In this screencast we are going to cover the three methods we use for approaching a reactive
material balance. The three main methods we will use when we have a reaction and are doing
a material balance are the molecular species balance, an atomic species balance, and lastly
the extent of reaction method. I will go through each method using the same reaction set up
to show you how to use them and provide some insight on which method is appropriate for
a given reactor system. We will have the reaction A + 2B goes to C. We will feed 10 moles of
A and 30 moles of B into our reactor. Once piece of information that we are going to
need is that the fractional conversion of A is 0.9. Given this information let's go
through each method and we will start with the molecular species balance. I will say
up front that this may be the least used method because the molecular species balance is kind
of cumbersome to use on multiple reactions. We start with the general material energy
balance where we have input minus output plus generation minus consumption is going to equal
accumulation. That is the case for all material and energy problems. For a steady state process
we get rid of the accumulation terms. For this approach we are left with the following
and I have written the output as the variable of interest since that is typically what we
are going to look at. The first thing we could do is use the fractional conversion of A as
0.9 to determine the moles of A leaving. Recall that the fractional conversion is the moles
of A that enter minus the moles of A leaving all over the moles of A entering. With the
information that we are given it looks like the following. We have 10 moles of A entering
so all we have to do is calculate the moles of A leaving and this gives us nA_out as 1
mole. This means our consumption term A is 9 moles. If we now look at B we can write
the species balance as nB_in nothing is generated but we do have a consumption of B equals nB_out.
We have to relate this consumption of B to our consumption of A. For the consumption term
we see from the stoichiometry of the reaction that two moles of B are consumed per mole
of A consumed. Since we had 9 moles of A consumed we can plug this into our equation. This will
give us the outlet rate of B. We solve for nB_out and we get 12 moles. Lastly we Can
do this for C. We have no input and we have no consumption so all we have is generation
is equal to our output. What we need to determine how many moles are generated so again we can
use the stoichiometry from our reaction we see that there is a ratio of 1 mole of C that
is generated per 1 mole of A consumed. We can calculate the moles of C out. We get an
output of C out of 9 moles. The take home message using molecular species balances is
that we need to use the stoichiometric ratios for the consumption and generation terms. To
use this method we need to be able to determine a consumption and generation term for each reaction involved.
We got the consumption of A given our fractional conversion of A. That is the molecular species
balance. Let's take a look at an atomic species balance. If we look at the general material
balance for our steady state reaction in terms of atomic species we know that we cannot consume
or generate atoms. This just becomes in = out. We do not really care what happens in the
reactor. We just care what we are putting into it and what comes out of it. Let's look
at the same reactor set up that we had before. We will write the balance on atomic species.
Since we used A + 2B goes to C we do not really know what the atoms that involved are so let's
make up a new reaction for this example. Let's say we are looking at the reaction C2H2 plus
2H2 goes to C2H6. It has that form A + 2B goes to C. We will use the same molar flow
rates that we had before. Now we want to write balances on carbon and hydrogen. Let's start
with a carbon balance. Here we are just writing a balance on the amount of carbon that goes
in and the amount of carbon that comes out. We know that we have 10 moles of our C2H2
times 2 moles of carbon per mole of C2H2. That is for our C2H2 term and we are just
going to do the same process for every molecule involved in our reaction. We have no carbon
in the second reactant, hydrogen, so this is just going to be equal to what comes out.
What comes out is not necessarily just C2H6. We may have some residual C2H2 so we have
to do the same thing. nA_out times the ratio of 2/1 plus n_C. I am going to use A and C
to keep our molecules easy to write in our equations. You can see that we have 2 carbon
per C2H6 or molecule C. This will also have the ratio 2/1. Let me simplify this equation
for our carbon balance. You see we have 2 unknowns. The amount of A that leaves the
reactor and the amount of C that leaves the reactor. Now we can do a hydrogen balance.
All we care about is what goes in and what comes out. We have 10 moles of C2H2 at a ratio
of 2 hydrogen per mole of C2H2. This is plus 30 moles of H2 that enters times our ratio
of 2 H per 1 H2. I do not care about the stoichiometry. I just care about the amount of atoms that
come in and out of our reactor. This is going to be equal to what leaves. Here we have some
unknown flow rate nA_out times 2/1 plus some unknown flow rate nB_out times 2/1 plus some
unknown flow rate nC_out times 6/1. Simplifying this equation looks like the following. You
can see from our two equations that we have 3 unknowns so we need another piece of information.
Before we used the fractional conversion of A. That still stands here. This told us that
the amount of moles of A leaving was 1 mole so we can plug that into our equations for
nA_out and solve for our other 2 variables. I get the molar flow rate nC_out is 9 moles
and this allows us to solve for nB_out which is 12 moles. Did we get the same answer that
we did before? We get 1 mole of A out, 9 moles of C out and 12 moles of B out. We do show
that our flow rates with the atomic species balance is the same as the molecular species
balance as we would hope. This method is one of the most straight forward methods since
it is really just in = out. It can be written for each independent atomic species. It is useful when
there are multiple reactions so this is one of the more common methods used. Let's look
at the last method. The extent of reaction method. This method writes expressions for
product species flow rates in terms of an extent of reaction and substitutes this information
with a known feed or product flow rate to determine the remaining unknown flow rates.
The extent of reaction is written as a symbol xi in some text books so I will use it here.
I think of the extent of reaction as how many reactions take place on a mole basis. Let's
say that we have A goes to B and 1 mole of A is reacted. Then the extent of reaction
is 1 mole of 1 mole per time period if it is a continuous process and we have a flow
rate. 1 mole of A reacts on a mole basis and our extent is equal to 1. It is important
that we realize that the extent of reaction has units and it is either moles or moles
per time. It is therefore important to use the appropriate stoichiometry for the species
that we are writing a balance for. Let's again take a look at the reactor set up that we
have and if we write a balance on A. We have the amount of A in minus the amount of A consumed
has to be equal to the amount of A out. This consumption term is important here. This is
where we replace our extent of reaction into the equation. This consumption term is just
the stoichiometric coefficient which in this case is 1 times the extent of reaction. We
get 10 moles of A in minus the extent of reaction has to be equal to the moles out. Again we
said that because we have the fractional conversion we know this to be 1. This means that our
extent of this reaction is 9 moles. For B we know we have 30 moles of B in. We are consuming
B as a stoichiometric coefficient of 2. We subtract 2 times the extent of reaction. This has
to be equal to what is coming out. We know the extent of reaction. We have 30 minus 18
equals nB_out which as we expect is 12 moles of B. We can do the same thing for C. This
time we do not have any C coming in and we do not consume any C so it is just our generation.
Just like our molecular species balance we relate the generation to the consumption of
the species using the extent of reaction. We have 1 mole of C per 1 mole of the reaction.
Basically we have nC_out is equal to the extent of reaction which is 9 moles. The extent of
reaction method is useful for multiple reactions. Let's say we have A plus 2B goes to C and
B plus C goes to D. We would just have an extent of reaction for the first equation
and an extent of reaction for the second equation and then write our balances using those two
extents of reactions. Hopefully this gives you an idea of each method and the differences
between the atomic species, the molecular species, and the extent of reaction method.