We have a reaction of the type A goes to some products and for the purpose of this problem
it doesn't matter what. The first question that we need to answer is what is the activation
energy for this reaction based on some experimental data that are shown here in the corner where
we've plotted what amounts to a straight line when we plot up the data points and we get
a particular slope and intercept for the log of the rate constant, which is 1/T where
T is in Kelvin and the rate constant is in reciprocal seconds. So to determine the activation
energy here, this is a very simple process, but we'll go ahead and review where this comes
from. For many process, although certainly not all, it can be assumed with pretty good
accuracy that the rate constant, which is purely a function of temperature, is equal to a
pre-exponential factor times the exponential of the negative of the activation energy over
RT. If we take the log of both sides here then we get that the log of k is equal to
the log of A - E/RT and that means if we plot log of k verses 1/T
we get that the slope is equal to -E/R and therefore the slope in this case is equal
to - 18 * 10^3 in units of Kelvin because 1/T is plotted in Kelvin and so therefore
we can use the fact that R is equal to approximately 1.99 * 10 ^ -3 kcal/ mol, K and we can
subsequently solve for the activation energy. I got that to be about 36 kcal/mol.
So that's the solution for part a. You can see that the Arrhenius assumption is a pretty
good assumption here. The data are quite linear. Of course these data are probably made up,
but the data are quite linear here which indicates that we do follow an Arrhenius type expression.
You're asked in part B to assume that in fact that Arrhenius expression does hold, alright and to use
it to calculate, given the value of the rate constant at 600 K, to calculate a new rate
constant at 700 K. So this also is a pretty straight forward problem. So here the question
is what is the rate constant at 700 K and to answer this we have exactly the same relationship
but we'll just use a different form as we used earlier and so I'm going to write it
in this form that the rate constant at any given temperature, say T_1 is equal to A,
the pre-exponential factor which is independent of temperature, times the exponential of - E/RT.
E is independent of temperature, R is independent of temperature, of course T depends on temperature
and that will be T_1 for k_1. We then can write an addition expression, I'll put it
up on top that k_2 is equal to the same A times the exponential of the same E and the
same R, over T_2 and now what I'll do is actually divide these two expressions by each
other and this should have already been done in your textbook depending on what textbook
you're using but I'll show how this works. The k_2/k_1 is equal to the exponential
of - E/R, I'm going to separate out these terms, times 1/T_2 divided by the exponential
of - E/R times 1/T_1. If you divide exponential terms it's the same thing as subtracting the
arguments and these arguments both have a - E/R in them so we can subtract that or we
can factor that out and so we then will get... this is often shown as exp * [E/R, (1/T_1
- 1/T_2)]. We're also factoring out a negative sign and switching the order of the arguments,
you'll often see it written this way. Now we just need to solve for our value of k_2.
E/R we've already determined is equal to 18,000 in units of kelvin. We know that T_1 is equal
to 600 K. T_2 is equal to 700 K and k_1 we have 2.15 s^-1. We plug all that in we can
solve for k_2 which is at 700 K and I get that's equal to approximately 160 s^-1., so an increase
of about 2 orders of magnitude when we increase the temperature by 100 K at that activation
energy. So now we can go back and take a look at part c of this problem. So we're going
to determine whether an existing 50 L CSTR for this reaction that follows these kinetics
will be suitable for the reaction and this is at a temperature, an isothermal temperature
of 600 K. The entering concentration of the reactant is 0.4 mol/L and the volumetric flow
rate through the reactor is 500 L/s. So we need to have conversion such that less than
5 mol/s of A is exiting the reactor in the outlet stream and we want to determine if
this is... I should have mentioned this earlier, this liquid phase reaction... to figure out
if this 50 L reaction will be sufficient. We need to show, we can't just answer yes
or no, we have to actual do some calculations here. Return here to part c of the problem.
This is a CSTR reactor in a liquid-phase reaction. So for this liquid-phase reaction we're actually
pretty dilute in A according to the problem statement. We're going to make some assumptions
such as that we're going to have constant volumetric flow rate but our balance will
be the flow rate of A coming in, our material balance will be the flow rate of A coming
in minus the flow rate of A coming out plus the net rate of generation of A times the
volume of the reactor is equal to 0. Here we have an algebraic expression because inside
this CSTR the net rate of generation of A is uniform everywhere in the reactor and so
we can write a simple algebraic balance for the entire contents of the reactor. Let's
plug in some more terms here. So we know that F_A0... so we'll keep these terms the same.
It will be useful to work in these kinds of terms, minus F_A. We can now substitute in
for the rate of reaction and that's, we'll go ahead and put the volume term first, that's
equal to a k times a concentration because if you remember we were told that we had an elementary
reaction of A going to products and so A, if it's an elementary reaction that's a simple
unimolecular decomposition as given in the problem statement, then we expect first order
kinetics and so we have k times the concentration of A and I'm going ahead and substitute this
in for variables that are given in the problem statement. That's equal to the molar flow
rate of A exiting the reactor over the volumetric flow rate which is constant through the reactor
for this liquid-phase reaction and that's equal to 0. Next I'm going to rearrange this
expression and that is, if we do the algebra, is equal to F_A0, 1 plus v k over the volumetric
flow rate. Now we'll furthermore express F_A0 in terms of C_A0. Now let's see what we know.
We know that the inlet concentration of A is 0.4 mol/L according to the problem statement.
The total volumetric flow rate is 500 L/s. It's about 200 mol/s that are entering the
reactor. Then our volume for this reactor we were told is 50 L. k in this case is 2.15
and again the volumetric flow rate is 500 L/s. Once we plug in all of that what we see
is that this reactor actually is not nearly large enough. We either need to do something
like increase the temperature up to 700 K or use a larger reactor because the outlet
flow rate that we get for this reactor that is our desired product is 165 mol/s which
is much, much greater than the 5 mol/s that we wanted leaving in the outlet stream. So
the answer to this question is no the reactor is not large enough. You can tell this, this
is sort of a quick trick in this problem, you can tell that what you're going to need
to do, this numerator is always going to equal 200 mol/s coming in. If you're going
to reduce that a lot going to the outlet you need this... combine this number v k over
v_0, which as we'll see in another context is also sometimes call the Damkohler number, to be
a lot greater than 1, so that you have a larger number that you're diving this by to get it
all the way down to 5 and we do not have a very large number here when we're just multiplying
50 times 2.15 over 500. Obviously that's a small number compared to 1 and so we don't
expect much conversion and this is what we see, that we don't get much conversion.
So to increase the Damkohler number, to increase this number we either need to increase the
volume or increase the temperature to get that rate constant up.