining it under standard condition. Enthalpy of formation. How

do we define the term enthalpy of formation? Try to understand. See guys, I'm writing it definition. The enthalpy change. It is defined as the enthalpy change. Enthalpy change means the amount of heat absorbed or released. Enthalpy change means the amount of heat absorbed or released. The enthalpy change when one mole of substance When one mole of substance is formed Is formed. When one mole of substance is formed from its constituent elements from its constituent elements. from its constituent elements which must be present in their standard states which must be present in their standard states which

must be present in their standard states. So before making you understand the meaning of enthalpy of formation let's try to understand exactly what Standard state of an element is all about. What is meant by standard state of an element? Let me tell you people standard state of an element. Standard state of an element. It is that state of an element in which element exists in nature. In which element exists in nature. Right? Standard state. It is that state of the element in which element exists in nature. Right? Or we can say standard state of an

element is the most stable is the most stable state Of the element as per thermodynamics. As per thermodynamics. Now try to understand. For example, if I talk about hydrogen, hydrogen in nature exists as H2 gas. If I talk about nitrogen, nitrogen in nature exists in the form of N2 gas. Chlorine in the nature exists in the form of Cl2 gas. Bromine in nature exists in the form of Br2 liquid. Iron in nature exists in the form of Fe solid. Right? So these are the these are what I call as standard states of Hydrogen, nitrogen, chlorine,

bromine, iron etc. If I talk about carbon, if I talk about for example carbon, right? What are the two main elotrops of carbon? Two main elot drops of carbon are your diamond and graphite. Two main elot drops of your carbon are your diamond and graphite. Now if I ask you among diamond and graphite among diamond and graphite which one is thermodynamically more stable which one is thermodynamically more stable? Graphite is thermodynamically more stable than diamond. So, so I would say I would say the standard the standard state of carbon is your graphite. Graphite is the

standard state of carbon. Why? Because graphite is thermodynamically more stable. And that state of the element that elotrop of the element that elotrop of the element which is thermodynamically more stable that is considered to be the standard state of the element. Right? That is Something which you call a standard state of the element. Right? So first thing standard state it is that state in which element exists in nature. Number one. Number two whenever you see an element which has got elot drops that elotrop which is thermodynamically more stable that will be considered as the standard

state of the element like like you have like you have carbon. It is two main elotrops are your diamond and graphite. But graphite is more stable. So graphite is something which I will be calling as graphite is something which I will be calling as the standard state of carbon. I believe this is clear to everyone. I believe this is clear to everyone. I believe this is clear to everyone. Now people let's go on. Let's talk about our actual thing that is standard enthalpy of formation. Look at it carefully. It is defined as the enthalpy change.

Enthalpy change means amount of heat Absorbed or released. Amount of heat absorbed or released. So the amount of heat absorbed or released when one mole of the substance is formed from its constituents which must be present in their standard states. Let's get to know what is meant by this particular definition. Try to understand people. Try to understand. Just try to understand. For example, for example, For example, I'm writing calcium carbonate here. I'm writing calcium carbonate here. If I ask you what are the constituent elements of calcium carbonate? It's calcium. It's carbon. It's oxygen. It's calcium.

It's carbon. It is oxygen. What is the standard state of calcium? It is calcium solid. What is the standard state of oxygen? It is O2 gas. Right? It is O2 gas. And if I ask you what is the standard state of carbon, it Is carbon graphite. It is carbon graphite. Perfect. Look at this particular reaction. Is this reaction balanced or not? Is this reaction balanced or not? There are a lot of people spammers we have. Just a second. Gone. Done guys. Concentrate in the class. Okay, concentrate in the class. Okay, I've blocked. Keep on keep

on telling me when The spammers come. Okay, keep on telling me. Perfect guys. If you look at this particular reaction, is this reaction balanced or unbalanced? This particular reaction right now is unbalanced. So, let me balance this reaction. Let me make it as 3x2. Now, is the reaction balanced? Now is the reaction balanced? Absolutely the reaction is balanced. My dear students, in this reaction, if you observe carefully, can I say one mole of Calcium carbonate is being formed? Absolutely one mole of calcium carbonate is being formed from its elements which are present in their standard

states. One mole of calcium carbonate is being formed from its elements which are present in their standard states. The amount of heat absorbed or released during this particular reaction I will be calling that as the enthalpy of formation of calcium carbonate. I told you enthalpy of reaction. Sometimes you Call it as enthalpy of formation. Sometimes you call it as enthalpy of combustion. Right? Okay. If you look here, if you look here carefully, one mole of calcium carbonate is getting formed from its elements which are present in their standard states. So whatever amount of heat will

be absorbed or released during the reaction that is something which you call as enthalpy of formation of calcium carbon. As simple as that. Let me take one more example. Let me take one more example. My dear students for example I'm writing NH3 gas. NH3 gas right? I'm writing NH3 gas. If I ask you what are the constituent elements of NH3? It's nitrogen and hydrogen. Now tell me what is the standard state of nitrogen? It is N2 gas. What is the standard state of hydrogen? It is H2 gas. Now is the reaction balanced? No, the reaction

is not balanced yet. So let me balance it. Let's keep it as 1x2. Let's keep it as 3x2. Now the reaction is balanced. Now the reaction is balanced. My dear students, if you look at this particular reaction carefully, can I say one mole of NH3 is being formed? One mole of NH3 is being formed from its elements which are present in their standard states. Right? One mole of NH3 is being formed from its elements which are present in their standard states. So whatever amount of heat will be absorbed or released during this particular reaction That

is something which you will be calling as standard enthalpy of formation of NH3. As simple as that. As simple as that. Right? Okay. Let me take one more example. Let me take one more example. After that, I'll give you a question. Let me take one more example. After that, I'll be giving you a question. For example, I'm writing here H2O liquid, right? H2O liquid. H2O liquid. Now, hydrogen oxygen. What is the standard State of hydrogen? H2. Perfect. What is the standard state of oxygen? O2. O2. Is the reaction balanced? The reaction is not yet balanced.

Now the reaction is balanced. Can I say one mole of water, one mole of water is being formed from its elements which are present in their standard states. Absolutely. So the amount of heat absorbed or released during this particular reaction is something which you call as the enthalpy of formation of Water. Perfect. And my dear students, my dear students, let me tell you the first reaction which you have over here. The first reaction is what you call as this reaction is what you call as the formation reaction of calcium carbonate. This reaction is what you

call as the formation reaction of NH3. This reaction is what you call as the formation reaction of water. Right? Formation reaction of water. Formation reaction of NH3. Formation reaction of Calcium carbonate. So whenever you have to write formation reaction of any substance one thing that should strike your mind. What is that? One mole of substance should get formed from its elements which should be present in their standard states. Which should be present in their standard states. Okay. Perfect. Perfect people. Perfect people. I believe this is clear. Now I have got a question for you. I

Have got a question for you. I'm writing a reaction. N2 gas plus P * H2 gas. It gives 2 * NH3 gas. This is the reaction which I have written over here. This is the reaction which I have written over here. I am telling you during this particular reaction, this is the amount of heat absorbed or released which is for example X KOJ. X kilogjles X kilogjles is the amount of heat which is absorbed because I'm writing the delta H is positive. Delta H Is positive means endothermic endothermic means heat absorbed. I believe during this

particular reaction X kJ of heat is being absorbed. X kilogj of heat is being absorbed. Now you tell me can I call this X as can I call this X kJ as the enthalpy of formation of NH3? What do you think? Can I call it as the enthalpy of formation of NH3? This is a general question from my side. A general equation from my side. I'm I'm writing a reaction over here. I wrote a Reaction. Perfect. I'm telling you that x kilogjles of heat is absorbed during the reaction. Now I'm asking you whether I should

call this x kilogjles as the enthalpy of formation of NH3. Is it? No. I'm not going to call it as the enthalpy of formation of NH3. Why? Why? Why? Remember the definition. One mole of NH3 should get formed over here. Is one mole of NH3 being formed? No. Two moles of NH3 are being formed. Two moles of NH3 are being formed. Can I say Can I say when two moles of NH3 are being formed, how much heat is absorbed? X kilojoules. Therefore, when one mole of NH3 will be getting formed, this will be the heat

absorbed. This will be the heat absorbed, right? This will be the heat absorbed. When one mole of NH3 will get formed and the amount of heat which will be absorbed or released when one mole of substance is getting formed, that is something which you call as Enthalpy of formation. So, can I say over here? Can I say my dear students can I say no doubt this was the total amount of heat absorbed during the reaction. But if in a question they ask you calculate the enthalpy of formation of NH3. So you will say enthalpy of

formation of NH3 is nothing but X by 2 kJ per mole. Right? Is it clear? Is it clear people? Is it clear to everyone? Quickly in the Charts. Quickly in the charts. Quickly in the charts. Right now over here one mole of calcium carbonate was getting formed. Right. And when one mole of calcium carbonate was getting formed whatever was the amount of heat absorbed or released that is something which I'm calling as enthalpy of formation of calcium carbonate. When one mole of NH3 was getting formed, whatever amount of heat was absorbed or Released during the

process, that is something which I was calling as enthalpy of formation of NH3. One mole of water was getting formed. Whatever heat was absorbed or released during this process, that is something which I'm calling as enthalpy of formation of water. But over here, two moles of NH3 was getting formed. At this point, was enthalpy of reaction equal to enthalpy of formation? No. So can I generalize a statement? Can I generalize A statement? Can I say enthalpy of the reaction and enthalpy of formation can only be equal can only be equal when one mole of the

substance when one mole of the substance would be getting formed? when one mole of the substance would be getting formed. When one mole of the substance would be getting formed from its elements which would be present in the standard states. Is this statement clear? Is this particular statement clear? Is this particular statement clear? Let me know in the chats quickly. Everyone, everyone in the chats. Everyone in the chats. So enthalpy of formation and enthalpy of reaction will be only equal when one mole of the substance will be getting formed. Okay, perfect. Now there is one

more thing which you need to remember directly from now on. Do remember the standard enthalpy of formation. The standard enthalpy of formation of an element of An element present in its standard state. the standard enthalpy of formation of an element present in its standard state is taken as zero. This is one more thing which you have to remember. What is meant by that? For example, I'm writing standard enthalpy of formation of H2 gas. Okay? For example, I'm writing standard enthalpy of formation of O2 gas. For example, I'm writing standard enthalpy of formation of graphite Of

graphite. Now look at these examples carefully. H2 gas it is the standard state of hydrogen. This is the standard state of oxygen. This is the standard state of carbon. And standard enthalpy of formation of an element in its standard state is taken as zero. Is taken as zero is taken as zero. Clear? I've got one question for you. I have got one question for you. I have got one question for you. Which of the following reactions define the standard enthalpia of formation? Which of the following reactions define the standard enthalpy of formation? Can you let

me know in the chats? Look at all these reactions carefully. Look at all these reactions carefully and let me know the answer of this question. It is an asked question, guys. It's an asked question. It's an asked question. Quickly, how do you define the enthalpy of Formation? In order to define the enthalpy of formation, one mole of substance has to get formed. One mole of substance has to get formed. Over here, one mole of substance is being formed over here. One mole of substance is being formed over here. One mole of substance is being formed.

But here, one mole of the substance, no, two moles of the substance are being formed. So, cross it first of all. Cross it First of all. Now one mole of the substance has to get formed from its elements which must be present in its standard states. What is the standard state of carbon? The standard state of carbon is graphite. But here you have used diamond. So no this not the case. Look at this one. What is the standard state of carbon? Standard state of carbon is graphite. Over here you have used carbon monoxide. This not

the case. Look at the second example. What is the Standard state of hydrogen? H2. What is the standard state of chlorine? F_sub_2. So this is going to be the correct answer of this particular question. Am I clear to everyone? Am I clear to everyone? Absolutely. It's going to be option B which is going to be the correct answer of this particular question. Yeah, I believe it's absolutely clear. Now my dear students, let me before showing you One more question, let me tell you one more important thing. What is that? That is calculation of calculation of

standard enthalpy of reaction from from standard enthalpy of formations. calculation of standard enthalpy of reaction from standard enthalpy of formations. Look at it carefully again. Look at it carefully again. For example, I'm writing a reaction. The reaction is like this. N1 A plus N_sub_2 B it gives N3 C + N4 D. Imagine this is a balanced chemical equation wherein N_sub_1 N_sub_2 N3 and N4 these are the stoometric coefficients of reactants and products. N1 N2 N34 these are the stoometric coefficients of reactants and products. Okay. Stoometric coefficients of reactants and products. Now my dear students for

example I need to calculate I need to calculate the amount of heat which is being absorbed or released during this Reaction. That means I'm asking you to calculate the enthalpy of the reaction under standard conditions. How do we do it? How do we do it? It is always equal. It is always equal. The standard enthalpy of formation of products minus the standard enthalpy of formation of reactants. It is always equal. Standard enthalpy of formation of products minus standard enthalpy of formation of reactants. How do I write it? Try to understand. Start with the product. It

Stoometric coicient is N3 multiplied by standard enthalpy of formation of C plustric coation here is N4 multiplied by standard enthalpy of formation of D. Right? This whole term is what I'm calling as standard enthalpy of formation of products minus standard enthalpy of formation of reactants. Start with a its stoometric coicient is n_sub_1 multiplied by the formation of a plus it is n_sub_2 multiplied by standard enthalpy of formation of b. My Dear students do remember this particular result as well. From this particular result you can solve lot of questions. From this particular result you can solve

a lot of questions. From this particular result you can solve a lot of questions. This is the general result by means of which you can calculate the standard enthalpy of reaction from the standard enthalpy of formations of products and reactants. And it is written as standard enthalpy Of formation of products minus standard enthalpy of formation of reactants. In the questions in the questions they will give you these values. In the questions they'll give you these values. Right? They'll give you these values. Okay? They'll give you these values. Number one. Number two, if you ask me

what is the unit of the standard enthalpy of the reaction, it is mainly majorly expressed as kilogj per mole. Right? It is majorly expressed in Kiloj per mole. Now I'll show you a question and this concept will be clear there only. I'll show you one question in which the concept will be clear. It'll be absolutely clear. One equation I'm going to show you. Okay. Look at this question carefully. Read the question carefully. Again, I'm telling you the same thing. Standard enthalpy of reaction is equal to standard enthalpy of formation of Products minus reactants. But take

into consideration the stoometric coefficients as well. Okay. Look at this question carefully. The standard enthalpy of formation of carbon dioxide, carbon monoxide and water are given as this this respectively. The heat exchanged by the reaction at constant volume. The heat exchanged by the reaction at constant volume at standard conditions in kilogjles for the reaction is For the reaction is see guys how exactly I'm going am I going to solve this question. First thing I'm going to write the reaction again. The reaction is like this. Carbon dioxide gas plus H2 gas. It gives carbon monoxide gas

plus H2O gas. This is the reaction that's given to me. Agreed? This is the reaction that's given to me. As per the question, if you look carefully, what am I supposed to calculate? I'm supposed to Calculate the heat absorbed or released during the reaction at constant volume. Heat absorbed or released during the reaction at constant volume. That means this is the thing which I'm supposed to calculate. This is the thing which I'm supposed to calculate heat absorbed or released during the reaction at constant volume. And what do you call this as I hope you remember

something which you call as internal energy change in the reaction. And since I'm supposed to calculate it under standard conditions basically this term I'm supposed to calculate internal energy change of the reaction I'm supposed to calculate basically right internal energy change of the reaction I'm supposed to calculate correct now try to understand carefully I told you already I told you already there's a relation between delta H and delta There's a relation between delta H and delta U. Do you remember that? Delta H is equal delta U plus delta NG RT. Now my dear students, can

you let me know what is the value of delta NG for the reaction? What is delta NG? First of all, delta NG represents number of moles of gaseous products minus the number of moles of gaseous reactants. Number of moles of gaseous products minus number of moles of gaseous reactants. Now this is the gaseous Product. This is the gaseous product. This is the gaseous reactant. This is the gaseous reactant. Number of moles of gaseous products. So 1 + 1 that makes it two. So number of moles of gaseous products are two minus number of moles of

gaseous reactants. 1 + 1 that's 2. So 2 - 2 is 0. So my dear students this is the reaction whose delta ng is zero. If the delta ng for the reaction is zero, you know delta h is equal delta u plus delta ng rt, right? Delta ng rt. And if You want to write it under standard conditions, you can write it too. Now since delta ng for the reaction is zero, so I would say delta h for the reaction is nothing but deltau for the reaction. Perfect. So basically I'm supposed to calculate delta U

for the reaction and I got to know delta U for the reaction is nothing but delta H. So indirectly I'm supposed to calculate delta H for the reaction. Indirectly I'm supposed to calculate standard enthalpy change of The reaction. And already I have told you standard enthalpy of the reaction can be calculated can be calculated with the help of standard enthalpy of formation. with the help of standard enthalpy of formation I I just need to write standard enthalpy of formation of products minus standard enthalpy of formation of reactants right so start with the products coation is

one right so leave it as such multiplied by standard enthalpy of formation of what Carbon monoxide plus 1 multiplied by standard enthalpy of formation of what standard enthalpy of formation of water perfect so this is your product side minus minus the reactant side minus the reactant side. Look here stochometric coicient is one multiplied by standard enthalpy of formation of what? Standard enthalpy of formation of carbon dioxide plus this is stoometric coicient is 1 multiplied by standard enthalpy of formation of H2. Yes, it's H2. Correct people? Now if you look at this particular equation carefully, if

you look at this particular equation carefully, what about this particular term? Standard enthalpy of formation of H2 gas. Hydro H2 gas is the standard state of hydrogen. And the standard enthalpy of formation of an element which will be present in its standard state it is taken as zero. So this term is zero. This term is zero. Right? This term is Zero. This term is given to you as per the question of carbon monoxide. Even this term is given to you as per the question enthalpy of formation of water. Right? Even this term of carbon dioxide,

this is given to you. Put the values here and get the answer. That's all. That's all. You will be getting the delta H of the reaction. And delta H of the reaction is nothing but delta U for the reaction. As simple as that. Right? As simple as that. And the final Answer of the question after solving this will come out to be 41.2 kiloj per mole. I believe this is clear. If this is clear, let me know once in the chats with some thumbs ups. Quickly guys, quickly, quickly, quickly. Everyone, everyone in the chats, everyone

in the chats, everyone in the chats, quickly, quickly, quickly. And I can see everyone has not liked the video yet. I want you guys to Smash that like button as well. Quickly, quickly. Perfecto. Perfect. Perfect. Perfect. I'm going ahead. I am going ahead. I believe the first the first enthalpy of the reaction which was called as enthalpy of formation that is clear. Now moving on now moving on to the second that is standard Enthalpy standard enthalpy of combustion. Now this is something which you'll understand easily guys. Easily very easily. Okay. So first thing I'm representing

this with delta H not C. I'm representing this with delta H not C. Now what this standard enthalpy of combustion is all about? Let me write its definition. My dear students, standard enthalpy of combustion it is defined as it is Defined as the enthalpy change. It is defined as the enthalpy change. What is meant by enthalpy change? The amount of heat absorbed or released. What is meant by enthalpy change? The amount of heat absorbed or released. Right? Okay. The enthalpy change when one mole of substance when one mole of substance under goes under goes combustion.

When one mole of the substance under Goes combustion in excess of oxygen in excess of oxygen. When one mole of the substance under go combustion in exc of oxygen. First of all, before making you understand this particular definition, let me tell you combustion, it is always an exothermic process. Combustion is always an exothermic process. If combustion is always an exothermic process, that means during combustion heat is always released. During combustion, heat is always Released. Therefore, that is the reason why enthalpy of combustion value always comes out be negative. That is the reason why delta H

not C is always negative because it's always exothermic because it's always exothermic. Right? Okay. Now people try to understand what it means. Try to understand what it means. Have a look carefully. Have a look. Carefully have a look. My dear students, for example, I'm writing a reaction. Understand it like this. CH4 plus O2 gas. It gives your carbon dioxide and water. This is the reaction a general reaction I have written. Tell me one thing is this reaction balanced or not. I don't think this reaction is balanced. This reaction needs to be balanced. This reaction needs

to be balanced. Right? So make this two times. Make this two times. Make this two times. Now the reaction is balanced. Now the reaction is balanced. This is one here. This is One here. Okay. If you look at this reaction carefully, first of all, it is the example of combustion reaction. You have taken a hydrocarbon, you are doing its combustion and combustion of hydrocarbon, it leads always to the formation of carbon dioxide and water. Its products are always carbon dioxide and water. Now people tell me one thing. Can I say in this particular reaction one

mole of substance is undergoing combustion. Do you see one mole of substance undergoing combustion? Right? During combustion heat is always released. So can I say whatever amount of heat will be released during this particular process that is something which you will be calling as enthalpy of combustion of CH4. Right? Okay. Whatever amount of heat will be released during this particular process that is something which you will be Calling as standard enthalpy of combustion of CH4. Correct? Because one mole of substance has undergone combustion. Look at the definition. The enthalpy change when one mole of substance

under goes combustion. Whatever heat is released during that process, that is something which you call as standard enthalpy of combustion. So this is standard enthalpy of combustion of your CH4. Agreed? Right people? For example, I'm taking let's say C3 H8 C3 H8 I'm doing its combustion. I'm doing its combustion. It always leads to the formation of carbon dioxide in water. Right? Balance this reaction. This is three times. This is three times. Right? Make it four times and make it five times. The reaction is balanced. If you look at this particular reaction carefully, can I say

one mole of propane is undergoing Combustion, one mole of substance is undergoing combustion. The amount of heat released, the amount of heat released when one mole of propane under goes combustion, the amount of heat released when one mole of propane under goes combustion, that is something which I'll be calling as standard enthalpy of combustion of your propane. Correct? Right. Now people tell me one thing. Tell me one thing. If I write the reaction like this 2 * CH4 + 4 * O2 + 2 * carbon dioxide plus 4 * water right + 4 * water

for example I'm telling you that standard enthalpy of this particular reaction is - 100 kgj I'm asking you to calculate the standard enthalpy of combustion of CH4 can you do it can you do it I gave you the reaction a balanced reaction Right? I'm telling you enthalpy change of this reaction is - 100 kgj. I'm asking you to calculate enthalpy of combustion of CH4. Can you Do it? Can you do it? See guys, in this reaction, look carefully. Two moles of methane are undergoing combustion. So when two moles of methane are undergoing combustion, how much

heat is released? 100 kg. So when one mole of methane will under combustion, how much heat will be released? 50 kg. Can I say enthalpy of combustion of methane will be minus50 because heat is being released. Right? If you look carefully Air one mole of methane was undergoing combustion air one mole of propane was undergoing combustion and in both these reactions where one mole of substance was undergoing combustion in both these reactions if you see enthalpy of reaction and enthalpy of combustion they are equal they are equal. But in the last case, is the enthalpy

of reaction and enthalpy of combustion equal? They're not equal. They're not equal. So, can I generalize a statement? Can I generalize a statement? Can I say enthalpy of the reaction will be only called as the enthalpy of combustion when one mole of substance when one mole of substance will be undergoing combustion? When one mole of substance will be undergoing combustion. If this is clear, let me know once in the chats with some fire emojis. Everyone people everyone means everyone. Is that clear? Is that clear? Is that clear? Is that clear people? Okay. One more thing

which I want to tell you over here. One more thing which I would want to tell you. Look at it very carefully. Look at it very carefully. Look at it very carefully. Okay. Right. Look at it very carefully. I'm writing the reaction H2 gas plus O2 gas. It gives me H2O liquid. H2O liquid. Let's assume the enthalpy change of the reaction is delta H not R. This is the amount of heat which is absorbed or released. This is the amount of heat which is absorbed or released. This is the amount of heat which is absorbed

or released. Right? It can be anything. It can be anything. Look carefully. My dear students, do you see one mole of H2 is undergoing combustion, right? One mole of H2 is undergoing Combustion. And when one mole of substance under go under goes combustion, standard enthalpy of reaction, this will be also called as standard enthalpy of combustion of H2. This will be also called as standard enthalpy of combustion of H2. Look at one more thing. Do you see one mole of water being formed? One mole of water being formed. One mole of water being formed from

its elements which are present in their standard states when One mole of substance is getting formed at that point of time delta HR is also equal to delta H not F. So can I say this is something which I'll be calling as standard enthalpy of formation of water as well. If this particular point is clear then you are fit enough to solve the questions. Let me know in the chats if this is clear. Let me know in the chats if this particular point is clear to everyone. Quickly in the chats everyone All clear all clear

I want everyone to say it guys this is something which very important this is something which is very important one mole of substance is undergoing combustion right and one mole of substance is being formed as well correct right and I have told you already when one mole of substance is formed from its elements present in their standard states at that time of at that point of Time. Standard enthalpy of reaction is called as standard enthalpy of formation. Formation of formation of water or you can say one mole of H2 is undergoing combustion. So you can

say standard enthalpy of reaction will be also called as standard enthalpy. I mean standard enthalpy of reaction will be also called as standard enthalpy of combustion of H2 over here. Okay, perfect. I believe this particular point is absolutely Clear to everyone and if this point is clear to everyone let me tell you one more thing from which you can get the questions right dear students take a note of this okay take a note of this for example I've got the reaction N1 A plus N_sub_2 B gives N3 C plus N4 D this is a balanced

chemical equation wherein N1, N2, N3 and four these are the stoometric coefficients of the reactants and products. Right? Now my dear students if I ask you how much amount of heat is absorbed or released during the reaction? How much amount of heat is absorbed or released during the reaction? One way one way of solving this is going to be enthalpy of formation of products minus enthalpy of formation of reactants. One way of getting this, one way of calculating standard enthalpy of reaction. What is that? Formation of products minus formation of reactants. But there is one

More way as well. It is going to be combustion of reactants minus combustion of products. It is going to be combustion of reactants minus combustion of products. How exactly you are going to write it here? Now you are going to start with reactants. This is your reactant. This is your reactant. Stoometric equation N1. So n_sub_1 * standard enthalpy of combustion of a plus n_sub_2 multiplied by standard enthalpy of combustion of b. This is What I'm calling as standard enthalpy of combustion of reactants minus standard enthalpy of combustion of products. So start with the product choice

coation is n3. So n3 * standard enthalpy of combustion of c plus n4 ult*lied by standard enthalpy of combustion of combustion of d. This is one more way of calculating the standard enthalpy of the reaction. This is one more way of calculating the standard enthalpy of reaction. First one Standard enthalpy of reaction is equal standard enthalpy of formation of products minus reactants. Also it can be calculated as standard enthalpy of combustion of reactants minus products. Okay, remember it directly and if you could remember it directly, it is the time to solve one question and this

is again the asked question. This has been asked in your examination. I don't remember exactly whether this was asked in means, J means or need. Look at this. Look at it carefully. Look at it carefully guys. The heat of combustion of ethanol is -300 kilo calories. If the heat of formation of carbon dioxide and water are this and this, calculate the heat of formation of ethanol. We are supposed to calculate the heat of formation of ethanol. We are supposed to calculate the heat of formation of ethanol. And we are already given with a Reaction. Reaction

is given to us. One reaction is given to us. No, Gokul Pankuri mam she did not leave an academy. She's the part of an academy. She's taking the batches. She's taking the internal badges, right? She's taking the internal badges. Look at this people quickly quickly quickly. If you look at this particular reaction Which is given to us, try to understand carefully. Can I say one mole of ethanol is undergoing combustion? Absolutely. One mole of ethanol is undergoing combustion. And when one mole of substance under goes combustion at that point of time standard enthalpy of reaction

is also called as standard enthalpy of combustion of what? Of ethanol. Is this standard enthalpy of combustion of ethanol given? Yes, it's given. It's Given as minus 300 kilo calories. Right? The question is done. It is given as -300 kilo calories. One mole of substance is under combustion. Correct? So standard enthalpy of reaction will be called as standard enthalpy of combustion of what? Of ethanol. Now people tell me one thing delta H not R standard enthalpy of reaction. Can I write it as standard enthalpy of formation of products minus standard Enthalpy of formation of reactants.

Can I do that? I can do that. Standard enthalpy of formation of products. Start with this. So it is 2 * the standard enthalpy of formation of carbon dioxide plus this is three times the standard enthalpy of formation of water. Right? So this is the standard enthalpy of formation of products minus now the reactance 1 * standard enthalpy of formation of ethanol perfect plus 3 * the standard enthalpy Of formation of O2 right right people now tell me one thing what is this delta h r all about it is equal to - 300 so this

term is equal to -300 is equal 2 * enthalpy of formation of carbon dioxide should be given to us. It is 2 * It is 2 * enthalpy of formation of carbon dioxide is given to us as how much - 94.3 - 94.3 okay plus 3 enthalpy of formation of water is also given to us how much - 68.5 perfect - 68.5 minus standard enthalpy of formation of ethanol we have to calculate Tell me one thing. What about this term? Can I say oxygen here is in standard state? Oxygen is present in its standard state

and standard enthalpy of formation. Standard enthalpy of formation of an element present in its standard state that has to be zero. Right? So my Dear students, all these things are given to us. All these things are given to us. Perfect. We are just supposed to calculate this particular term. Can't we do a bit of calculation plus minus here? Right? Can't we do a bit of plus minus here? Is it clear people? Is that clear? Is that clear to everyone? Quickly, quickly, quickly, quickly. And when you solve this, when you solve this, this is the answer

which you get. - 941 kilo Calories. One simple question for you. Let me see if you can solve this or not. Look at this question carefully. Look at this question carefully, guys. The heat of combustion of sucrose. The heat of combustion of sucrose is 1350 kilo calories. How much of heat will be liberated when 17.1 g of sacrosse is burnt? What do you think? What do you think? How you are going to solve this question? It's simple again. It is again simple guys. It is again simple. Look at it carefully. Look at it carefully. As

per the question is concerned the heat of combustion of sacrosse the heat of combustion of sucrose is 1350 kilo calories. What does that mean? That means when one mole sucrose when one mole sacrosse that means when one mole sucrose under goes combustion how much heat is released? As per the question it is These vicular caps heat of combustion of sacro is given that means when one mole sucrossse under go combustion this much heat is released but as per the question 17.1 g of sucrose are undergoing combustion 17.1 g of sucrossse are undergoing combustion that means

how many moles of sacros are undergoing combustion mass of sacrosse divide by molar mass of sacros So 1x 20 moles 1x 20 moles of sacrosse are undergoing combustion as per the Question we know when 1 mole sacrosse under go combustion this much amount of heat is released. So when 1 upon 20 moles of sucrose when 1 upon 20 moles of sucrossse under goes combustion how much heat will be released? I believe this will be 1350 multiplied by 1 upon 20. The value comes out be how much in the charts people are saying 67.5. So 67.5

kilo calories of heat will be released. Isn't this something which I was Supposed to calculate? Isn't this something which I was supposed to calculate? This is the amount of heat which is released when 1 by 20 moles of sucrossse under go combustion. This is the amount of heat which is released when 17.1 g of sucrossse are undergoing combustion. I hope this is clear to everyone. I hope this is clear to everyone. Right? Okay. The question is saying how much Heat will be liberated? This much amount of heat will be liberated. Right? this much amount of

heat will be liberated. Okay? Right? Is it clear people? Let me know once in the chat if it is clear. If it is super clearly, have you heard about something called as calorphic value? In bio, you would have used this term many a times. In bio you would have used this term many times. Calorific value. Calorific value. PV CV caloropic value. How do you define the term calorific value? How do you define the term caloropic value? My dear students, let me first of all write its definition. Then I'll make you understand what this calorific value

exactly is all about and where it's used. Where it's used. Calorific value. It is Defined as the amount of heat released. The amount of heat released or liberated. The amount of heat liberated when 1 g of fuel. When 1 g of fuel under goes complete combustion. When 1 g of fuel under goes complete combustion, when 1 g of fuel under goes combustion, when 1 g of fuel under goes combustion, the amount of heat released at that Particular point of time is something which you call a caloropic value is something which you call a scaloric value.

And I say more the calorific value of the fuel. And I say more the calorific value of the fuel more the heat released more the heat released. And if more the heat released, I'll say better is the quality of the fuel. better is the quality of the fuel. So basically on the basis of CV value, Basically on the basis of CV value, you can check the quality of the fuel. You can check the efficiency of the fuel. Better is the quality of the fuel or you can say more is the fuel efficiency. More is the

fuel efficiency. More is the fuel efficiency. Now let me make you understand this particular thing here. See understand During combustion heat is released. When 1 g of fuel under goes combustion, whatever amount of heat is released, that's something which you call as caloric value. More the calorfic value of the fuel, more will be the heat released when the when one gram of fuel under goes combustion. More the heat released that means more is the efficiency of the fuel, better is the quality of the fuel. Right? Now people try to understand It a bit more in

detail. try to understand it a bit more in detail. For example, for example, for example, for example, my dear students, I'm writing the standard enthalpy of combustion of methane is equal. I'm giving the random value. It is minus 100 kiloj per mole. It is - 100 kgj per mole. If I ask you what it means, if I ask you what it means, standard enthalpy of combustion of methane is Minus 100 kgj. What it means? It means that as per definition, it means that when one mole methane under goes combustion, 100 kgj of heat is released.

When one mole methane under goes combustion, 100 kgj of heat is released. Right? When one mole methane under go combustion 100 kgj of heat is released. That is the meaning of this particular thing. Right? That is the meaning of this particular thing. Now can you let me know what is the mass of one mole of Methane? Mass of 1 mole of methane means molar mass of methane. How much that is? 16 g. So instead of 1 mole of methane I can write when 16 g of methane under go combustion 100 kgj of heat is released.

Right? Now people you tell me when 16 g of methane under goes combustion this much is the amount of heat released. Now tell me when 1 g methane under go combustion. When 1 g methane under go combustion use the unitary use the unitary method. When 1 g under goes combustion, the heat release will be 100 divid by 16 kilogjles. 100 divid by 16 kiloj. And people if you look at the statement carefully when 1 g methane under goes combustion this is the amount of heat released. As I told you the amount of heat released when

1 g of fuel under goes combustion that determines the CV value of the fuel. So indirectly I calculated the CV of methane. I calculated CV of methane. This is the calorfic value of meth 100 by 16. This is the amount of heat which is released when 1 g of methane was undergoing combustion. Okay. Now guys, do we have to do all the process every time? No. Why I did it? Just to get you just to get you a formula. just to get you a direct formula. Okay. Caloric value of a fuel is always equal. If

you look at this, this is basically calorific value. This is caloric value Of methane. What was 100? 100 was magnitude of enthalpy of combustion of methane. What was 16? 16 was 16 was the molar mass of methane. So the formula of caloric value will be mod of standard enthalpy of combustion of a fuel divided by molar mass of the fuel. This is the general expression by means of which you can calculate the calorific value of the fuel. Right? General expression. Thank you. There we go. Thank you. Thank you so Much. This is the general expression

by means of which you can calculate the caloric value of the fuel. Standard enthalpy of combustion of fuel magnitude divide by molar mass of the fuel. Do you know more the CV more the caloric value of the fuel lighter is the quality of the fuel. More is the efficiency of the fuel. More is the heat released when one gram of fuel underos combustion. As simple as that. As simple as that. Now my dear students, If caloric value is clear, you should be in a position to answer this question as well. You should be in a

position to answer this question as well. Isn't it very simple? Enthalpies of combustions of ethane, ethine, butane, octane are given. Arrange them in the order of their fuel quality. Arrange them in the order of their fuel quality. So what do you do? I will be calculating the calorific value of Everyone. How do you get calculate the caloroporic value? How do you calculate the calorfic value? So first of all for example I'm calculating the caloric value of ethane right it will be equal standard enthalpy of combustion of ethane right divided by divide by its molar mass

right sorry okay what about this term this term it's given to us as minus so I'll take 20 kilj divided by molar mass of Ethane 12 to 24 + 6 30. So this is basically your calorific value of EA. Similarly, you'll calculate calorific value of every fuel, right? Then you'll compare whose calorific value is more. The one which has got more calorific value that is going to have the better quality. I hope this is clear to everyone. If this is clear, let me know once in the chats quickly. If it is clear, let me know

once in the Chats quickly, people. Quickly, quickly, quickly. I'll give you on my telegram t.me/ M E/ W A S S I M S I R C H E M right I'll send you the uh PDF of this particular session in this telegram okay all right people look at this particular question carefully calculate the enthalpy of combustion of C2H6 calculate the enthalpy of combustion of C2H6 the enthalpy of combustion of C2H6 oh my God what did I do calculate the enthalpy of combustion of C2H6 Okay. So, first of all, since we have to calculate the

enthalpy of combustion of C2H6, tell that one thing. Can we write the combustion reaction of C2H6? We can. Can we write the combustion reaction of C2H6? We can write the combustion reaction of C2H6. Right? And combustion of your hydrocarbon, it always leads to the Formation of what? It always leads to the formation of your carbon dioxide. And with carbon dioxide, you will write H2O as well. Water as well. Right? You'll write water as well. Okay. This is your combustion reaction of what? This is your combustion reaction of C2H6. Perfect. Perfect. What am I supposed to

calculate? What am I supposed to Calculate? I'm supposed to calculate the enthalpy of combustion of C2 and six. First of all, is the reaction balanced? No. Balance the reaction. This two times. This has to be two times. This is six. So, this has to be three times. And this will be this will be how much? This will be how much? 2 + 6 by 4. This will be seven times. I believe you check it once. This is 7 to 14. No, not seven times. I have done some mistake. Just a second. This won't be 7

times. How much this will be? It is x + y / 4. 4 to 8 8 + 6 / 4. So it's 14 / 4 which is 7 by2. Okay, this is 7x2. Now you must be thinking how did I get the stoometric coicient here? You must be thinking about that right? If you do not know, let me quickly tell you. If you do not know, let me quickly tell you. Whenever you have got the hydrocarbon in this manner, CX HY and You are doing its combustion. When you're doing its combustion, it will first of

all give you carbon dioxide and water. Now there's a general way of balancing it. General way. If this is X, take this X here. This is Y. Make Y by 2 here. And what is the coefficient of oxygen here? It will be X + Y / 4. This will be the stoometric equation of oxygen. If you do not know, this is how you balance it. Now if you do not know one more thing For example you have got something like this CX HY O Z you are doing its combustion right you are doing its combustion

it will give you carbon dioxide and with carbon dioxide it will give you water as well right how do you balance this X make this X here this is Y keep this Y by2 here right and what what about the coefficient of O2 it's X + Y / 4 - Z by 2 this is how you are going to balance it Right. This is how you are going to balance it. That's all. That's all. That's all. That's all. Perfect. Perfect. Do remember this particular thing. Do remember this particular thing. Now people, let's have a look

on the question now. Let's have a look on the question. If you look at the reaction, can I say one mole of ethane is undergoing combustion? Right? One mole of ethane is undergoing combustion. And when one mole of Substance under under goes combustion I'll say enthalpy of reaction is what you call as enthalpy of combustion of C2H6 right this something you can write because one mole of substance is undergoing combustion one mole of substance is undergoing combustion now my dear students look at all the data I'm given with standard enthalpy of combustion of H2 combustion

of CH4 combustion of C3 H8 that's given To me that's given to me uh Wait wait wait wait wait wait wait wait. I hope this particular point is clear. This perfect. So guys I hope this point is clear because this is something which is very important. Now look at the reaction. The enthalpy change of the reaction. The reaction is given as C3 H8 plus H2. It gives C2H6 plus CH4. This is the reaction that's given to me. The reaction is given in the balanced format. Right? The reaction is given in The balanced format. As per

the question the enthalpy change of the reaction the enthalpy change of the reaction is given how much it's given it's given to me as -55.7 kJ per mole now my dear students how do I write enthalpy of reaction I can write enthalpy of reaction as I can write enthalpy of reaction as enthalpy of combustion of reactants minus enthalpy of combustion of products so start with the reactants it's going to be enthalpy Of combustion of C3 H8 because it's stoometry equation is one right plus enthalpy of combustion of H2 because its stoometric equation is also one.

So this is the enthalpy of combustion of reactants minus enthalpy of combustion of products. So this is your enthalpy of combustion of your C2H6 plus enthalpy of combustion of your C H4. Right? And my dear students this particular term its value as per the question is - 55.7. Now look at this particular equation. Look at this particular equation. Enthalpy of combustion of C3 H8 is given to me. This term is given. Enthalpy of combustion of H2. Enthalpy of combustion of H2 is given to me. Enthalpy of combustion of CH4 CH4. It is given to me.

Can't we calculate from this particular equation? Can't we calculate the enthalpy of combustion of C2H6? Absolutely. That is something which I was supposed to calculate. Basically, That is something which I was supposed to calculate. I believe this is clear. I believe this is clear. Perfect. I believe this is clear guys. Let me repeat it again. I'm going to cover each and every chapter in this particular 45 days crash course which has started today. Yes. Perfect. So this sort of equation I hope you can easily solve from now on. Now have you heard about Hessa's law?

Have you heard about Hessa's law? Tell me that. Say it in the chats. Have you heard about Hessa's law? You would have heard, right? You would have. I'm pretty much sure. From Hessa's law confirmed equation comes only few things you should know about Hess's law you're done you're sorted right look at the statement according to this law the net enthalpy change will be the same the net enthalpy change will be the same if The reaction is carried out in a single step or more than one steps. First of all, tell me whether enthalpy is a

state function or path function. Do you know about that? Enthalpy. Enthalpy is a state function. If enthalpy is a state function, that means it is path independent. It is path independent. You should know it. Enthalpy that's a state function. It is path independent. Now people see what h's law is all about. For example, for Example, I have got a reaction which is carried out in a single step. Imagine this reaction is carried out in a single step. Right? Imagine this reaction is carried out in a single step. A gives B. Enthalpy change over here. For

example, is delta H1. Now the same reaction the same reaction is carried out in more than one steps. For example, first A gets converted into C. Then C gets converted into D. Then D gets converted into B. Perfect. My reaction was A to B. This my reaction A to B. First it is a single step. Then a series of steps. I made the reaction happen through two parts. I made the reaction happen through two parts. Single step path, multi-step path. Right? Now A gives C. Let's say its enthalpy changes delta H2. C gives D2 delta

H3 delta H4. What is Hess law all about? Look at the statement. Hess's law says that you know enthalpy is a state function. It is path independent. So the Net enthalpy change during the process will remain same. The net enthalpy change will be the same. If the reaction is carried out in one step or more than one step, what does that mean? That means whatever will be the value of delta h1 that will be equal delta h2 plus delta h3 plus delta h4 that's all isn't it simple people isn't it simple whether you want to

carry out this reaction in one step or more than one step right enthalpy change Will always remain the same enthalpy change will always remain the same Yeah, enthalpyic change will always remain the same perfect it is independent of the path. Now few more things which I want you guys to remember. For example, for example, take a note of these things. I'll use these things in a while. For example, you got the reaction like this. A gives b. Let's say it's delta h is Some x kj. It delta h for example is x kj. My dear

students, if you multiply or leave multiplication aside, if you reverse this reaction, if you reverse this reaction, when you reverse this reaction, you have to change the sign of delta H. Now the delta H of this reaction will be minus X kJus X KOJ. For example, I'm multiplying this reaction by some number. For example, I'm multiplying this reaction by N. N A gives NB. I multiplied this reaction by N. This n has to be multiplied with its delta h value as well. So I'll say delta h for this reaction will be n * x n

* x okay n * x okay so first thing when you reverse the reaction delta h sign has to be inversed I mean delta h sign has to be changed when you multiply a reaction by some number that number has to be multiplied with its delta h as well. Okay. Now my dear students when you add some Reactions when you add some reactions when you add some reactions their delta H values are to be added to their delta H values are to be added to. I'll write it over here. I'll write it over here. For

example, you have got a reaction A + B gives C plus D. Then you have got the reaction A + F gives X + Y. Right? For example, the first reaction it delta H is delta H1. Here it's delta H2. If I add these two reactions, after adding these two reactions, I'll get a net Reaction. If I ask you what will be the enthalpy change of the net reaction, since you added these two reactions, so the enthalpy change of the last reaction will be delta H1 plus delta H2. Is it clear? When you add two

reactions, when you add the reactions, their delta H values are to be added to. That's all. Their delta H values are to be added to. Now, for for example, I'm giving you a very small question. Let me see if you can solve this or not. Let's say I've Got the reaction carbon graphite. Carbon graphite plus half of O2 gas. Let's say it gives carbon monoxide gas. Let's say it gives carbon monoxide gas. I'm giving you some random values. Random values. Let's say delta H for this reaction is equal - 100 kiloj for example. For example,

now carbon monoxide plus half a photo it gives you carbon dioxide it gives you carbon dioxide. Imagine it's delta H is for example some random Values - 200 kJ - 200 kgj - 200 kgj now now 2 * carbon dioxide 2 * carbon dioxide gives 2 * carbon dioxide gives 2 * carbon graphite 2 * carbon graphite plus 2 2 * O2. You're supposed to calculate delta H for this particular reaction. Can you give it a try? Can you give it a try people? Can you give it a try? This will be the Type of

the question which will be asked. A very simple and basic question. A very simple and basic question. My dear students, the reaction whose delta H is to be calculated. Remember it in short. The reaction whose delta H is to be calculated. You need to make you need to make that reaction out of the given reactions out of the given reaction. Right? The reaction whose delta H is to be calculated. You have to make that Reaction out of the given reactions. Now, how will you how will you make this reaction? How will you make this reaction?

Tell me that. How will you make this reaction? Let me call this as equation one. Let me call this as equation two. Let me call this is equation three. How will you make equation three? My dear students, the first thing which I'll do for example, if I add first and second, when I add first and second, this and this gets Cancelled. So I'll be getting carbon graphite. Carbon graphite plus half O2 half of O2 makes it 1 * O2 gives carbon dioxide. Since I added these two reactions, so their delta H values are to be

added to so it'll be -00 plus plus - 200 right and this will be - 300. This will be - 300. Perfect. Perfect. Now guys think over it. If I multiply this reaction by number two, if I multiply this reaction by number two, That two has to be multiplied with this as well. So it'll be - 600. it'll be - 600. Now if I reverse this reaction, if I reverse this reaction, it'll be something like this, right? And when I reverse the reaction, the sign of delta H has to be changed. So it'll be plus

600 kilogjles. Look at this reaction carefully. My dear students, look at this particular reaction carefully. Where we supposed to make the same reaction? Just try to Compare these two reactions. Absolutely these two reactions are same. Right? So can I say I made I made the reaction by certain operations right and I got the delta H which I was supposed to calculate. I believe it's clear. I believe it's clear. For example, one question I'm giving you. Look at this question guys. We are given with certain reactions their delta H values are given. What are we supposed

to calculate? What are we supposed to calculate? We are supposed to calculate the enthalpy of formation of H2O2. We are supposed to calculate the enthalpy of formation of H2O2. Look at it carefully. First of all, we are supposed to calculate the enthalpy of formation of H2O2. Will you be able to write the formation reaction of H2O2 on your own? Will you be able to write the formation reaction of H2O2 on your own? In order To write the formation reaction of H2O2, one thing has to be in my mind that one mole of H2O2 has to

get formed from its constituent elements which must be present in their standard states. Right? Look at this reaction. One mole of H2O2 is getting formed from its elements which are present in their standard states. So this is the formation reaction of H2O2 whose delta H is to be calculated whose delta H is to be calculated. Now Guys try to understand. Call this is equation one. Call this as two. Call this as three. Call this as four. Can I say I need to make I need to make the fourth reaction. I need to make the fourth

reaction out of the given reactions. Out of the given reactions. Now how will I make this particular reaction out of the given reactions? I have to do some algorithm here. Right? Okay. So what do I do? What do I do? My dear students, just if I look at All the reactions carefully, if I reverse the first reaction, let's see what do I get. Let me reverse the first reaction. When I reverse the first reaction, N2 plus 4 * H2O, it gives what? It gives your N2H4 with N2 H4, you get 2 * H2O2 as well.

Since you reverse this reaction, so what about this delta H? Now you have to change the sign. It'll be plus 800. It will be plus 800 kiloj. Right? It will be plus 800 kgj. It will be plus 800 kgj. Right? Let me keep the second reaction as such. Second reaction I'm keeping as such. N2 h4 plus o2. It gives n2 +2 * h2. Second reaction I kept as such. It's delta h is how much? - 600 kgj per mole. The third reaction which I have the third reaction which I have for example I'm multiplying this

reaction by two when I multiply this reaction by two it becomes 2 * H2 plus 22 gets cancelled So it's O2 gives 2 * H2 since I multiplied this reaction by two that two has to be multiplied by this number as well so delta H over here will be equal - 600 kJ right right people till here I believe everything is clear till here I believe everything This clear? Now my dear students, look at these three equations carefully. Look at these three equations carefully. Look at these three equations carefully. Tell me if I add these

three reactions. If I add These three reactions, N2H4, N2 H4 cancel, right? 2 * water, two 2 * water, 2 * water makes it four times water. That four * water with this four * water is cancelled. Right? this N2 this N2 is canled. So what am I left with? I'm left with something as 2 * H2 plus O2 makes it 2 * O2 it gives what? It gives what? It gives 2 * H2O2 right I made this particular equation how when I added Them when I added them since I added them so I have

to add their delta H as well. So delta H for this final reaction will be this plus this plus this. So it will be 800 plus - 600 plus - 600. So it is delta H will be equal -400 KJ. Okay. When you add them delta H will be -400. Now people is this reaction and this particular reaction same? They're almost same but only difference is stoometry coefficient. Right? So if I multiply this reaction With 1x2 when I multiply this reaction with 1 by 2 it becomes something like this that 1x2 has to multip has

to be multiplied with this number as well 1x2 into 200 makes it what it makes it -200 kJ per mole so dear students look at this particular equation carefully did we make did we make the reaction whose delta h was to be calculated yes we made the reaction whose delta h was to be calculated and delta attach came out to be how much - 200 kJ per mole. I believe This is again clear to everyone. Yes. Is it clear people? Is it clear? Is it clear? Is it clear? Now again one important concept enthalpy of

neutralization. Enthalpy of neutralization. So before starting the enthalpy of neutralization let me know once in the chats if all the things are clear till here or not. Quickly Enthalpy of neutralization. Let me know once in the chats if all the things are clear guys. One and a half hour more. That's it. One and a half more. That's it. We are almost done. Let's say it quickly. Okay. First of all, before discussing the concept of enthalpy of neutralization, what is neutralization? First of all, what is neutralization? Simple. What is Neutralization? Whenever you see acid reacting with

a base leading to the formation of salt and water, this particular reaction is what you call as neutralization reaction. This is what you call as neutralization reaction. Let me tell you neutralization is always exothermic. Whenever you see acid reacting with a base, always heat will be released. It's always exothermic. It's always exothermic. Right? It's always exothermic. Neutralization is always exothermic. Now before making you understand, before making you understand the enthalpy of neutralization, there are few things which I want to tell you. Do you know how do we calculate number of gram equivalents of any substance?

Do you know that number of gram equivalents of a substance is always equal to number of moles of the substance multiplied by its n factor. Right? Okay. Point number one. Number of gram equivalents of a substance is also calculated by this formula. Normality multiplied by volume of solution in liters. Just remember them directly. Remember them directly for now. Okay. This is the second formula to calculate gram equivalent. Right. Right. Now what is the relation between mill Equivalence and gram equivalent? Number of millie equivalence of the substance is always equal. Number of gram equivalents of the

same substance multiplied with thousand. One more result. Remember it. Next result. Normality is always equal marity multiplied by n factor. Just remember these results. Just remember these results. Why you need to remember these results. There's a logic behind that which you'll get in some time. First Remember these results. Number of gram equivalents of any substance is always equal to moles of it multiplied by its n factor. Right. Similarly, normality multiplied by volume in liters. Millie equivalence is gram equivalence multip,000. Normality is marity multip. I believe these results you would have somehow studied before also right

in your redox or some other chapters. Okay. Now guys try to understand imagine I'm taking 1 g equivalent of an Acid. Imagine I'm taking 1 g equivalent of a base. Imagine I'm taking 1 g equivalent of an acid. Imagine I'm taking 1 g equivalent of a base. Imagine 1 g equivalent of an acid is completely neutralized by 1 g equivalent of a base. Imagine 1 g equivalent of an acid is completely neutralized by 1 g equivalent of a base. The amount of heat released at that point of time is something which you Call as enthalpy

of neutralization. So what is enthalpy of neutralization? Enthalp since neutralization that's always exothermic. So heat is always released right? So heat is always released. So that's why enthalpy of neutralization value is always negative. Now how do we define the enthalpy of neutralization? When 1 g equivalent of an acid is completely neutralized by 1 g equivalent of a base at that point of time whatever amount of energy is Released that is something which you call as enthalpy of neutralization. Okay. Now understand very carefully. Understand very carefully. I just gave you this definition. The heat evolved when

one equivalent of an acid is completely neutralized by one equivalent of a base in dilute solution is called as heat of neutralization or enthalpy of neutralization. Right? Now guys try to understand very carefully. For example, I'm taking HCL. For example, I'm taking HCl. Is HCl the example of strong acid? Yes. HCl aquis. It is the example of a strong acid. I'm taking a strong base NaOH. It's a strong base. Strong base. Let them react. They will lead to the formation of salt. They will lead to the formation of salt. And with the salt you get

water as well. With the salt you get water as well. Okay. Imagine that I have taken 1 g equivalent of HCl. Imagine that I have Taken 1 g equivalent of NaOH as well. Okay. Now HCl is a strong acid. If it is a strong acid, it's a strong electrolyte. Strong electrolytes they get completely dissociated into their ions. Strong base strong base again strong electrolyte which gets completely dissociated into their ions. So this HCl in its aquis solution can I say it would have got completely dissociated into its ions. Its ions would be H positive aquis

plus Cl negative aquis. Similarly NaOH It's a strong base again strong liquide completely dissociated. So it would have got dissociated as Na positive aquis plus O negative aquis right this is a salt this is a salt again completely dissociated Na positive aquis plus what plus CL negative aquis and what do we have with it with it we have got H2O liquid as well try to understand very carefully try to understand very carefully guys na positive Na positive done and dusted CL Negative CL negative done and dusted what are we left with We are left with

H positive aquis plus O negative aquis. What does it give? It gives one mole of H2O liquid. It gives one mole of H2O liquid. Are you understanding what is happening? Are you understanding what is happening? I took an acid and a base. So what is happening at the end? What is happening? What is the end result? End result is from acid H positive is coming. from base O Negative is coming and at the end H positive is interacting with O negative and leading to the formation of water. This is your end result. So first of

all you should know when acid reacts with a base what happens actually from acid H positive comes from base O negative comes that H positive and O negative interacts and forms water forms water and whenever there are interactions whenever there is attraction H positive to O negative Attraction whenever there's attraction energy is released energy is released right so during this particular process when H positive and O negative interacts will be released will released. Do you see? I took 1 g equivalent of an acid of a strong acid. I took 1 g equivalent of a strong

base. And I told you when 1 g equivalent of a strong acid, when 1 g equivalent of an acid is Completely neutralized by 1 g equivalent of a base, whatever heat released is something which you call as enthalpy of neutralization. Perfect. What happens in reality? I'll say from 1 g equivalent of acid we got one mole of H posit2 from 1 g equivalent of base we got one mole of O negative and they interact they are leading to the formation of water right so can I define enthalpy of Neutralization in one more way as well

I'll say enthalpy of neutralization is the is the amount of heat is the amount of heat released is the amount of heat released when one mole H positive interacts with interacts with one mole O negative and leads to the formation of one more water Right so during this particular process whatever heat is released that's what I call as enthalpy of what that's what I call as enthalpy of neutralization so either Do you define enthalpy of neutralization like this or you define enthalpy of neutralization like this? One or the same thing. And let me tell you

experimentally it has been observed when one mole of H positive interacts with one mole of O negative forms 1 mole of water heat released during this Particular process is 57.2 KJ right or in calories you can say 13.7 kilo calories. This is the amount of heat released when one mole of water is formed from one mole of H positive and one mo of H negative right perfect since heat is released so I'll be using the minus sign so I can say enthalpy of neutralization in this particular case is nothing but -5 - 57.2 2 kgj

or -3.7 kilo calories right Guys now I need to do certain variations over if you look carefully I took strong acid and strong base I took 1 g equivalent of strong acid which gave one mole of H positive this 1 g equivalent of strong base it gave one mole of O negative and one mole of H positive one mole of O negative interacted and they gave us this much amount of heat they gave us this amount of heat. This is what I call as this what I call as enthalpy of Neutralization over here in this

particular case. Right? Perfect. Now guys, there are certain things which I need to tell you. Strong acid and strong base. Strong acid strong base. When 1 g equivalent of strong acid when 1 g equivalent of strong acid is completely neutralized by 1 gram equivalent of strong base the amount of heat released at that particular point of time is something Which you call as what which you call a standard enthalpy of neutralization which will be - 57.2 2 kilo ghou per equivalent or you can call it as -3.7 kilo calories per equivalent. Correct. Correct. This was

strong acid strong base case. Now try to understand. Imagine imagine instead of strong acid I've got a weak acid. The base is strong. Imagine I'm taking one gram equivalent of this. I'm taking one of one g equivalent of this as well. Be careful with this. I'm taking 1 g equivalent of weak acid. And here I'm taking 1 g equivalent of strong base, right? Definitely there'll be salt formation and water formation. Can you tell me what will be the amount of heat released over here? What will be the amount of heat released over here? Will it

be same as that of will it be same as that of 57.2 2 kilogj or 13.7 kilo calories or it'll be different will it be same Over here I took 1 g equivalent of weak acid 1 g equivalent of strong base right so will the heat release be this much what do you think what do you think will the heat release be this much no heat released will be less than this why is that why is that since we have taken 1 g of strong 1 g equivalent of strong base so from 1 g equivalent

of strong base definitely you will get one mole of O negative let's say this is your NaOH This is your CH3 CO perfect from 1 g equivalent of strong base you will get one mole of O negative but CH3 CO it's a weak acid will it be completely ionized no it'll be partially ionized it'll be partially ionized Right? It won't get completely dissociated into its ions. So what do you get? Do you get do you get one mole of H positive from here? No. You get less than one mole of H positive over here. Less

than one mole of H positive. We know when one mole of O negative reacts with one mole of H positive, heat released is 57.32. But here no doubt you have got one mole of O negative. But here you have got less than one mo of H positive. So will heat release be this? No, it'll be less than this. Right? It'll be less than this. Or I can I can say it another way as well. I can say it another way as well. I can say it another way as well. I can say it another way

as well. What was what was I expecting? How much heat should have got released? This much heat should have got released. It should have got released. But in reality, how much heat will be getting released? Less than 57.2. So if I take the difference of these two, if I for example this is 50 50.2, 2 let's say 50.2 Jew of heat is being released let's say 50.2 kiloj of heat is Being released how much should have got released this much if I take difference of these two how much the value comes out be 7.2 kiloj

I will say 7.2 2 kgj this much amount of heat would be utilized to ionize this weak electrolyte right so I would say enthalpy of ionization of your weak acid here in this case is 7.2 kgj I hope I'm clear how much I was expecting to get formed to get released 57.2 how much heat actually got released 50.2 If I take the Difference of the two, I got the value 7.2 kg. That means this much amount of heat will be utilized ionize the weak component. I hope this is clear. I hope this is clear. I'm

giving you one question here. I'm giving you one question here so that it'll be absolutely clear to you. Look at this question carefully. Carefully you look at this question from this. From this it'll be all clear. See guys the enthalpy of neutralization of NaOH and NH40 by HCl. So first of all NaOH plus HCl in the second case NH40 plus HCl right what do you get? You get salt and water here absolutely salt water. Similarly in the second case what do we get again? We get salt water. Perfect. Now understand carefully. This is your strong

base. This is your strong acid. This is your weak base. This is your strong acid. Okay. Enthalpy of Neutralization. Enthalpy of neutralization when NaOH reacts with HCl is how much? Enthalpy of neutralization in the first case is equal -13680 calories. In the second case, how much it is? In the second case, it is -12270 calories. Right? Can I say in the first case 13680 calories of heat is released. Right? In the second case 12270 calories of heat is released. This was strong acid strong base case. Strong acid strong base. How much heat was released here?

13680. Now this is strong acid weak base. How much should have been? How much should have been? How much should have been? 13680 calories of heat should have been released. But how much is getting released? 12270 calories of heat is getting released. If you take the difference of the two, if you take the difference of the two, 1 1410 calories, This much amount of heat would have got utilized to ionize the weak component. And the amount of heat which is utilized to ionize the weak component that's what I call as enthalpy of ionization of NH4

over here which comes out be 1410 cal this is the final answer of the question done and dusted right am I clear am I clear people this much amount of heat would have got utilized to ionize the weak I believe I'm clear am I am I clear people how to find it weak or strong see all your weak acids weak bases they are your weak electrolytes okay strong acid strong bases they are always your strong electrolyte okay Okay, one more similar sort of question. One more similar sort of question. Look at this particular equation. We

have got dchloro acidic acid with NaOH. Heat of Neutralization is this much. So this much amount of heat is released. HCl by NaOH. This much amount of heat is released. NH40 with HCl. This much amount of heat is released. We have to calculate the heat of neutralization of dchloric acid by NH40. Look at this question carefully again guys. Look at this question carefully. We have got dicchloric acid CH Cl2 CO with NaOH. This is strong base. This is weak acid. How much heat is being released in this Case? 12830 calories of heat is released in

this case. In the second case, it is HCl with NaOH. HCl is your strong acid. This is your strong base. How much heat is released in this case? 13 680 calories of heat is being released in this case. Third one NH40 this NH4 NH4 plus HCl this is strong acid and this is weak base. How much heat is released in this case? 1 270 calories of heat is released in this case. Now tell me one thing. Tell me one Thing guys. Strong acid strong base. How much heat is released? This much. Make it as a

reference. This is your reference. This much amount of heat is released and strong acid. Strong base interacts. Now this is strong acid but weak sorry this is strong base weak acid. How much heat should have been released? 13 680 calories of heat should have been released. But how much is being released? 1 2 830 calories of heat is Being released. If I take the difference of two, this is 0. 8 - 3 5. This will be four. Right? Correct. This will be 16 - 8 6. So I would say 650 calories of heat has been

utilized ionize the weak acid. So what did I get? I got the enthalpy of ionization of dchloric acid. Right? How much I got it as 650 calories. Number one. Number one. Number two. Number two. This is your reference. Strong acid strong base. Now here weak Base strong acid weak base strong acid. How much heat should have been released? 13 680 should have been released because that is the case of strong acid strong base. But how much is getting released? 1 270 is getting released. If I take the difference of the two zero this is 1

1 410 calories of heat this much amount of heat will be utilized. See this much amount of heat should have been released but this much is getting released if I take the difference of the two. So this Much will be used to ionize the weak base. So I got the enthalpy of ionization of your weak base NH40 as well. How much that is? 1410 calories. Correct. What am I supposed to calculate? What am I supposed to calculate? Calculate the heat of neutralization. Calculate the heat of neutralization of your CH2 CO when it interacts with NH4.

Now this is your weak acid. This is your weak base. How much heat should have Been released? How much heat should have been released? 13680. Right? But some part of this heat would be utilized to ionize this and this. So how much is the enthalpy of ionization this and this? It is 650 + 1410. When you solve this, you'll get you'll get the enthalpy of neutralization of weak acid and weak base from here. I believe this is clear. I believe this is clear. I believe this is clear. Okay. Is it coming out to be 850?

Uh, it Is 850. Yeah. Sorry. This is 850. Okay. This is 850. Right. Am I clear people? Am I clear? Am I clear to everyone? Am I clear to everyone? So, what is the correct answer? What will be the what is this value coming out to be? Can you let me know in the chats? It'll be coming out to be I think 1 420 calories. So this much amount of Heat will be released. So enthalpy of neutral is to be minus of this. Okay. It has to be then option B. It has to be option

B. I hope these sort of questions you can easily solve from now on. Now one more type of question. One more type of question. Look at this particular question carefully. Calculate the amount of heat. Calculate the amount of heat that will be released when 300 ml of 0.2 Molar HCl is mixed with 200 ml of 0.2 molar NaOH. See guys how simple this question is. See how simple this question is. As per the question your HCl you are treating it with NaOH it will lead to the formation of it will lead to the formation of

salt and water right salt plus water correct correct strong acid strong base now my Dear students if I ask you marity of HCl is given 0.2 to you know normality is equal marity multiplied by N factor n factor of HCL is one so normality since this is one so normality of HCL is its marity right so 0.2 2 and you know normality multiplied by volume in liters that gives you the number of gram equivalents. So if I ask you initially at time t is equal to 0 how many gram equivalence of HCl do I have?

So it will Be 0.2 which is normality 0.2 which is normality of HCl multiplied by volume of HCL in liters. volume of HCl in liters which will be 300 divid by,000 correct so 0 cancel it will be 0.6 so 0.06 so I have got 0.06 g equivalence of HCl in the beginning right how many gram equivalence of NaOH in the beginning what is the normality of NaOH marity of NaOH is 0.2 its N factor is 1 so its Normality is 0.2 2 as well. So 0.2 multiplied by volume in liters. Correct? So it'll be 0.04

g equivalence. So initially at time t is equal to 0 I am geared with 0.06 g equivalence of HCl 0.04 g equivalence of LOH. Now tell me one thing over here I'm going to use the equivalent concept. I'm not going to use a mole concept. I'll be using the equivalent concept and equivalent concept says that the one whose limit The one whose gram equivalents are less the one whose gram equivalents are less present that is your limiting region. So whose gram equivalents are less? Na na is my limiting regent. Okay. And equivalent concept says whatever

since it is your limiting reagent so it would have got completely consumed. So 0 g equivalence of limiting reagent will be left. Now equivalent concept says that gram equivalence of limiting Reagent reacted is same as that of gram equivalence of excess reagent reacted. So I got to know I got to know I got to know I got to know 0.04 g equivalence of NaOH have reacted with 0.04 04 g equivalence of HCl of HCl. This is something which I got to know right. These are the gram equivalent of acid and base which have reacted. These

are the gram equivalent of acid and base Which have reacted. Now guys if you look carefully if you look carefully enthalpy of neutralation of HCl and NaOH is given as this much. What does that mean? That means when 1 g equivalent of acid is completely neutralized by 1 g equivalent of base, how much heat is released? 57 kJ of heat is released. But as per the question 0.4 g equivalence 0.04 g equivalence of acid, right? Is neutralizing 0.04 g equals of base. How much heat will be Getting released? 57 * 0.04. This will be kilogjles.

So this much heat will be released during this process. I hope you got it. I hope you got it. See enthalpy of neutralization of HCl and NaOH is this much. What does that mean? That means when 1 g equivalent of acid is completely neutralized by 1 g equivalent of base, this much heat is released. But we already know It is only 0.04 g equivalence neutralizing 0.04 g equivalent. Right? When 1 g equivalence of acid and base neutralize this much heat is released when 0.05 g when 0.04 g equalence of acid and base neutralized so multiplied

by 0.04 04. So this much amount of heat is getting released. Right? I'll give you one more similar sort of question. I'll give you one more similar sort of question. I'll give you one more similar sort of question and I'll write the reaction as well. H2SO4 as per the question is interacting with NaOH. It leads the formation of Na2SO4 and with this you get water as well. Right? Strong acid strong base and you know enthalpy of neutralization when strong acid and strong base interacts is equal - 57.2 kJ you know it already you know it

already now guys try to understand very carefully first of all What is the n factor of H2SO4 it is two right n factor of H2SO4 is two if n factor of H2SO4 is two so multiplied marity with two so it becomes 0.4 4. So normality of H2SO4 is 0.4. So normality multiplied by volume in liters. Normality multiplied by volume in liters. So divide with,000 because it was in ml, right? So this value will be how much? 1.2 divid by 10 0.12 0.12 g equivalence of H2SO4 we have got in the beginning. Normality multiplied By volume

in liters. So 0 cancel it will be 1.2 1.2. 2 divid by 10. Okay. Similarly, NaO how many how many gram equivalence of NaO do we have? Normality since NaO it's N factor is one. If it is N factor is one. So it's normality will be equal to marity multiply N factor that's one. So that will be 0.2 only. So 0.2 normality is 0.2 volume is 200 ml but make it in liters. So this 0 cancel 0.4ide by 10 0.04 G equivalence. So as per the question we got to know that initially we had 0.12

glo of acid and 0.04 gloons of base. Equivalent concept says that the one which has got lesser equivalence that is your limiting reagent. So which one is limiting reagent? This NaO which is the limiting reagent and equivalent concept says that gram equivalence if this is the limiting reagent it would have got completely Consumed. So I'll be left with zero gram equivalence. So how many gram equivalents of limiting reagent got consumed? 0.04 and equivalent concept says that equivalent concept says that equivalent concept says that whatever g equivalence of limiting reagent would have got consumed. Same gram

equivalence of excess reagent also would have got consumed. So 0.04 g equivalence of acid would also have got consumed. So gram equivalence consumed Right. So these many gram equivalent of acid and base are getting consumed. Correct? Now guys already we know when 1 g equivalent of strong acid is neutralized by 1 g equivalent of strong base how much heat is released? This much amount of heat is released. Now in this case 0.04 g equivalent of acid when getting neutralized by 0.04 04 g of base how much heat will be getting released multiplied by 0.04 04

whatever the value comes out be that is the amount of heat Which is getting released. I believe this is clear to everyone. Yes. Is it is it clear to everyone people? Quick be very quick. Be very quick. We are almost done. We are left with only few topics. There are some theoretical topics which I want to tell you now. Some theoretical topics. some theoretical topics people latice enthalpy I believe every one of you would be knowing latis enthalpy It is defined as the enthalpy change when one mole of an ionic latice is formed from its

constituent gaseous ions do you see in this case one mole of NaCCl is getting formed absolutely ly one mole of NaCl is getting formed from its constituent gaseous ions. These are the constituent gaseous ions. These are the constituent gaseous ions of NaCCl. One mole of NaCl is getting formed from its constituent gaseous ions. Now how this One mole of NaCl would have got formed when they would have attracted Na positive would have attracted with Cl negative formed NaCl. Whenever there are attractions energy is released, right? energy is released. Perfect. So when one mole of ionic

latice is getting formed from its constituent gaseous ions, the amount of heat released in this particular case is something which I call as latis enthalpy of what? Latis enthalpy of NaCCl. And in this Particular case, is the value going to be positive or negative? The value is going to be negative because energy is released in this case. Because energy is released in this case. Have a look. One mole of Mgl2 is getting formed. One mole of ionic lattice, one mo of Mgl2 is getting formed again from its constituent gaseous ions. These ions would have attracted

during attractions. Energy is released. When one mole of ionic latice is getting Formed from its constituent gaseous ions, whatever heat is released during this particular process, that is something which I'll be calling as lice energy of what? of Mgl2 which will come out to be negative in this particular case because because they're interacting they're attracting right yes perfect anyways latice energy you would have Studied in chemical bonding as well okay there you might have studied in different way you would have studied it the amount of energy needed to break one mole of ionic solid into

its constituent gaseous ions You can you can define it like that as well. No issues. Okay. So amount of energy amount of energy if I define it in this way the amount of energy Released when one mole of ionic latice is formed from its constituent gaseous ions that amount of energy released is something which I call as latis enthalpy of that ionic solid. Okay. Perfect. Perfect guys. There is something called as heat of hydrogenation. Again one theoretical topic. Again one theoretical topic. Heat of hydrogenation. Look at the Definition. The heat released The heat released during

the complete hydrogenation during the complete hydrogenation of one mole of unsaturated organic compound into its saturated organic compound. For example, you have got C2H2 C2 H2 C2 H2 this is your C2H2. So I have taken one mole of C2 H2 one mole of C2 H2. Okay, I'm doing its hydrogenation. I'm doing its hydrogenation. It is Getting converted into C2H6. It is getting converted into C26. Can I say I'm doing the hydrogenation of 1 mole of unsaturated hydrocarbon. I'm doing the hydrogenation of one mole of unsaturated hydrocarbon. Right? The amount of energy which is released the

amount of energy which is released during the hydrogenation of one mole of unsaturated organic compound into a saturated one. This is a Saturated one. Now it's a saturated right. It is saturated. It does not contain double or triple bond. The amount of heat which is released when one mole of unsaturated hydrocarbon when one mole of unsaturated organic compound gets converted into saturated organic compound during this particular reaction whatever amount of heat will be released that is something which I call as enthalpy of hydrogenation and over here it's going to be the enthalpy of Hydrogenation of

C2H2 nothing else the amount of heat released when one mole of unsaturated organic compound right under goes hydrogenation and gets converted into the saturated organic compound. That amount of heat released is something which I call as enthalpy of hydrogenation of that unsaturated organic compound. So basically in your mind there should be one thing you have to convert unsaturated into saturated due to Hydrogenation with the help of hydrogenation. Okay. And remember it has to be one mole. One mole of unsaturated organic compound should get converted into one mole of saturated organic compound. Okay. Perfect. One more

term which you must be knowing that is heat of atomization. What is Heat of atomization? What is heat of atomization? The amount of heat which is required amount of heat which is required to dissociate one mole of a gaseous to dissociate one mole substance to dissociate one mole substance into its gaseous atoms. If you look here I had one mole of Cl2. I had one mole of Cl2. Right? Perfect. I'm breaking this Cl2 into its gaseous atoms. it gets converted into two CL atoms. Right? So since I have to break this bond, so heat is

needed, energy is needed in this process. The amount of energy which is needed, the amount of energy which is needed, the amount of energy which is needed to dissociate to break one mole of substance into its gaseous atoms. That is something which you'll be calling as that is something which you which you'll be calling as enthalpy of atomization. And here I'll be calling it as the Enthalpy of atomization of Cl2. Right? Enthalpy of atomization of Cl2. Perfect. For example, let's say I have to dissociate one mole of this substance into its gaseous atoms. One mole

of this substance into its gaseous atoms. Energy is needed to dissociate one mole of H2 into its gaseous atoms. So whatever amount of energy is needed in this case that is something which I'll call as enthalpy of Atomization of H2 right isn't it simple isn't it simple people perfect so in short what is enthalpy of atomization it is the amount of heat which is needed to dissociate one mole of substance into its gaseous atoms right Since heat is needed, so the process is endothermic. Since heat is needed, required, so the process endothermic. Again, one simple

thing. Again, one simple thing. Now look At this conversion diagram. Look at this conversion diagram carefully. See, first of all, for example, this ice, a solid, right? Liquid and vapor. This is gas. Conversion of solid into liquid at its melting point. Conversion of solid into liquid. Conversion of solid into liquid at its melting point. The process overall is called as fusion. The process called as fusion. Conversion of liquid into its gas at its boiling Point. Right? The process called as vaporization. The process called as vaporization. Conversion of solid directly into its gas into gas at

its at its sublimation temperature. At its sublimation temperature, the process is called as sublimation. These are the three three things which I believe everybody of you would be knowing already, right? Solid into liquid at this melting Point is called as fusion. Liquid into gas at its boiling point, vaporization. solid into gas directly at its sublimation temperature. The overall process called as sublimation. Now here I'm going to define enthalpy of fusion, enthalpy of vaporization, enthalpy of sublimation. Now tell me how you guys are going to define it? How you guys are going to define enthalpy of

fusion? How do you define enthalpy of fusion? When one mole of solid when one mole of solid Gets converted into one mole of liquid at its melting point. When you convert solid to liquid, heat is required, right? Heat is required. So whatever amount of heat is required to convert one mole of solid into one mole of liquid at its melting point, that is something which you call as enthalpy of fusion. Right? The amount of heat which is required to convert one mole of liquid into one mole Of gas at its boiling point at its boiling

point that amount of heat required is called as enthalpy of vaporization. The amount of heat which is required to convert one mole of solid directly into one mole of gas at its sublimation temperature that is called as enthalpy of sublimation. Simple done and dusted. Right? Am I being clear? Am I clear to everyone? Am I clear to everyone? You can see it over here. The amount of heat required to convert one mole of solid into liquid at its melting point is called as one mole of solid into one mole of liquid at its melting point

is called as enthalpusion. Right? The amount of heat which is required to convert one mole of liquid into one mole of vapors, one mole of gas at its boiling point. That heat required is called as enthalpy of what? Enthalpy Of vaporization. Same is for enthalpy of sublimation. Same is for enthalpy of sublimation. Okay. Then guys there is something called as bond dissociation energy. There is something called as bond dissociation energy. Right? These are few last topics. I don't think I'll take more than half an hour to complete this chapter. Half an hour more. That's it.

Half an hour more. That's it. Okay. The amount of energy Two questions will come from this. Okay, two questions. The amount of energy required to dissociate. Be careful with the terminologies. The amount of energy required to dissociate one mole of a gaseous bond. one mole of a gaseous bond into separate gaseous atoms into separate gaseous atoms. What does that mean? For example, let's say I have got one mole of H2 gas. I have got for example one mole of H2 gas, right? See H2 means it will be like this. Okay. So can I say see

in one molecule of H2 there is one bond present there is one bond present can I say in one mole of H2 there will be one mole of these gaseous bonds present in one mole of H2 in one mole of H2 there will be one mole of these gaseous bonds present in order To break the bond energies required the amount of energy required to break one mole of gaseous bond bonds so that it gets converted into its gaseous atoms. So that it gets converted into gaseous atoms. The energy required in this particular case is something

which I call as bond dissociation energy. So this is the bond dissociation energy of H2. Bond dissociation energy of H2. Right? For example, I have got one mole of Cl2. Can I say one mole of Cl2 contains one mole of these gaseous bond? These gaseous bonds in order to break the bond. Energy is required. The energy required to break one mole of gaseous bond into its atoms into its gaseous atoms. Right? The amount of energy needed in this particular case is something which you call as bond dissociation energy of Cl2. Bond dissociation energy of Cl2.

Right. Bond decision energy of Cl2. Now guys, if you look at it very carefully, I need to basically tell you one important thing. For example, I have got one mole of CH4 gas. I have got one mole of CH4 gas. Okay? CH4. Tell me one thing. In one mole of CH4, in one mole of CH4, how many moles of CH bonds are there? Can I say in one mole of CH4? There are four moles of CH bonds. Right? In one mole of CH4, there are four moles of CH bonds. Correct? Now for example, Now for

example in one mole of CH4 we have got one mole of this bond, one mole this bond, one mole this bond, one mole this bond. For example, you are breaking one mole of this bond. What do you get? You get CH3 you get CH3 plus H, right? Perfect. The amount of energy needed in this particular case, let me call that As bond dissoci energy one. Let's say that's exclusion. Let's say that's exclusion. Now what do you have? CH3 CH3 gas. Right? In CH3 gas in one mole of CH3 gas, how many CH bonds are there?

There are three moles of CH bonds. Let's say out of those three moles of CH bonds, I'm breaking again one more mole of CH bond. It gets converted into what? It gets converted into CH2 gas plus H. The energy needed in this process is Something which I call as bond dissoci energy 2 which is for example x2 kj right now I'm left with one mole of CH2 gas in one mole of CH2 gas there are two moles of CH bonds out of those two moles of CH bonds imagine I need to break one mole of

CH bond imagine I need to break one mole of CH bond what do I get plus H CH plus H amount of energy needed in this case at bond energy 3 that is X3 that is X3 now what do I have C ch 1 mole of CH gas what does it have it Contains one mole of CH in order to break that one mole of CH so that it gets converted into C gas plus plus plus H gas let's say bond dissociient energy in this particular case is X4 kJ is X4 kJ right is X4

kJ now my dear students Look at all these reactions. How many reactions do we have? How many reactions do we have? How many reactions do we have? Quick. How many reactions do we have? Four. Right? Bond dissoci energy of the first one is x1. Second one x2. Third one x3. Fourth one, X4. Which bond was I breaking? I was breaking CH bond. Right? If I take the average of these X1 + X2 + X3 + X4 divided by 4, this is something which I'll be calling as bond energy of C. So there's a difference between

bond dissociation energy. These were bond dissociation energies of CH. This is something which you call as bond Energy of CH. Bond energy is basically the average of all these bond dissoci energies of CH. Did you get all these things? Did you get it people? Did you get it? Let me know in the charts quickly. Did you get it? I'm giving you one more case. Let me give you one more case so that you can understand carefully. For example, I have got one mole of NH3 gas. One mole Of NH3 gas contains three moles of NH

bonds. Out of those three moles of NH bonds, imagine I'm breaking one mole of NH bond. So I get NH2 gas plus H. Right? Energy needed in this particular process is bond dissoci energy one which is X1 KJ. Right? Right. People look at this particular case. Now we have got one mole of NH2 gas. 1 mole of NH2 gas contains two moles of NH bonds. Out of those two moles of NH bonds, I'm breaking one. I get one NH gas plus H Gas right energy needed for this particular process is something which I call as

X2 kilogjles. Right? Now NH gas to convert into N gas plus H gas. N gas plus H gas energy needed in this particular case is your X3 what is this X1 X2 X3 what is this X1 X2 X3 these are the respective bond dissociion energies of NH if I ask you what will be the bond energy of NH I was breaking this bond Only bond energy of NH it will be X1 + X2 + X3 / 3 average Am I clear guys? Am I clear with all these things? Am I clear with all these things?

Now guys, now at the end, one thing from which surely question comes in your exam. One thing from which question surely comes in your exam. What is that? Calculation of enthalpy of reaction From bond energy data. There's a result which needs to be derived but I won't be deriving that. I'll give you the result directly and I'll show you how to apply that result from which you'll be getting the questions. Understand? Understand? My dear students, remember enthalpy of the reaction is always equal to your bond energy of Reactants minus bond energy of products. This is

the result. How do we write it? How do we use it? That is important. That is important. My dear students, for example, for example, I'm writing the reaction. Reaction is like this. N2 gas plus P * H2 gas it gives 2 * NH3 gas it gives 2 * NH3 gas look at it very carefully look at it very carefully n2 the structure is like this n triple mon 3 * H2 the structure is like this H2 NH3 NH3 this is N H and H Perfect. N H Now my dear students, if I write enthalpy of

the reaction is equal bond energy of reactants. I'll start with the reactant. I'll start with the reactant. What about its stoometric question? One one and then energy need to break and triple bond. Perfect. This one is done. Now go towards your H2 stoometric option Is three. And this bond we have to break. energy need to break H bond. So this is your bond energy of reactants minus bond energy of products. So first I'll write two. First I'll write two. How many NH bonds I need to break? Three, right? So three times energy needed to break

NH bond. Perfect. In the question this data this data and this data it'll be given when you put it here you'll be getting the Enthalpy of this particular reaction. you'll be getting the enthalpy of this particular reaction. Right? Let me give one more example. Let me give you one more example. For example, you have got C2 H2 C2 H2 plus 2 * H2 gives C2H6. This is the reaction. For example, this is the reaction. Imagine I need to calculate enthalpy of this reaction. I need to calculate for example enthalpy of this Reaction from bond energy

data. How I'll be doing it? Bond energy of reactants minus bond energy of products. First of all know the structure. C2 H2 C2 H2 is like this. This your C2 H2. Now two times H2 H2 is like this. C2H6. This is your C2H6. This is your C2H6. Right? This is your C2H6. Now enthalpy of the above reaction is equal bond energy of reactants. Stoometric question here is one right? So nothing to do with it. So how many types of bonds do we have here? Two types of bonds. One type is CH then CC. So how

many CH bonds? Two CH bonds. Two CH bonds you have to break. And you have to break one C triple bond C bond one. One this bond you have to break. And apart from this glossometric coation here is two and there is one bond. So energy need to break this H bond. So this is your bond energy of reactance Minus. Talk about your products in your products stoometric equation here is one. Okay. Now how many types of bonds do we have? Two types of bonds. One is C single bond one is C single bond H.

Right. How many PH bonds do we have? Six. So six CH bonds we have to break. And how many CC bond you have to break? One C CC bond you have to break. Right? One CC bond you have to break. In the question all these things will be given to us. When you put it here in the Question you'll be getting the enthalpy of the reaction. Guys are you getting are you getting all these things? Are you getting all these things? Are all these things clear to you? Can you let me know once in the

chats quickly? If all the things are absolutely clear to you, then only I can show you certain questions. Look at this question. Calculate the enthalpy of formation of HCl. Calculate the enthalpy of formation of HCl. So, first of all, you have to write the formation reaction of HCl. And this is the formation reaction of HCl. One mole of HCl is getting formed from its elements which are present in their standard states. This is the formation reaction of HCl. Right? Now calculate the enthalpy of formation of HCl. Right? So since one mole of HCl is getting

formed. So enthalpy of reaction here what it will be called as you can call It as enthalpy of formation of HCl as well. Because one mole of HCl is getting formed. And I have told you already when one mole of substance gets formed from its elements present in their standard states enthalpy of reaction is called as enthalpy of formation. Right? So first of all this is half of H2. This your H2. This is half of Cl2. Right? And this 1* HCl. So enthalpy of the reaction is equal bond energy of reactance. So half of This

is HH plus half of this is Cl bond minus product equation one. So this is H CL correct you're done this is half of bond energy of HH the bond energy of gaseous H is given as 104. So this is 104 plus half of Cl2 is 58. It is 58 minus HCl. HCl is 103. Right? Solve it. Get the enthalpy of the reaction in Kilo calories per mole. Right? It's a matter of calculation now. Nothing else. It is a matter of calculation now. Nothing else. Nothing else. Can you solve this question? Calculate the heat of

hydrogenation of ethine. First of all, take ethine. This is your ethine. You are doing its hydrogenation, right? What you'll be getting? You'll be getting ethane. So first of all one mole of substance one mole of unsaturated organic compound under goes hydrogenation and gets converted into saturated compound. Enthalpy of this particular reaction is something which you call as enthalpy of hydrogenation of what? Of ethine. Correct. Basically we have to calculate enthalpy of this reaction. Now look at this particular reactant. Enthalpy of reaction is equivalent. Look at this particular reactant in this one. How many different types

of bonds we have? Two types of bonds. CH and CC. How many CH bonds? Four CH bonds, right? How many CC bonds? One CC bond. Perfect. H2 its interation is one, right? So HH we need to break here. Now comes to the products. On the product side, on the product side, what do we have? On the product side, we have got two types of bonds, right? Out of which six are your CH and one is your CC. One is your CC. Now, I think all the parameters are given. This is given to me. This is

given to me. This is given to me. This is given to me. This is given to me. Put the values and get the answer. That's it. Right. Right people, I hope I'm clear with all these things. Let's do one more question. The dissociation energy of CH4. The dissociation energy of CH4 is 360. Right? And dissoci energy of ethane is 620. Calculate the bond energy of CC. So dissoci energy of CH4 is how much? 360 kilo calories per mole. Dissotion energy of C26 ethane is how much? 620 kilo calories per mole. Right? So over here how

many bonds we have to break? Four CH bonds. So in order to break four CH bonds how much energy is needed? 360 kilo calories. So in order to break one CH bond 360 divid by 4 is 90 kilo calories. So this much heat is is needed In order to break one CH bond. Right over here C2H6. In case of C2s6, how many types of bonds do we have? C2H6. So we have got six CH bonds we need to break. And there is one CC bond we need to break. Perfect. CH bond is 90. 6 *

90 plus CC bond is equal 620. So energy need to break CC bond here will be 620 minus 94. So this is 514. This will be how much? 080 80 kilo calories per mole. Right? That's Something which I was supposed to calculate. So this is the bond energy of your CC. I hope I'm clear. I hope I'm clear. I hope I'm clear. Right. Okay guys and one last question with that we'll end the session with that your thermo chemistry part is clear and trust me guys two questions sure shot short shot short shot short shot

short Shot short shot short shot short shot short shot short shot short shot you'll get from the session short two questions should I give you this as a homework should I give you this question as a homework will you do it on your