 Let's look at how the enthalpy of atomization is defined and there's a clue in the name here. It's got something to do with breaking bonds to form atoms. So the enthalpy of atomization is defined as the enthalpy change in breaking the bonds of some amount of substance to give one mole of atoms in the gaseous state. So if we look at this example, here we have a hydrogen hydrogen single bond which is broken to give one mole of hydrogen atoms in the gaseous state and the enthalpy for this reaction is the enthalpy of atomization which is this 217.5 kJ per mole and we know that to break this bond we need to provide energy. So this is an endothermic reaction which is why the enthalpy is positive. Now in some older textbooks and data sources, the enthalpy of atomization is defined as the enthalpy change when the bonds of one mole of a substance are broken to form atoms. So here we are forming one mole of atoms and here we are breaking one mole of a substance or another way to see that is if we multiply this reaction by two, we get this reaction and so the enthalpy in this case is also double this value. Now this is usually just a difference in the way the data is reported. So the data source that I'm using defines it in this manner. So for this video, everywhere that I've written down the enthalpy of atomization, it is for one mole of substance that is being broken down and this difference can become important in numericals. So you need to be careful about this. Now let's take one more example. So here we have methane in gaseous state which gives one atom of carbon and four atoms of hydrogen after these CH bonds are broken. And for this reaction, the enthalpy of atomization is 1665 kJ per mole. If we compare both of these, we can make some observations about bond energies. So in this case, we have a single bond between two hydrogen atoms which is being broken down. So the enthalpy of atomization will be equal to the bond enthalpy of the hydrogen-hydrogen single bond. And this is true only when both of these are in the gaseous state. Like if one of these was a liquid, this wouldn't be true. And in the same manner, if we look at methane, what we're doing here is we are breaking four CH bonds. So we know that the enthalpy of atomization is the energy required to break all these bonds. And we know that here there are four CH bonds. So can we just divide this by four to get the bond enthalpy of one CH bond? To answer that, let's look at this reaction in a series of steps. So let's say I start with methane and at each step, I'm breaking only one bond. So here I've broken one hydrogen bond. Then successively, I've broken one more and one more. And finally we have broken all four CH bonds. So what I've done is I've taken this reaction and I've written it down as a sequence of four separate reactions in which in each of them I'm breaking one CH bond. Now if we look at the enthalpies corresponding to these reactions, we find that these are not the same, which means that for every step the energy required to break the CH bond is different. And why this happens is that after we break the first CH bond, the remaining hydrogens may rearrange around the carbon, making it more difficult or more easier to remove the next hydrogen atom. So if we just take the enthalpy of atomization and divide it by four, what we get is the average bond energy of the CH bond, which is 416.25 kilojoule per mole. So the point to note here is that in case of diatomic molecules the bond enthalpy and the enthalpy of atomization are the same. But for polyatomic molecules, like in the case of methane, the energy required to break the CH bond at each step is different. So although here we are breaking four bonds, when we divide the enthalpy of atomization by four, what we're getting is the average bond enthalpy of a CH bond. And also a quick check here, after breaking this reaction into different steps, by Hess's law we know that if we add these four reactions, we can strike out everything in common and we get back this reaction. And correspondingly when we add all these four values of enthalpies, we get a total enthalpy of atomization, which is the 1665 kilojoules per mole. Now let's go through an exercise problem connecting all these ideas. So in this problem we are given this reaction where CCl4 in gaseous state is broken down into one carbon atom and four chlorine atoms. And we are asked to calculate the enthalpy of this reaction and the bond enthalpy of the carbon-chlorine bond. And to calculate these, the values are given here. First is the enthalpy of formation of CCl4, then we have the enthalpy of atomization of carbon, enthalpy of atomization of chlorine, and enthalpy of vaporization of CCl4. And here we are going to assume that the enthalpies of atomization that are given for one mole of substance of which the bonds are broken. And so maybe you can pause the video here and give this a try before we continue. So the first thing to note is that in this reaction, you can see that all three of these are in the gaseous state. And because we have these enthalpies of atomization and this enthalpy of vaporization, if we write down the reactions for which these enthalpies are calculated, we can look at how we can get this final reaction. First, based on this information that is given to us, we write down all their corresponding reactions. So we have carbon going from solid to gaseous state. And for this, the enthalpy of atomization is given. Then we have chlorine, which gives two chlorine atoms. And the enthalpy of atomization for this is also given. Then we have the enthalpy of formation of CCl4, which can be written like this. And finally, the enthalpy of vaporization of CCl4, which is written like this. Now, what we want to do is we want to look at these four reactions for which we know the enthalpies. And we want to see how we can rearrange them and modify them to get this reaction for which the enthalpy of the reaction and the CCl bond enthalpies are. Looking at these two reactions, the first thing maybe to try is to multiply this reaction by 2 and add it to this reaction above. So we have carbon plus 2Cl2 giving carbon and gaseous form plus four chlorine atoms. So now we can see that the left hand side of this reaction matches the left hand side of this reaction. So if we were to reverse this reaction and write it here, so now if we add both of these reactions, the carbon and the chlorine from both sides will go off. But what we get here is the CCl4 solid and the problem has this CCl4 in gaseous state. So to solve this, we can use this last reaction and we can reverse it and write it here. So now if we were to add all three of these, we can take off the solid CCl4 from both sides and we can write the final reaction like this, which is the same as asked in the question. So now to calculate the enthalpy of this reaction, we can use Hess's law and we can say that the enthalpy for this reaction will be equal to the summation of the enthalpies of these multiple steps and so we can write this as our enthalpy of the reaction will be equal to twice the enthalpy of atomization of Cl2 because we multiplied by 2 here first, then we added it to this reaction. So we add the enthalpy of atomization of carbon, then we reversed these two reactions and added them. So from here, we subtract the enthalpy of formation and the enthalpy of vaporization and if we plug in all these values that are given in the question, we get the enthalpy of this reaction to be 1304 kJ per mole. Now the second part of the question asks for the bond enthalpy of the carbon-chlorine bond and if we look at this reaction, we can see that here there are four carbon-chlorine bonds which are broken to get this one atom of carbon and four atoms of chlorine. So just like we saw before, we can write the bond enthalpy of the carbon-chlorine bond as the enthalpy of this reaction divided by the number of bonds broken, which is four in this case. So if we use the value of enthalpy from here and we divide it by four, we get the bond enthalpy of the carbon-chlorine bond to be 326 kJ per mole.