Fick's law and equimolar counter diffusion

Fick's law

Mass transfer by molecular diffusion is the movement of a species because of a concentration difference. For a binary mixture, the molar flux of A in the $x$ direction (relative to the mixture as a whole) is

$$J_A = -D_{AB}\frac{dC_A}{dx}$$

where $D_{AB}$ is the diffusion coefficient (m$^2$/s), $C_A$ the molar concentration (kmol/m$^3$). The flux is toward lower concentration.

Typical $D_{AB}$: gases about $10^{-5}$ m$^2$/s, liquids about $10^{-9}$ m$^2$/s, solids $10^{-10}$ to $10^{-14}$ m$^2$/s.

Equimolar counter diffusion

Two vessels, at the same total pressure $P$ and temperature $T$, are connected by a tube of length $L$. Gas A diffuses from vessel 1 (partial pressure $p_{A1}$) to vessel 2 ($p_{A2}$), and gas B diffuses in the opposite direction at the same molar rate: $N_A = -N_B$.

 [ vessel 1 ]=======L=======[ vessel 2 ]
   pA1 high        A -->       pA2 low
   pB1 low         <-- B       pB2 high

Since there is no net bulk flow, the flux is only diffusion:

$$N_A = -D_{AB}\frac{dC_A}{dx}$$

At steady state $N_A$ is constant. Integrate from $x = 0$ ($C_{A1}$) to $x = L$ ($C_{A2}$):

$$N_A = \frac{D_{AB}}{L}\,(C_{A1}-C_{A2})$$

For ideal gases $C_A = p_A/\bar R T$, so

$$N_A = \frac{D_{AB}}{\bar R T\,L}\,(p_{A1}-p_{A2})$$

with $\bar R = 8.314$ kJ/kmol K (or 8314 J/kmol K, with $p$ in Pa). The concentration profile of A is linear in $x$, as for conduction through a plane wall.