This screencast goes over how to choose and use a reference state if you have a system
where the reference state isn't already determined by the kind of data that you need to look
up. What we are going to look at is a vapor liquid equilibrium with benzene and toluene.
If we have 1 mol of benzene vapor and 1 mole of toluene vapor and it comes in at 60 degrees
C and exits in a liquid vapor form at 20 degrees C and you have 1 mole of liquid that is 40%
benzene. So you have 0.4 moles of benzene liquid. 0.6 moles of toluene liquid. In addition
you have a mole of vapor that is 60% benzene. So you have 0.6 moles of benzene vapor and
0.4 moles of toluene vapor coming out. These are all coming out at 20 degrees C. The question
is what reference state can you use. You can use any reference state you want where the
reference state is a state at which the enthalpy is 0 but there are certain ones that are more
convenient than others. In this particular problem the 3 possible reference states would
be benzene and toluene vapor at 60 degrees C, or benzene and toluene vapor at 20 degrees
or even benzene and toluene liquid at 20 degrees C. Why would any of these or all of these
be a choice for possible reference states? What it means is that in part of your problem
when you are looking for enthalpies of all of these difference substances because of
how you set the reference state you can actually have the enthalpy as a 0. Let's put together
our enthalpy chart. In this particular table we are going to set as our references benzene,
toluene, vapor at 60 degrees C. Let's fill in what we got from our material balances
and because we set our reference to vapor phases at 60 degrees C both of these enthalpies
are 0. So you can see why you may want to choose that as a reference state. Let's figure
out the rest of the enthalpies and we will start with the first one, H1, and our reference
is 60 degrees as a vapor so what we have to do is start at the reference temperature which
is 60 degrees and take the integral to our final temperature which is 20 degrees. This
is the heat capacity of the vapor dT. If we calculate that then that is -2.96. Let's put
that into our table. If we figure out H2 it is exactly the same as H1 except the C sub
P is different. When we calculate that we end up with -5.95. Our H3 ad our H4 are going
to be a little more complicated because we are going to have a phase change in there.
So remember we have for H3 we have benzene liquid at 20 degrees. The first thing we do
again is start at our reference which is 60 degrees as a vapor and now we take this to
the boiling point of benzene and then this is C sub P of the vapor dT. The reason we
take it to the boiling point is that is where the heats of vaporization of condensation
are measured. In this case we are actually condensing some of the vapor so we have to
subtract the heat of vaporization. Now it is a liquid at 80.1 degrees. We need it to
be a liquid at 20 degrees. We take this from 80 degrees to 20 degrees. Now the heat capacity
that we are going to have to use is that of the liquid because that is the state that
this is in. When we put these numbers together. This is equal to 1.5 minus 30.765, this is
all in kJ/mol, minus 7.6 which ends up being -37 kJ/mol. We can go back up and put that
into our table and for out final calculation we are going to do the exact same thing only
instead of going to the boiling point of benzene we are going to go to the boiling point of
toluene. So 60 degrees C up to 110.62. Then the C sub P of toluene vapor dT. Now we are
going to subtract the heat of vaporization of toluene. Finally, we start from 110.62
to our final state 20 degrees C, C sub p of the liquid of toluene dT. When we calculate
these numbers we get 7.5 minus 33.47 minus 8.5 which is -34.47 kJ/mol. Now let's go back
and put that in our table. Minus 34.5 kJ/mol. If you look at your table now to find our
Q all we have to do is look at the enthalpies multiplied by their moles coming out because
there are no enthalpies coming in. The reason for that is that is how we set our reference
state. Let's find our Q. Our Q is just going to be delta H which in this case is just going
to be the final enthalpies. When we multiply the right side of the table we have 0.6 moles
times -2.96 plus 0.4 moles times -5.95 plus 0.4 moles times -37 plus 0.6 moles times -34.5.
That gives us a Q of -40 kJ. Look at what a mistake you would make if you did not include
the phase change in your energy balance. That is something to be aware of any time you are
doing an energy balance with a phase change.