In this screencast I'll show you how to use the heats of combustion to calculate the enthalpy
for a reaction. So in order to do so what we are going to need is for all species to
be combustible with O2 and they have to form CO2 and H2O. Here I am assuming our species
are going to be hydrocarbons. For this particular screencasts we are going to ignore any combustion
with sulfur or nitrogen. In order to use the heats of combustion to find this heat of reaction
we take the sum of the heat s of combustion at standard state of all of the reactants
and we subtract from that the sum of the heats of combustion at standard state of the products.What
we mean by standard state is that the heats of combustion reference state is 25 degrees
C and 1 atmosphere. One thing that you might now is that in other ways to figure out the
heats of reaction it's the sum of the products minus the sum of the reactants. I will show
you here why we use the reactants minus the products and this is going to be a result
of Hess's Law. Let's start with the following reaction. Here we have liquid methanol that
will produce formaldehyde gas and hydrogen gas and what we are looking for is what is
the heat of reaction for this particular reaction. The first thing we do is we write the combustion
reactions for all three species and look up their heats of combustion. First the combustion
of liquid methanol and if we look that up in a table we find that the heat of combustion
for this is 7266 kJ/mol. So we will have in this column the heats of combustion. The next
thing we will look at is the combustion of formaldehyde. Here the heat of combustion
formaldehyde is -563.46 kJ/mol. Finally we are going to look at the combustion of hydrogen
gas and this has the heat of combustion of -285.84 kJ/mol. One of the things you might
notice in all of these combustion reactions is that they are measured as if the water
as a product is liquid not gas. Now we need to rearrange our 3 reactions so that our intermediates
cancel out and our products and reactant are on the correct sides. Let's do this starting
with reaction 1. Here we are going to write it the way it's been given because we want
the methanol to be on the left hand side or as a reactant and since we write this in the
order it's given we just write the heat of this particular reaction the same. Now let's
look at the second reactions. Here we are combusting formaldehyde however we need the
formaldehyde as a product not as a reactant so we have to reverse the reaction and by
doing that we reverse the sign of the heat of the combustion. Note instead of this being
-563.46 kJ/mol it is going to be positive 563.46. Since we are reversing the reaction
instead of getting heat out of the reaction we have to put heat in. finally let's look
at our third reaction which is the combustion of hydrogen gas. Again we need the hydrogen
gas to be the product not the reactant so we are going to have to reverse the reaction
which means we are going to have to reverse the sign of our heat of combustion. Therefore
instead of it being -285.84 kJ/mol it is going to be positive 285.84 kJ/mol. Now let's add
all of these us. First let's check that all of our intermediates cancel out. We have a
Carbon dioxide here which cancels with here. We have 2 H2O here which cancel these 2 H2O.
Finally we have 1 1/2 Oxygen gases here and 1 1/2 O2 on the right. When we have canceled
everything we get our original reaction back. Methanol liquid goes to formaldehyde gas plus
hydrogen gas. Since we have summed up the reactions we can also sum up the heats of
combustion. So the heat of reaction for methanol forming formaldehyde and hydrogen gas using
the heats of combustion add up to be 1,576 kJ/mol.