In addition to diffusion through a stagnant gas, there are a variety of other scenarios from which non-equimolar counter diffusion can arise (i.e., where we might obtain a bulk flow). A good example is heterogeneous catalysis problems (i.e. reactions on a surface).

Consider the reaction:

$2A \rightarrow 2B+C$

If this reaction occurs between gas molecules on a solid surface:

we clearly have a net molar flux away from the surface as two
moles/molecules of A is making a total of three moles of product
(although our net *mass* flux would still be zero).

In solving the problem for the total flux of species A, we again start with the definition of the total flux:

$\displaystyle{N_A = -c_{total}D\nabla y_A+c_AV}$

In this case our *V*is given as

$\displaystyle{V = \frac{c_Av_A+c_Bv_B+c_Cv_C}{c_{total}} = \frac{N_A+N_B+N_C}{c_{total}}}$

where we can use stoichiometry to relate N_{A} to the
other N's to get:

$\displaystyle{\frac{1}{2}N_A = -\frac{1}{2}N_B = -N_C}$

Be careful not to trick yourself into saying 2N_{A} =
-2N_{B} = -N_{C}. Since there are *already*
2 moles of A forming 1 mole of C, this would actually be backward
and make our answer really wrong.

Plugging this into *N*_{A} we get

$\displaystyle{N_A = -c_{total}D\nabla y_A + y_A\left( N_A-N_A-\frac{1}{2}N_A\right)}$

$\displaystyle{N_A = -c_{total}D\nabla y_A - \frac{y_A}{2}N_A}$

solving for *N*_{A} gives

$\displaystyle{N_A = -\frac{c_{total}D\nabla y_A}{1+\frac{y_A}{2}} }$

which in one dimension becomes

$\displaystyle{N_A = -\frac{c_{total}D\frac{dy_A}{dz}}{1+\frac{y_A}{2}} }$

Calculate the magnitude of diffusive mass flow/flux for systems with non-zero bulk flow