OK, so in this problem, we're dealing with actually a pretty simple application of solving
the CSTR equation. But maybe you have a slight twist on it is that it's for Michaelis-Menten
kinetics, so enzyme catalyzed kinetics. And again, I'll let you read the problem statement,
but you're given some Michaelis-Menten parameters here. So a maximum velocity and a Michaelis
constant. And in this problem, you need to use these in order to solve for the space
time, in part A, that's required to get an 80 percent conversion of the substrate under
conditions where it enters at 1 mole per liter. Alright, so this again requires, as always,
for part A, starting off with a material balance. So the amount of substrate that comes into
the reactor minus the amount that comes out is equal to the amount that's reacted times
the volume of the reactor. And we can solve, in this case, for the flow rates of
the substrate. And it's going to be more convenient, it turns out, to express them in terms of
volumetric flowrate times the concentration of the substrate going in minus the concentration
of the substrate leaving. And that will be equal to the rate of reaction. And we'll just
go ahead and divide through by a volume, and if you remember, tau is equal to the volume
divided by the volumetric flowrate. And so this is the number that we want to solve for, and
then we can take the reciprocal of that number, and that will give us tau, the space time.
OK, we now need to express this reaction rate in the Michaelis-Menten form. So that's equal
to the maximum velocity times the concentration of the substrate times the Michaelis constant
plus the concentration of the substrate. Alright, and now we really just have one more thing
we need to solve for before we can solve for the unknown, which is the concentration of
substrate. And this is fairly straightforward in a liquid phase system - if we start 1 mole
per liter concentration and we convert 80 percent, you get 80 percent conversion here,
then the concentration of substrate leaving our system in the final state is just 0.2
moles per liter. OK, so now we can look at our equation. This again, I've circled variables
I want to solve for. We know the initial concentration of 1 mole per liter. We have the final concentration,
and the concentration in the tank of the substrate, is 0.2 moles per liter. We know vmax and Km,
which are given in the problem statement. And again, we can plug straight in with the
concentration of the substrate. The units here are pretty straightforward that everything's
in moles, liters and minutes. And so finally we can solve for v naught over v and take
the reciprocal of that number. And that should give us a tau of around 4.6 minutes. OK, so
now in part B of this problem, we have some kind of upstream process upset where the feed
stream becomes less pure, and there's a component that inhibits the enzyme, that is present at
a concentration of 0.1 moles per liter, that affects the enzyme through an uncompetitive
mechanism. Alright, the required space time to achieve 80 percent conversion is found
to increase by a factor of 2. And so now our new space time is going to be equal to 9.2
minutes. So it's doubled the amount of time. And so we're going to use that to solve for
the equilibrium constant for binding of the inhibitor onto the enzyme. OK, so now a lot
of features of this problem are going to stay the same - it's still 80 percent conversion
- and so everything on the left hand side of our material balance is just going to stay
the same. This is just in minus out. The only that's going to change is our reciprocal
space time is going to go from being 1 divided by 4.6 minutes to 1 divided by 9.2 minutes.
And it's the right side of the equation that's going to change, and it's going to change
because the kinetics will be a Michaelis-Menten type form but with uncompetitive inhibition,
also in the form of the rate law. And so we can basically rewrite the expression that we originally
had. And now it's going to be v naught over v - and I'm going to express this as a 2 v,
because again the space time is double, and so this becomes half, and so we just take
our 1 over 4.6 minutes and 9.2 minutes here, and that's going to be our reciprocal space
time being in this term. And then, just as in the previous case, again we multiply by
Cs naught minus Cs. And then this is equal to the new reaction rate, where we have an
inhibitor present. And that will be equal to - again, the numerator in this kind of
expression stays the same, vmax times the concentration of substrate - in uncompetitive
inhibition, the substrate binding term, the Michaelis constant, doesn't change, but now
the concentration of the substrate is multiplied by 1 plus CI divided by KI, where KI is the
binding constant for the inhibitor. And that's really what we want to solve for. OK, and
now everything again is pretty straightforward. So the maximum velocity at zero inhibitor
concentration is still the same as that given in the problem statement. The Km is still
the same as that given in the problem statement, that's just describing the binding of the
substrate to the enzyme, unaffected by the inhibitor, and so that remains the same. The
concentration of the substrate that we want to achieve at 80 percent conversion is still
0.2 moles per liter, so we know that. And then we're given the concentration of the
inhibitor, which I believe was 0.1 moles per liter, in the problem statement. Alright,
we have this equation and only a single unknown, which is KI. Again, unit conversions aren't
really an issue here, so I'm not going to go through the details. Everything's still
in moles, liters, and minutes. And so we can solve for a KI being equal to 0.026 moles
per liter. OK, the only, maybe slight tricky thing about this problem, is that you're
asked for a - to determine sort of the binding equilibrium constant for an inhibitor. And
if you remember, in the Michaelis-Menten format, you usually actually describe inhibitors - and
substrates, for that matter, these constants - in terms of the reverse reaction that's
asked for. So, for example, if we have an inhibitor being bound here, that's one way
to describe the reaction, and that's actually what the problem statement asks for. The problem
statement is asking for the equilibrium constant for binding of the inhibitor onto the enzyme.
Alright, so that's what's being asked for. But in this equation, what was used in order
to define the true KI - I'm going to call this KI prime for the equilibrium for this
- is just the reverse of this. So it's how easily the enzyme frees itself up to become
ES plus I. And so that's actually the KI that is here in this problem. And so in a sense,
this is just a reverse reaction, and this is a simple enough relationship that we can
just take the reciprocal of one equilibrium constant to give us another one. And so therefore,
KI prime is equal to 1 over 0.026, which is equal to about 38 liters per mole. OK, and
so this problem began as it involves fairly simple balance equations, and hopefully the
main point of this problem is really to illustrate that the procedures that you use to set up the
material balances and solve them are not really changed at all by the fact that you have Michaelis-Menten
kinetics even with the presence of an inhibitor, and instead, the only thing that really changes
is maybe the complexity of some of the algebra that's involved. But otherwise, the procedures
are very much the same.