This problem involves a system with multiple reactions for which we need to be able to
use both material and energy balances, and in an integrated fashion.
And so we'll take a look at the multiple parts of this problem, which sort of increase in
complexity.
Alright, the first part of the problem, part A, asks that we find the volume of our reactor
system, and we notice that we're given information on how much A is converted in this reactor,
and we also have information on the kinetic rate parameters for the reaction involving
A, which is reaction 1 shown above.
So because we're given this information, we can use a material balance on A to determine
the volume of the reactor, so we'll proceed to set that up now to show how this works.
And so if we write the material balance on component A, for a CSTR we have this form,
that the number of moles of A that enter the reactor minus what comes out is equal to that
which is produced from reaction, so that's the overall form of the balance.
Next we need to plug in for the rate of reaction, the rate of, in this case, consumption of
A, and it's consumed through a reaction that's elementary as written, and so we can write
those kinetics out as follows.
We have a rate constant for that first reaction times the concentrations of each of the reactants,
and this reaction is given as being irreversible, and then we still keep the reactor volume
in there, and that's what we want to ultimately solve for.
Now we can take another simplifying step, which is replacing the molar flow rate of
component A with the overall volumetric flow rate of the stream times the concentration,
and so this allows us to solve the problem in terms of variables that are given, which
include the concentrations of A and B, rather than working with the molar flow rates.
Ultimately we're interested in solving for the volume, so we'll just rearrange our equation
as follows.
And now we can start plugging in the values for these various constants, and because this
is a liquid phase reaction system as given in the problem statement, then the volumetric
flow rate does not change as a result of the reaction, and so we treat that as a constant.
We know the outlet concentration of A is half a mole per minute, and because we feed an
equimolar amount of B, and B is consumed at the same rate as A based on the stoichiometry,
alright then we know that we have half a mole per minute of B as well, and so half a mole
per minute squared in the denominator, and this lets us solve for the volume of the reactor,
206 liters.
Now in part B, we're asked to determine the overall selectivity of component D to component
U in the product stream, and in order to determine this, the overall selectivity is defined as
the molar flow rate of D out the reactor divided by the molar flow rate of component U out
of the reactor, and so we need to be able to solve for those two molar flow rates.
What we'll do for simplicity is solve for the molar flow rate of U out of the reactor
first, and component U is the easier one to solve for because there's only one reaction
that involves component U.
That's going to make the math simpler.
It turns out that we could solve this problem first for component D, but we'll start out
for component U and see how that maybe makes our life a little bit simpler.
The CSTR mole balance on component U has a very similar form to what we wrote for component
A, it's just that the molar flow rate out of our product U is equal to the volumetric
flow rate times the concentration of U, and that's equal again to the rate at which U
is formed times the volume.
And the rate at which U is formed, we're told that it's an elementary reaction, and so we
can express that rate as follows, where the reaction is given as being elementary and
reversible, alright and so we have the forward reaction of D going to U as well as the reverse
reaction of U going to D, and so that rate is forward rate constant times the quantity
concentration of D minus 1 over the equilibrium constant times the concentration of U, and
all that multiplied by the volume.
In order to solve for this, and we ultimately want to be able to solve for the concentration
of U in this expression, we're going to need to both evaluate Kc, and we're going to need
to eliminate this concentration of D, so let's take those two items in turn, starting with
evaluating the equilibrium constant, Kc.
For an elementary reaction like this, the concentration based equilibrium constant is
equal to the exponential of the negative of the Gibbs free energy of reaction for this
reaction 2, over RT, and we are given information in the problem statement on the Gibbs free
energies of formation of the different components, and here we're interested in particular in
the difference between component D and component U, and we see that overall there's an increase
in the Gibbs free energy of reaction of +2 kilojoules per mole for this reaction that
we can plug in for that delta G. So now we can plug in that value of 2 kilojoules per
mole, and then plug in an appropriate value for the gas constant and the temperature.
That allows us to evaluate the equilibrium constant at this temperature being equal to
0.45.
Alright, so we've evaluated this term in the equation.
We know the value of the volume, we solved for that in part A.
We know the value of k2, which is given in the problem statement.
We know the value of the volumetric flow rate, and so now we just have two unknowns in our
equation, so C_U and C_D, and we need another equation that relates those two to each other,
and so that's going to be another form of the material balance, but here we'll take
a simpler step and we'll actually use the stoichiometry of the reactions in order to
evaluate what C_D should be in terms of C_U, so how do we do this?
Well, we can say that the flow rate of component D out of the reactor, we can track in terms
of the extents of the two reactions.
So for every mole of A that reacts, we'll call that the extent of reaction 1, we form
two moles of D, and for every mole of U that we form, we consume one mole of D, and so
we'll call that second reaction the extent of reaction 2.
This squiggle sub 2, is just equal to the molar flow rate of U out of the reactor, so
it's how much U we form in reaction 2, and this squiggle sub 1 is the amount of A that
we reacted, as we've already said.
So what this means is that we can write the molar flow rate of D as follows, where we've
just collected the terms for the expressions of the two extents of reaction, and if we
just divide this equation for the molar flow rate of D by the volumetric flow rate through
the reactor, which is just a constant term, then we can convert the molar flow rates to
concentrations, and so we see that we have an expression now for the concentration of
D in terms of this difference in concentrations, and since we know that C_A0 is two moles per
liter and C_A is 1/2 mole per liter from earlier in the problem, then we have a second equation
that relates the concentration of D to the concentration of U, and we just need to solve
these two equations simultaneously.
Now I'll skip the algebra for that step, and just note that we can plug in these values,
and we plug in just the C_A0 minus C_A difference in this expression for C_D that depends on
the stoichiometry, and then the mole balance that we wrote earlier for component U, so
the mole balance shown here on the top line for the solution to B, if we just plug in
those values then we get this expression, and so we see we have these two equations
and two unknowns for C_D and C_U.
We can just plug in this 2 times 1.5 minus C_U for C_D here, and solve for C_U, and what
we get when we do all that simple linear algebra is that we get a C_U equal to 0.90 moles per
liter and a C_D equal to approximately 2.1 moles per liter, and so therefore the selectivity
for forming D compared to U is just that 2.1 over 0.9, so we get an overall selectivity
of 2.3.
Alright, now here instead of doing this stoichiometric balance, where we solved for F_D in terms
of the extents of reaction, we could have written the full kinetic material balance
on component D in which we had a balance that looked very similar to what we wrote for U
up here in this top line, that would depend on the rate of reaction of D times the volume.
D is consumed in the second reaction and formed in the first reaction, and so we would have
a much more complicated algebraic situation.
The rule of thumb in dealing with these systems of multiple reactions is that you will have
to do one of those full material balances, kinetic material balances, for each reaction
that you have.
You have to do at least one of those, and so we've already done in part A one of those
balances for A, we did in part B that balance for U, and all the others, so for component
B and D we can get just from stoichiometric relations, so from extents of reaction.
So what we've done in this problem was we've chosen the two easiest components to do our
kinetic balances on, so that would be component A and component U, because they're each involved
in only one reaction, and then we did our stoichiometric relationship in terms of the
extents of reaction for component D, in order to solve parts A and B of the problem.
Next we'll move to component C.