This screencast is going to demonstrate how to do an energy balance on a reaction using
the heat of formation method. There are 2 methods that you can do an energy balance
on a reaction. Previous screencast showed how to do the heat of reaction method. This
is the heat of formation method and if you compare the two you will see that the answer
is exactly the same. We are going to use this reaction all in gas. CO plus H2O goes to CO2
plus H2. The way to approach this method is first to do the material balances and that's
should be your first step whenever doing an energy balance. Then we are going to put together
an enthalpy table and it is going to include the references and since we are using the
heat of formation method the reference have to be the elemental states of whatever make
up your molecules. That would be like carbon solid, O2 gas, H2 gas etc.. The reason for
that is what the method does for the enthalpies is it starts with the elemental species, forms
the molecules which is where we put in the heat of formation and then takes them from
a reference temperature to a final temperature. Finally, for each enthalpy the way we calculate
it is the heat of formation of that particular molecule and then the integral from the reference
temperature which is 25 degrees C, and this is always 1 atmosphere, to the temperature
that you are looking for, the Cp or the heat of the capacity dT. If you are doing some
kind of reaction that uses ideal gases you can often look this up in a table and that
means that you don't have to do as many calculations. We are going to have 1 mole of CO coming in.
It is going to have 50% excess H2O which means that there is going to be 1.5 moles of H2O.
We are going to have an 80% conversion of the CO which means coming out we have 0.2
moles the CO2 and the H2 coming out are going to be 0.8 moles and the H2O coming out is
going to be 0.7. The next thing we want to do is to draw our enthalpy table. One thing
to make sure that you do is that you put your references on and as I said it's carbon in
its solid form, H2 in its gas, O2 in its gas at 25 degrees C and 1 atmosphere. Your n_in
is usually in moles per hour or second and your H_in and H_out as well are in kJ/mol.
What do we have? We have CO, we have H2O, we have CO2, we have H2, we have none of this
coming in and we have 1 and 1.5. Coming out 0.2, 0.7, 0.8, and 0.8. Our next job is to
figure out the enthalpies that go in the table. Let's start with our first one for carbon
monoxide. Our H is going to equal the heat of formation for carbon monoxide plus the
sensible heat it takes to go from 25 degrees which is the reference temperature to your
final temperature. In this case our reactants come in at 300 degrees and our products come
out at 500 degrees. Since our carbon monoxide comes in at 300 degrees right now when we
put the heat of formation our enthalpy is at 25 degrees C. In order to raise it to 300
degrees C we have to add some sensible heat and in this case we go to an enthalpy table
of ideal gases whose reference are also 25 degrees C. We look up at 300 and we find that
that is 8.17. We do the same thing for H2O. The heat of formation is -241.83 and then
using the same table we look at what it takes to bring it to 300 degrees. Now the other
4 enthalpies are products so they are at 500 degrees. Again our heat of formation for carbon
monoxide and this time we add 14.38 because it is at 500 degrees. our H4 again is -241.83
and we add 17.01. Our H5 is -393.5, that is the heat of formation of CO2 and then we look
it up and find that to take it to 500 degrees is 21.35 kJ/mol. Finally, here H2 has 0 heat
of formation because it is already in it's elemental state so all we have to add to that
is the sensible heat. When we put these all together let's put them back into our enthalpy
table. So this is -102.36, this is -232.36, this is -96.15, -224.82, -373.16, and finally
13.82. Since we have included the heat of formation in our enthalpies now all we have
to do to find our Q is the sum of the enthalpies times their number of moles out minus the
sum of the moles times their enthalpies in. What does that look like? That is everything
that came from the right side of the table which are the products so now we subtract
from that and that is going to equal -12.6 kJ.