In this example, we're going to look at the case where we have diffusion in a catalytic
reaction. We have a porous, non-catalytic layer on top of a flat catalyst surface, and
the reaction is A goes to 2B in the gas phase, and the reaction is extremely fast, essentially
instantaneous on the catalyst surface. The product B that forms then diffuses back through
the porous layer and into the bulk, and in the bulk, diffusion is fast. So the questions
we want to answer is, what's the concentration profile for the reactant A in this porous
layer, and then what is the flux of A into the layer? So the first thing we want to do
is look at a diagram that represents the system. So we have this porous layer where our reactant,
A, is diffusing through this non-catalytic layer and react on the catalyst surface. And
then our product, B, diffuses back out into the bulk. We want to calculate what is this
concentration profile on the porous layer, and to do this, we need to account for the
fact that there's a mole change in this gas phase reaction which causes a bulk flow in
this porous layer. So first thing we want to do is indicate our notation for flux. So
this is the molar flux of A. Now the molar flux is consisted in this case of two parts. We
have one part due to diffusion and the other part due to the fact we have bulk flow. And
we have this bulk flow because of the mole change. So the diffusion term for component
A is a diffusivity and change in the concentration of A with respect to z, and the minus sign
is here because the diffusion is from high concentration to low concentration. And this
is just Fick's law. And we can write this in terms of a total concentration times a
mole fraction, just makes it a little easier to do the manipulations. The total concentration
from the ideal gas law would just be the pressure over RT. Pressure and temperature are constant
in the system, so this total concentration is constant. We're also going to assume the
diffusivity is a constant independent of concentration. So now we want to determine the bulk flow.
So if we know the molar flux of A plus the molar flux of B, and they're going to have
opposite signs, that's the total molar flux. And then I take the mole fraction times that. It
would give me what's the bulk flow of component A. wB is just -2 times wA from the stoichiometry,
and we can substitute this into here and we can get that the bulk flow of A as the mole
fraction times -y of A. Now we determined up here that the total flux of A is this diffusion
term plus the bulk flow term, so we're going to substitute the two terms in that we determined
here and here into the flux of A. So we end up with this equation, and now all we're going to
do is rearrange. So now we have the flux of A in terms of mole fraction of A, total concentration,
diffusivity, and distance. And we now need to take advantage of a mass balance, so we
want to take advantage of a mass balance to further determine how does the concentration
change with z, or how does wA change with z. So again, we'll look at our system and look
in particular at z and at z plus delta z, and the flux then across these boundaries. In other
words, we have the flux in at z minus the flux out at z plus delta z. That's equal to zero.
Now keep in mind, flux has sign and so remember that A is actually moving in this direction.
But this equation is the steady-state equation, that if we divide by delta z, take the limit
which says the flux is a constant since there is no reaction in the porous layer. The boundary
conditions for this system is that at z=0, the concentration of A is essentially zero
because the reaction's instantaneous. And at z equals delta, the concentration is just CA at
delta, a number that we have to know to obtain this profile. So we can use this equation
and apply it to the equation we derived here, and so we're going to make that substitution.
So I've made the substitution, I've multiplied both sides by -1 to eliminate the minus sign.
So we can take this equation and integrate it, which says that the term inside the derivative
is a constant. And we'll eventually need to find the value of this constant. And then
we can separate variables where, we have the mole fraction on one side and we have constants
and dz on the other side. Now we can integrate this, the integral of dy of A by the 1 plus y of A is the log
of 1 plus yA. K1, c, and D are constants, z, and then we have a constant of integration. So
we have these boundary conditions. So when z equals zero, CA equals zero, which means yA equals zero. We can make this
substitution into the equation, the log of 1 equals k2, and the log of 1 is zero, so therefore
k2 is 0. So I have written the equation now with k2 as zero, and now I can use the other boundary
condition, namely at z equals delta, cA is cA delta, or yA is cA delta over the total concentration. We can
make that substitution into the equation. This means we now have solved for k1. So I can substitute
k1 into this equation and obtain an equation that relates the concentration to z, which
is our objective. I'm going to rewrite this. The reason I did that, because now I can take
exponential of both sides. And now I can take and just rearrange this equation and solve
for the concentration of A in terms of z and constant. And this was our original objective,
is to determine the concentration as a function of z. The second part of the problem is then
to determine the flux of A. So I've just reproduced here an equation that we developed earlier.
And now I can substitute in. But first, I can simplify this by writing this as d of the
log of 1 plus yA with respect to z. And I'm doing that because, here I have the log of 1 plus yA, so
I can substitute in this term on the right of that equation. So most of these are constants
and I have dz/dz, essentially, which is 1, so that means the flux of A minus CD
over delta times the log 1 plus - I'm going to write this in terms of yA again.
So this is the flux of A to the surface. So this is the reaction rate because this is
a steady-state system, and that flux to the surface is the amount that reacts.