Okay in this problem, we're dealing with an elementary reversible reaction of the form
A goes to B in a liquid phase CSTR, and that CSTR is adiabatic.
In part A we want to use the given information about the feed and the outlet streams in order
to determine the heat of reaction.
We're given that the conversion is 50%, the outlet temperature is 325 Kelvin, and the
inlet temperature is 300 Kelvin, and because there is a rise in temperature for this adiabatic
reaction, we expect that the reaction is exothermic.
So if we do a little sketch of our reactor here, we have the outlet at 325 K, the inlet
at 300 K, and a conversion for part A of 50%.
We can write an energy balance for an adiabatic reactor in terms of the following form.
Here we have the sum of Theta i CPi over the negative heat of reaction times the temperature
difference between the inlet and outlet stream.
Heat of reaction of course is what we want to solve for.
We have a 25 K temperature difference between the inlet and outlet streams, and then that
leaves us with this sum of theta i CPi term, which is basically the total heat capacity
of the inlet stream, so our stream consists of both our reactant A and an unreactive solvent,
and we have to include the heat capacities of both of those terms, corresponding to their
relative amounts of A and the solvent.
So we can express that as follows.
We have 85, which is the heat capacity for component A, and then we have 3 times, or
45 divided by 15, based on the concentrations of the solvent compared to component A times
that heat capacity for the solvent, again because we have 3 times as much solvent, and
that gives us a total heat capacity of the stream of 235 joules per mole Kelvin.
We're able to treat this as a constant based on the information given in the problem statement,
and furthermore we're able to treat the heat of reaction as a constant given the fact that
the heat capacities of components A and B are equal to each other, and so therefore
the heat of reaction is invariant with temperature.
So we can solve for it as our only unknown and get about -12 kilojoules per mole, which
makes sense because we expect an exothermic reaction here where the temperature is rising
in the adiabatic reactor.
So that's part A, and part B, we want to determine what the equilibrium conversion would be at
325 Kelvin.
Now, we recognize that we've actually achieved a conversion of 50%, so the equilibrium conversion
should be what we would get in an infinitely sized reactor, which should be something greater
than 50%.
To determine this equilibrium conversion, we need the equilibrium constant at the temperature
of 325 Kelvin.
We're given the value of 12 at 300 Kelvin, and so we need to determine what the value
would be at the higher temperature.
We can use the expression that's shown here to determine that, where basically the equilibrium
constant at the new temperature is given by the one that we know, which is equal to 12
at 300 Kelvin, times this exponential of delta H of reaction over R. Now we multiply that
by a reciprocal temperature difference according to the Van't Hoff relationship.
For the heat of reaction we plug in the value we just calculated, for R we plug in the value
making sure we keep track of joules versus kilojoules per mole to keep the units straight,
and then we can solve for the equilibrium constant at the higher temperature, and we
find that it is indeed lower as we expect for an exothermic reaction.
So it's gone down from 12, and it is equal to 8.35 now.
Now that we've determined the value of the equilibrium constant, we can use that to determine
the value of the equilibrium conversion using a very straightforward equilibrium relationship,
that the equilibrium constant is equal to the ratio of the product to the reactant at
equilibrium, and furthermore that the concentration of the product is equal to the initial concentration
of the reactant times the conversion.
That will be divided by the initial concentration of A times 1 minus X for the reactant concentration.
The C_A0's can of course cancel from the numerator and denominator, and we set that equilibrium
conversion, or that expression with equilibrium conversion equal to 8.35 at 325 Kelvin, and
we can solve for X to be equal to about 0.89.
And now here comes the hardest part of the problem, which is part C, so we want to determine
the volumetric flow rate at which conversion will be 65%, and recall from problem A that
in the initial configuration with the volumetric flow rate of 10 liters per minute that the
conversion is 50%.
So now there's some new volumetric flow rate at which the conversion will be 65% instead
of 50%.
Okay, we have a little sketch of our process right here, for the new process for part C,
that again we have an entering fluid at 300 Kelvin, we achieve a conversion of 65%, we
expect that there will now be some new outlet temperature corresponding with this higher
conversion, which has released more heat from the reaction, and so we should have a higher
temperature, but to determine the relative volumetric flow rates, which are related to
the residence time or space time of the reaction, we'll need to actually do a material balance
on both the case where we have 50% conversion and the case where we have 65% conversion
in order to determine how the volumetric flow rate would change between these cases.
So here's one form of the material balance that's in terms of the space time, tau.
This just says that tau over the inlet concentration is equal to the conversion over the reaction
rate inside the reactor.
The next step, once we've specified this form of the material balance, is to write down
the rate of reaction, and that rate for an elementary, reversible reaction of A going
to B is shown here.
Now we just want to point out that it has two temperature dependent terms, which is
the reaction rate constant, lowercase k, and the equilibrium constant, uppercase K.
So we'll need to take that into account later in this problem.
Alright, the next step in the problem solution is to relate the concentration to conversion,
and so we can do that in just the same way we did above, that the concentration of A
is equal to C_A0 times 1 minus X, concentration of B is equal to C_A0 times X, and so we get
this expression when we collect terms.
And again, that lowercase k and capital K are functions of temperature, and we can cancel
that C_A0.
I'll give this an asterisk, because it will be a key equation in our solution.
First, we've see that there are a couple of temperature dependent terms, and so it would
be good to know for this new case where we have 65% conversion, what is the temperature
in the reactor, and fortunately we can use the same energy balance approach that we developed
before.
Here is that form of the energy balance just as it was above.
In this case we know that the conversion is 65%.
We have the same feed stream in terms of its ability to carry heat, so the same amount
of solvent relative to the reactant A, and so our theta i CPi sum is still the same.
The heat of reaction is as we solved for it above, and the inlet temperature is still
300 Kelvin, so that our only unknown is going to be the temperature T inside the reactor.
We can solve for that T, and it ends up being roughly 333 Kelvin.
So we can plug that into our little diagram above.
Now, as we've mentioned, one variable that the temperature will affect is the equilibrium
constant that's in our material balance, and so we can solve for that exactly as we did
above, that the K at 333 Kelvin will be Equal to 7.57, where the only number that we really
change is that temperature of 333 Kelvin in the reciprocal temperature difference.
Now the rate constant, we're not given information on the pre-exponential factor, so we'll just
leave that as an A for now and see if it cancels later, but we are given information about
the activation energy, that it's equal to 50 kilojoules per mole, and we'll divide by
RT, where we'll use the same R value that we used before of 8.314 times 10 to the minus
3 kilojoules per mole, and we can plug in whatever temperature we're interested in for
that reactor situation, so we have something that's a function of temperature here.
So we'll go ahead and insert that function for k into our material balance, which is
the starred equation.
Then we get this simple expression from doing that substitution, and now what we'll do is
we'll evaluate tau at the conversion of 50%, where we want to go to a temperature of 325
Kelvin, and also at a conversion of 65%, where we know we go to a temperature of 333 Kelvin.
So at 325 Kelvin, we can plug in the conversion of 0.5, we again leave A in the expression,
we can plug in values of E and R, of 50 and 8.314 times 10 to the minus 3 as indicated
above.
We can also plug in the 0.5 for the other conversion term and the 8.35 for the equilibrium
constant.
We also can write this same expression at the higher conversion and higher temperature,
and all that changes is that temperature and the K values and the conversion values, or
the capital K values, excuse me, goes form 8.35 to 7.57.
And we adjust the temperature accordingly, and if we simply take a ratio of these two
equations, so that we get the tau at 325 K over the tau at 333 K, then we actually have
numbers for everything except A, and A cancels in this case when we take this ratio.
So we can plug in E and R and just take this ratio, and what we find is that it's equal
to roughly 0.70.
Now that ratio of space times is actually directly related to the ratio of volumetric
flow rates at the two conditions.
It's actually reciprocally related.
We expect that it's going to take a longer space time, or we see that it will take a
longer space time at the higher conversion and higher temperature, so that means that
the volumetric flow rate has to be lower, and if we're starting with a volumetric flow
rate of 10 liters per minute at 325 K, we can multiply that by 0.7, and we'll need 7
liters per minute to achieve the higher conversion, so it ends up that we need a longer space
time, and therefore a lower flow rate in order to achieve that lower conversion.