This problem deals with the use of stoichiometry and solving for how the concentration changes
as a function of the number of moles in a gas phase reaction, so this is key, in which
we have a loss in the total number of moles occurring with the reaction. So we have 3
moles on the left hand side going to 1 mole on the right hand side and it turns out as
we'll see that this can affect our volumetric flow rate through the reactor and thus our
concentration. So we want to be able to account for that and we'll follow
the procedure that is outlined by Fogler and Fogler has some particular notation that we'll
use, but of course the general philosophy behind how to solve this problem is important
regardless of the text, it's just the variables are called particular things by Fogler.
So I've written down some of the given information here. So we're told that the conversion
is 75%, we're given temperature and pressure, and we're told that the temperature doesn't
change in this problem because it's isothermal and we'll also assume that the process pressure
doesn't change with the reaction. What we want to solve is the outlet concentration.
I've assumed a 10 mole basis for the feed and based on the given compositions of 60%
and 40% we end up with 6 moles of B and 4 moles of A in the feed.
The first thing we need to do is we're told that the conversion is 75% for our limiting reactant,
so the question is, what is the limiting reactant? Well we can actually see just by a quick thought
experiment that it has to be A, that if we consume all 4 moles of A, so we consume 4
moles of A, we're at this 2 to 1 stoichiometry and so only 2 moles of B will be consumed
and we'll therefore have 4 left over. So that means we'll have B left over when we've consumed
all of the A and therefore A is the limiting reactant. So we want to use that to find
out the concentration of A. The concentration of A is just equal to the molar flow rate
of A divided by the volumetric flow rate of the total stream. We can express the molar
flow rate in terms of our given variable, the conversion. So F_A is just equal to F_A0
times 1 minus x based on the definition of conversion, and we're going to divide that
by the volumetric flow rate which we don't know. The volumetric flow rate is equal to
the total molar flow rate times RT over P by the ideal gas law, since we're told that
ideal gas conditions apply. And then we can write, we'll do a little trick here to simplify
the algebra and we'll also write that the volumetric flow rate of the inlet stream can
be expressed based on the inlet properties and the reason we'll do this is we'll take
a ratio and that will end up making our life a little bit easier. So we'll divide the first
expression by the second one and then multiply though by v_0. So then we get that v is equal
to the inlet volumetric flow rate and so we'll be able to see how the volumetric flow rate
at any given conversion is a function of the volumetric flow rate at the outlet. If
we take that ratio we get F_T/F_T0 times T/T_0 times P_0/P. As
I've mentioned, in future sections of the text, we'll learn how to deal with pressure
drop, for example in a packed bed reactor, we'll have an expression for P over P_0 but
for now we're going to assume that goes to 1, that there's no change. And we'll end up
developing expressions for T as a function of T_0 when we're doing energy balances, but
for now we're going to assume that goes to 1 and so we only have to deal, for today,
with this F_T over F_T0 ratio. How are we going to end up doing that? We have in this
case, this reaction of 2 A plus B goes to C and this is a decrease in number of moles
from 3 moles written here to 1 mole on this side and I've written out a couple of facts
here so that for every 2 moles of A that react on this side of the expression, we lose 2
moles of total stuff in our reactor, because we only have A, B and C in our reactor.
So we lose 2 moles of stuff for every mole of A that reacts, therefore we lose 1
total mole of stuff in our reactor. Therefore F_T, if we're keeping track of
this total number of moles that we need to know, that's equal to the number of moles that
we start with times the change in the total molar flow rate as a function of the A that
reacts, as a function of the fact that we have a reaction in this case. So we can track
the reaction in terms of the number of moles of A that react and that's equal to F_A0 times
X and then what we need to multiply that by because we're interested in the total number
of moles, a change in the total number of moles and this is just the change in A, is
the ratio of, in the denominator, the moles of A that react to the moles in the numerator,
the total moles, the total change in the number of moles as a result of that reaction of A
and so for every 2 moles of A that react, we lose 2 total moles of stuff in this reaction.
We have a negative here because we're losing moles and we expect the total molar
flow rate to decrease from the initial value of F_T0 as the reaction proceeds.
So we have this negative sign and it ends up being a - 1. So if we now take the
step of dividing this expression through by F_T0 because we see
that it's going to be useful to have this ratio here. Then we can
get this expression that F_T over F_T0 is equal to 1 plus F_A0 over F_T0 which is just
the mole fraction of A in the feed, Y_A0, times the conversion X times this variable
that Fogler calls delta A, which is the change in the number of moles in the reaction expression,
divided by the stoichiometric coefficient of the component that you're tracking, which
for us is A. So now we can express, we can plug in numbers for all of this. We
actually have all the numbers and so I've just rewritten it here in terms of Y_A0 delta
A times X, so collective terms, plugged in our numbers of 40% for Y_A0 based on a 10
mole basis we have 4 moles of A so 40%, this negative 1 for our delta A value and we're
told in our problem statement that we have a conversion of 75% and so when you plug in
all those numbers and do the simple algebra we see that the volumetric flow rate is equal
to 0.7 times the initial volumetric flow rate so we get a 30% drop in the volumetric flow
rate because we're losing moles in this reaction at a constant temperature and pressure. Now
we have the expression for v in terms of v_0 that we needed right here,
so we can plug in right here for v in terms of v_0, and so that will just give us
that what we need is F_A0 over v_0 times 0.7, what we just solved for, times one minus x and we
know the value of x. Next what we're going to solve for, this value is equal to
C_A0. I'm just going to go ahead and solve for the volumetric flow rate and show how
this can work out and so if we do that on another page that I've already written up.
We're again using the ideal gas law, so the volumetric flow rate is equal to this
RT_0, F_T0 over P_0. So R, the gas constant, times the temperature times the total molar
flow rate over the pressure. I can just use this value of the gas constant in terms of
atmospheres, liters, moles and Kelvin. Temperature and pressure are given in the problem statement,
I'm just going to let the F_T0 float for a minute. Here's our expression for C_A now,
as just recopied off the other page, where X is equal to 0.75. We now have an
expression for v_0 that's equal to all of this stuff, so we just plug it all in so that
we can do the necessary algebra. When we do that we get this term F_A0 over F_T0 which
again is Y_A0, which is the inlet mole fraction of A that's 40%, 0.4. Now we have everything
that we need in terms of the constants we've already filtered in at a conversion of 75%
and when I plug in all those numbers I get an outlet concentration of 0.028 mol/L
So the algebra of this problem is fairly straight forward. But again what
I hope that this symbolizes is what these Y_A0 and delta terms really come from and
what you're really doing here, is you're keeping track of how the total molar flow rate changes,
how the total molar flow rate changes as a result of the reaction of the limiting reactant
A. So when 2 moles of A react, we lose 2 moles of total stuff and so that
gives us this negative 1 and we're ratioing this and so that's where the delta A comes
from and we're determining this based on the number of moles of A that we react but we're
interested in the total moles and that's where the Y_A0 comes from. You really don't have
to use this kind of notation and so as a short cut and in a lot of problems we'll use this
Y_A0 and delta A notation as Fogler does, but you can see that you can arrive at it pretty
quickly from a simple mole balance following the stoichiometry of the reaction rate expression.