Many energy balances involving reactions specify both the inlet and outlet conditions however,
there are cases when the outlet temperature isn't known. So for these problems we have
to use the same energy balance such that Q= delta H assuming negligible work, potential
and kinetic energy. Instead of finding Q we are given the Q and we need to find the outlet
temperature so that we can solve for the enthalpy of our products. We are going to consider
the reaction of n-heptane to toluene and hydrogen in a continuous vapor phase reactor. So we
have 10 moles per second of pure heptane and these are sent to the reactor at 400 degrees
C and 1 atmosphere. The fractional conversion of heptane is 0.7. The reaction requires 150
kW of heat to proceed and we want to know what is our outlet temperature. Again we are
going to assume negligible work, kinetic and potential energy. The first thing we start
with is our material balances. We will start with the fractional conversion such that 0.7
equals the moles in minus the moles out. That I will call n-heptane. This is the conversion
of heptane divided by the moles in. And solving for this we find that the moles out of heptane
is 3 moles per second. Next we are going to use a C balance and we know that there are
7 carbon atoms in n-heptane and then there are 10 moles per second of heptane coming
in and this has to equal the 7 carbon atoms that come in with the toluene as well as the
7 carbon atoms in the heptane that comes out. This allows us to solve for the toluene which
equals 7 moles per second. Finally we have to solve for the hydrogen coming out so we
do a hydrogen balance. coming in we have 16 hydrogen atoms per moles of heptane and there
are 10 moles of heptane. This equals the 16 hydrogen atoms of the heptane that come out
plus the 8 hydrogen atoms that are in the toluene plus there are 2 hydrogen atoms in
every mole of hydrogen gas. Now we solve for the hydrogen gas that comes out and we get
28 moles per second. Now that we have done our material balances we can put together
an enthalpy table. Note that our reference are elemental species at 1 atmosphere and
25 degrees C. We choose that reference because the heat of formation that we are going to
need to solve this problem for the species are measure there.Now we are going to look
up those heats of formations in a physical properties table. So the heat of formation
for the heptane is -187.8 kJ/mol. The heat of formation for toluene is equal to 50 kJ/mol.
THe heat of formation for the hydrogen gas is 0 because it is already in it's elemental
form and if you notice our reference is H2 gas at 25 degrees C. We are also going to
need the heat capacities. Note that I have truncated each heat capacity after the first
term in order to simplify the calculation. First thing we are going to do, let's go back
up here, is calculate that H1. Our H1 is going to equal the heat of formation of the heptane
plus the sensible heat, the integral from the reference temperature to the temperature
of the heptane coming in which is 400 degrees C of the heat capacity. So our H1 is going
to equal -96.8 kJ/mol. We do not know the final temperature so we are going to have
to write these out going enthalpies as a function of that temperature. So for our H2 is going
to equal that heat of formation plus our heat capacity of the heptane times our final temperature
minus our reference temperature. We are going to do this for all 3 of our enthalpies and
note that they are all a function of that final temperature. Also because hydrogen which
the fourth enthalpy doesn't have a heat of formation we write this only in terms of it's
heat capacity. Now let's put these enthalpies back into the table. Next we enter these values
in our energy balance. So Q which equals our delta H equals the sum of the enthalpies coming
out minus the sum of the enthalpies coming in. Here are the enthalpies of the products
coming out. So this first on here is the heptane, the second one here is the toluene and finally
here we have the hydrogen. So we sum all of these up and we subtract from i the enthalpy
in which is simply the 10 moles per second of heptane that comes in times its enthalpy
and this has to equal our Q which is 150 kW or 150 kJ/s. We then solve for out T final
and we end up with 250 degrees C. This is an example of how you would solve a problem
when you are given the heat but you don't know what the outlet temperature is.