Understanding the Standard Enthalpy of Combustion: A Comprehensive Guide

so we'll be talking about the standard enthalpy of combustion which is denoted by this Delta HC naught and there are

some Clues to what this means here itself the Delta H is for enthalpy C is for combustion and the not or the circle

here is the standard part and we know that when we talk of combustion we're 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 C4 h10 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 as well

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 1 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'm turning these units of kilo Joule per mole into kilojoule 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 to the molar mass of methane is 16 so I'm going to divide this by 16 and for hydrogen the molar mass is 2 so I'm

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 kilojoule 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 see 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 a standard

enthalpies of combustion and calculate the enthalpy of formation so for each of these there 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 this a try and we'll 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 2 h 2 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 the CH4 on the right hand

side so let's try this out let's first multiply this reaction throughout by two so this will become 2 h 2 this half

multiplied by 2 will become 1 here and there will be a 2 H2O 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 an 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're 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're going to subtract the standard

enthalpy of combination 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 kilo Joule per mole