Equimolar Counterdiffusion (EMD) Video Lecture | Mass Transfer - Chemical Engineering

In this screencast we are going to talk about
a specific case of diffusion and we are going
to stage this by asking a question first.
So let's take a look at these two tanks in
which we have a species A, and species B.
And the way I have drawn them, blue doesn't
necessarily represent species A and yellow
doesn't represent species B, those are just
the two mixtures in tank 1 and tank 2. And
they are closed off by a valve here in the
middle that separates them from making contact.
We know that the temperature is constant for
the system, that it is a closed system and
that the pressure is going to remain constant.
So I have defined positive Z from left to
right and we have two points of interest,
z1 and z2, which are the entrances to both
tanks. The mole fraction of A and B in tank
1 is written as yA1 and yB1 and the same for
tank 2, yA2 and yB2. We are told that based
on our mixtures the mole fraction of A in
tank 1 is greater than that in tank 2 and
the mole fraction of B in tank 2 is greater
than that in tank 1. When this system is opened
and the mixtures come in contact with each
other, which statement is false? A- there
will be a net flux of A to the right, B- the
net flux of B is zero, C- the magnitude of the
net flux of A is equal to that of B or D-
the net flux of A is opposite in direction
to the net flux of B.
So we know that since we
have a concentration gradient for A going
from Z1 to Z2 that we are going to go from
a higher concentration to a lower concentration.
So we would expect A to move to the right.
So we know that part A of the question must
be true and therefore is not one of the answers.
We also know that the same must be true for
B, that since we have a higher concentration of B on the right side, that we are going to have a flux of B from
the right to the left. So therefore we know
that at least, B must be false. Now the magnitude
of the net flux of A is equal to that of B-
so we are saying here that Na must equal the
magnitude of Nb. So what do we know about our
system? Since temperature and pressure are
constant, we are working with gases, we know
that the overall concentration must be constant
in our system. So if the net flux of A was
not equal to the net flux of B we would get
some kind of buildup of these gases in part
of our system and therefore our concentration
couldn't be kept constant, we would have a
pressure gradient and we know that that can't
happen here because it is a closed system.
So the magnitude of the A must
be equal to B and not only that, it
must be in the opposite direction of B. Therefore,
we are saying that the net molar flux of A
is equal and opposite in direction to the
net molar flux of B. Now this special case
where we have the net flux of A equal and
opposite to the net flux of B in a closed
system is called equal molar counter diffusion
or EMD for short. So what does this mean for
our two constituent equations that we worked
out in previous screencasts for our net fluxes.
Recall that the net molar flux of A was equal
to our diffusive flux plus our mass transfer
due to bulk flow. Now in this case we are
saying that our net flux N, is equal to Na
plus Nb since we have a binary system. And
Na is equal to negative Nb so this must be equal to
zero. This means that the net molar flux of A
is just equal to the diffusive flux of A,
in this case, JA. This also means the same
for B and therefore, by relationship, the
diffusive fluxes must be equal and opposite.
So what does this mean? If we were to look
at the molar flux of A and we have a binary
mixture in which the total concentration is
constant, the pressure, temperature
constant, and the mole fractions are constant
so we have a steady-state process but different
between two points like we have above, between
points Z1 and Z2, then we could write this
simply as the following. Now this becomes
the equation that we use to characterize
steady-state equal molar counterdiffusion. Now if
we wanted to plot what the mole fraction versus
Z looks like, we go from Z1 to Z2 we know
that the mole fraction of A is going to decrease
from point Z1 to Z2. The same is true for
xb, but we are starting on the Z2 side and
going down. Now one last thing to point out
with this binary system is if we know that
the flux of A must be equal and opposite to the
flux of B does this also mean that the diffusivity
of AB is equal the negative diffusivity of
BA? Well we know that the concentration is
constant and this is equal to the concentration
of A plus the concentration of B and we could
differentiate this. We know that there is
no change in concentration so we could write
it as the following. This implies that the
differential of the concentration of A is
equal to the negative differential of the
concentration of B and we know what our flux
equations look like. In this case we know
that JA is equal to negative JB so therefore that
means that we have the following relationship.
And if the distances we are looking at then
we can just remove them and therefore show
that the diffusion coefficients are not equal
and opposite as we had with the fluxes, but
in fact, the equality is that the diffusion
coefficient for A and B is the same as the
diffusion coefficient is the same for B and
A. And this will be true every time we have
equimolar diffusion for a binary system. And
in another screencast we will apply this
to an example problem.