In this screencast we are going to use the atomic species balances method to solve for
a reaction. So we have a 100 moles of ethane, and they are sent to a reactor, and they react
to form ethylene and hydrogen gas. So let's write out that reaction. So we have ethane,
and that goes to ethylene plus hydrogen gas. However some of the ethylene, our C2H4, reacts
with ethane, and it forms propylene and methane, and given the information that the fractional
conversion of ethane is 0.7, and the selectivity of ethylene to propylene is 5, and we want
to find the composition of the product gas.
Coming in as we said, we have 100 moles
of ethane, and coming out in the product gas. We have our ethane, since that was not completely
converted. We have our ethylene which is our desired product. We have hydrogen gas, we
have propylene and finally we have methane, So let's do a degree of freedom analysis on
this, and make sure that this can actually be solved. So how many unknowns do we have?
Well we have the 5 unknowns coming out in the product gas, and we have 2 atomic species.
That is the carbon and the hydrogen. We have the selectivity as well as the conversion,
and finally if we look at this second equation. We have the fact that methane and propylene
come out in the same molar ratio. So given that information we see that our degrees of
freedom are 0. So that means that this can be solved. So lets look at some of the equations
that we are going to use to solve this. So the first one we look at is the fractional
conversion. So 0.7 equals the moles fed in of ethane minus the moles of ethane that come
out of the reactor divided by the moles of ethane, and what that allows us to do is solve
for the moles of ethane, which is 30 moles. The next thing that we know is that the selectivity
of our ethylene to propylene is 5, and finally we know that our number of moles of propylene
have to equal the number of moles of ethane. Now let's do our atomic species balances, and
in atomic balances species balances. We balance the number of atoms on each side. So the first
thing that we are going to do is balance the carbon atoms first, which come in. So we have
100 moles of our ethane, and there are 2 carbons for every 1 mole of ethane, and that is all
the carbon that comes in. Let's see what comes out. Well we know that there are 30 moles
of ethane that come out, and again, there are 2 carbons for every 1 mole of ethane,
and then we add to that the number of moles of our ethylene and we see again. 2 carbons
for every mole of ethylene. In addition we have our propylene. So in every mole of propylene
we have 3 carbons per mole. We also have our methane, and in our methane we have 1 carbon
for every mole of methane. So this is what is known as a carbon balance. The next atomic
species that we will do is the hydrogen balance, and we do it the exactly same way. So we have
our 5 unknowns, and we have 1,2,3,4,5 equations. 5 equations 5 unknowns we can solve this problem.
Now since we were asked for the composition we are going to need to find the total number
of moles, and that equals 160 moles. Our composition, which is our mole fraction,
is going to be the number of moles of one of our substances divided by the total number
of moles. For example, if we want to look at the mole fraction of the ethane in the
product gas. That would be 30 divided by 160, which equals 0.18. So here is our composition
for the product stream coming out of the reactor. Hopefully this will help you in trying to
use the atomic species balances method when you are dealing with 2 reactions.