 So we will be talking about the standard enthalpy of combustion which is denoted by this delta H C naught and there are some clues to what this means here itself. The delta H is for enthalpy, C is for combustion and the naught or the circle here is the standard part and we know that when we talk of combustion we are referring to a reaction with oxygen. So what the standard enthalpy of combustion means is that we are calculating the enthalpy per mole or per unit amount when the reactant reacts with oxygen and all the reactants and the products are in their standard states. So let's take an example to understand this better. So here we have one mole of butane which is C4H10 undergoing a combustion that is reaction with oxygen giving CO2 and water. This is a reaction with oxygen and also the reactants and the products are in their standard states and because this reaction is balanced we know that the enthalpy here will be for one mole of butane. So I can say that the standard enthalpy of combustion for this reaction is minus 2658 kilojoules per mole. Now a couple of more things to take note here. The first thing is that here we have one mole of butane that is undergoing combustion and the standard enthalpy of combustion that is given here is given for this one mole. So let's say if we multiply the entire reaction by 2 we get 2 moles of butane, 13 moles of oxygen, 8 moles of CO2 and 10 moles of water and the enthalpy also is multiplied by 2 and it becomes negative 5316 kilojoules per mole. But this enthalpy is not equal to the standard enthalpy of combustion because the standard enthalpy of combustion is defined for one mole. So that's one thing. Now the other thing to note is that this value is negative. Now when a substance undergoes a combustion reaction we know that energy is released like when we are burning something and because in combustion energy is released it is an exothermic reaction and so by our sign convention we have the enthalpy to be negative and usually this value is negative but sometimes like in the case of nitrogen it can be positive because nitrogen does not easily undergo combustion and it only does so at very high temperatures. If we have the data for standard enthalpies of combustion of different substances one very useful thing we can find out from this data is we can compare different fuels and find out which is better. So let's say I look up the values for methane and hydrogen. Now I want to know which of these is a better fuel or in other words what I want to find out is for the same amount let's say one gram of methane and one gram of hydrogen which of these will release higher energy on combustion. So for the same amount of fuel the one that releases more energy is a better fuel and from these values we know that the energy released in combustion is expressed as kilo joule per mole. So if I divide this whole thing by molar mass I am going to write it as let's say m is the molar mass. This m has units of gram per mole and so I can get rid of the mole and on dividing by molar mass I am turning these units of kilo joule per mole into kilo joule per gram and by using this number after division by molar mass I can compare the two fuels. So let's do that here and see what happens. So the molar mass of methane is 16 so I am going to divide this by 16 and for hydrogen the molar mass is 2 so I am going to divide it by 2. So if I calculate this value I get minus 55.6 kilo joule per gram and for hydrogen it is minus 142.9 kilo joule per gram and this value that we get by dividing the standard enthalpy of combustion by the molar mass is called calorific value and so by comparing the calorific value we can say that hydrogen gives more energy per gram than methane and so it is a better fuel. Apart from the calculation of calorific value there is one more way in which these standard enthalpy of combustion values are useful. Let's see what that is. Let's say we want to calculate the value of standard enthalpy of formation of methane and we know that methane is formed by this reaction. So to calculate this instead of actually calculating this value we can use our standard enthalpies of combustion and calculate the enthalpy of formation. So for each of these that is graphite, hydrogen and methane we know what will be the values of standard enthalpies of combustion so if we look at the combustion reactions of these three we can see that here carbon, hydrogen and methane undergo combustion to give products which are carbon dioxide and water. So we know that these are the combustion reactions and these are their corresponding enthalpies. So what we want to know is can we rearrange these reactions to get this reaction here? You can pause the video here and give us a try and we will continue in a bit. One way to think about how to rearrange these reactions to get this reaction is to look at the left hand side and the right hand side individually. So let's take these two reactions first. Now what if I multiply the second reaction by 2 and add it to the first one? So I will get something like C plus 2H2 here which is the left hand side of this reaction. And here on the right hand side we have this CH4. So if we reverse this reaction and add it to these two above we can get this CH4 on the right hand side. So let's try this out. Let's first multiply this reaction throughout by 2 so this will become 2H2 this half multiplied by 2 will become 1 here and there will be a 2H2O here. And now let's just add this reaction and this reaction and now that we can see that the products are the same in both of these reactions. Let's reverse this combustion reaction for methane and write it down here so we have and now if we were to add both of these reactions so we can strike off whatever occurs on both sides and what we get by addition is the same reaction for which we wanted to find the enthalpy of formation. So now because of Hess's law we know that the enthalpy change for this reaction will be equal to the summation of enthalpies of the different steps that we took. So we can write this as enthalpy of formation is equal to 2 times the standard enthalpy of combustion for hydrogen because we first multiplied by 2 here then we added it with this reaction. So we are going to add the standard enthalpy of combustion of graphite and then because we reverse this reaction and then added it to this sum we are going to subtract the standard enthalpy of combustion of methane from this. So because these three values are given to us to find the enthalpy of formation of methane all we have to do is plug these in so we have 2 times this value plus this value minus the enthalpy of combustion for methane which is this value. So if we calculate this keeping in mind the negative sign here we get the enthalpy of formation of methane to be minus 74.8 kilojoule per mole.