---
title: "Fick's law and equimolar counter diffusion"
subject: "Heat Transfer"
chapter: 7
key: "semester_5/heat_transfer:ficks-law-equimolar-counter-diffusion"
source: https://www.bctengineer.com/q/semester_5/heat_transfer/ficks-law-equimolar-counter-diffusion
site: BCT Engineering
---

# Fick's law and equimolar counter diffusion

Heat Transfer · Chapter 7: Introduction to Mass Transfer

- Asked in: year not given, 6 marks
- Video: [Equimolar Counterdiffusion (EMD)](https://www.youtube.com/watch?v=IKmE3Z3zzBo) — LearnChemE
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/heat_transfer/ficks-law-equimolar-counter-diffusion
- Whole chapter: https://www.bctengineer.com/md/semester_5/heat_transfer/chapter-7

## Question

State Fick's law of diffusion and compare it with Fourier's law. Derive the expression for the molar flux in steady equimolar counter diffusion of two ideal gases A and B between two large vessels connected by a tube of length L.

## Short answer

Fick's law: JA = –DAB dCA/dx (molar flux proportional to concentration gradient), analogous to Fourier's q'' = –k dT/dx. For steady equimolar counter diffusion through a tube of length L, with no net bulk flow, NA = –DAB dCA/dx constant. Integrating: NA = DAB (CA1–CA2)/L. For ideal gases, CA = pA/RT, thus NA = DAB (pA1–pA2)/(RT L). Concentration profile linear.

## Answer

#### 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.

| Quantity | Heat conduction | Mass diffusion |
|---|---|---|
| Law | $q'' = -k\,dT/dx$ | $J_A = -D_{AB}\,dC_A/dx$ |
| Driving force | Temperature gradient | Concentration gradient |
| Property | $k$ (or $\alpha$) | $D_{AB}$ |
| Units of property | W/m K (m$^2$/s) | m$^2$/s |

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: $N_A = -N_B$.
```text
 [ 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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How does Fick's law compare with Fourier's law?
   A. Both relate flux to a gradient; the driving force is concentration for Fick and temperature for Fourier
   B. Fick's law uses a temperature gradient
   C. Fourier's law uses concentration
   D. They are unrelated; Fick's law has no property
   Answer: A. J = -D dC/dx is the mass analogue of q = -k dT/dx. (Key point: Fick's law analogy)

2. Why is there no bulk flow in equimolar counter diffusion?
   A. Because the pressure differs between vessels
   B. Because only A moves
   C. Because N_A = -N_B, so the net molar flux is zero at constant total pressure
   D. Because the tube is closed
   Answer: C. Equal and opposite molar fluxes cancel, so the flux is purely diffusive. (Key point: Equimolar counter diffusion)

3. The steady molar flux of A in terms of partial pressures is
   A. D (pA1 - pA2) R T L
   B. D L (pA1 - pA2)/(R T)
   C. D (pA1 - pA2)/(R T L)
   D. D (pA1 + pA2)/(R T L)
   Answer: C. Integrating gives N_A = D (C_A1 - C_A2)/L, and C = p/(R T). (Key point: Flux expression)

4. The concentration profile of A along the tube is
   A. logarithmic
   B. parabolic
   C. exponential
   D. linear
   Answer: D. N_A is constant, so dC_A/dx is constant and the profile is linear, like conduction through a plane wall. (Key point: Concentration profile)

## Related in Heat Transfer

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- [Diffusion of water vapour through stagnant air](https://www.bctengineer.com/question.md?key=semester_5%2Fheat_transfer%3Aevaporation-water-test-tube-stagnant-air) — Heat Transfer, chapter 7
- [Water heating in a tube, Dittus-Boelter](https://www.bctengineer.com/question.md?key=semester_5%2Fheat_transfer%3Adittus-boelter-water-heating-tube-length) — Heat Transfer, chapter 3
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