Mass transfer, diffusion and interfacial transport

Name the transfer and its direction

Molar flux is amount transferred per area per time, such as mol m⁻² s⁻¹. Multiplying a uniform flux by area gives an amount rate. Diffusion is transfer relative to the bulk motion; advection carries species with the moving fluid. A measured total flux can contain both. For a simple dilute, one-dimensional model with fluid velocity u, one can write total flux as−D dC/dz+u C.

Fick’s law J=−D dC/dz is a useful constitutive model for an isotropic, appropriate dilute mixture at fixed conditions. D is a diffusion coefficient in m²/s. The minus sign makes positive flux point toward decreasing concentration. More general mixtures can require chemical-potential or coupled transport models; the simple concentration law is not a universal description of every interface.

Connect a gradient to conserved transfer

Our original exercise is a steady, motionless, planar layer with constant D=2×10⁻⁹ m²/s, thicknessδ=1.00 mm and no generation or consumption inside it. Reservoirs maintain C(0)=2.00 and C(δ)=0.50 mol/m³. Constant flux and constant D make the concentration linear: C(z)=2.00−1.50 z/δ, with z measured in metres when used with δ.

Convert the thickness before evaluating the gradient:1.00 mm=0.00100 m. The gradient is−1.50/0.00100=−1500 mol/m⁴. Then J=3.00×10⁻⁶ mol m⁻² s⁻¹ toward increasing z. The plot labels distance in millimetres for readability; its calculation still uses metres with the supplied D.

A boundary can add another resistance

At a phase interface, concentrations on the two sides can differ even at equilibrium because the species partitions between phases. Transfer between bulk phases can include a fluid layer on each side and the relevant partition relationship. Combining resistances requires expressing the driving force on a common basis. Subtracting unrelated gas and liquid concentrations without that relationship can produce a misleading result.

Increasing contact area can increase the total transfer rate if the local flux and conditions remain comparable. Increasing mixing can alter a layer coefficient, but it also costs energy and can change flow patterns or fragile material. Fouling and changing composition can alter transfer over time. Use a balance, an applicable local law and evidence for the boundary conditions rather than assuming area alone fixes performance.

Flux is continuous even when concentration jumps

Extend the single-layer idea to two contacting liquid phases. For this separate dilute model, assume a steady planar interface, no reaction or accumulation, constant coefficients and negligible additional interfacial resistance. Concentration is amount per volume of its own phase. Supply the interfacial equilibrium relation C1i = m C2i, where the distribution ratio m is dimensionless because both concentrations use mol/m³. Equal chemical tendency at the interface does not require equal concentrations.

Let the bulk concentrations be C1b = 4.00 and C2b = 0.50 mol/m³. Supply m = 2, k1 = 0.002 m/s and k2 = 0.004 m/s. Positive flux points from phase 1 to phase 2. With no material stored at the interface, the same J must satisfy k1(C1b−C1i) = k2(C2i−C2b). Substituting C1i = 2 C2i gives 0.002(4−2 C2i) = 0.004(C2i−0.5). Thus C2i = 1.25 and C1i = 2.50 mol/m³, and either film gives J = 0.003 mol/(m²·s).

The two film drops are 1.50 in phase 1 and 0.75 in phase 2. The concentrations jump from 2.50 to 1.25 at the equilibrium interface; that jump is not itself an extra resistance. These are phase-specific reference values. Subtracting the two bulk concentrations and multiplying by an arbitrary one-side coefficient would discard both the partition relation and the second film.

To write the account using the phase-1 basis, add the first-film drop J/k1 to m times the second-film drop J/k2. This gives C1b−m C2b = J(1/k1 + m/k2). Here the driving difference is 4−2×0.5 = 3 mol/m³ and the converted resistance is 500+500 = 1000 s/m. Their quotient again gives 0.003 mol/(m²·s). The next phase-equilibria lesson develops why a composition relationship and an amount balance remain separate pieces of a separation model.

Check area, duration and thickness separately

For supplied uniform area 0.020 m², the amount rate is J A=6.00×10⁻⁸ mol/s. Over 600 s with the same maintained boundaries it transfers 3.60×10⁻⁵ mol. That is an accumulated amount, not a concentration or a claim that the reservoirs will remain unchanged without replenishment.

A separate layer twice as thick, with the same D and boundary concentrations, gives half the gradient and half the flux:1.50×10⁻⁶ mol m⁻² s⁻¹. The orange line shows that counterfactual over 2 mm. If a closed finite reservoir were depleted, the boundary values would evolve and this constant-flux multiplication would no longer predict the full history.

Check the common flux directly: phase 1 delivers 0.002×(4−2.5) = 0.003 mol/(m²·s); phase 2 carries 0.004×(1.25−0.5) = 0.003 in the same units. A coefficient in m/s multiplied by a concentration difference in mol/m³ produces the required mol/(m²·s). The interface relationship converts bases; it does not create or destroy transferred A.

Try the reasoning

1. Why does the flux have a positive sign when concentration decreases with z?

   Show the answer and reasoning

   Fick’s law includes a minus sign, so the net transfer is toward increasing z, down the concentration gradient.

2. Can mol m⁻² s⁻¹ be reported as the total amount transferred?

   Show the answer and reasoning

   No. Area and duration must also be accounted for, with any changes in flux over time.

3. Does a concentration jump at a phase interface prove the interface is out of equilibrium?

   Show the answer and reasoning

   No. Equilibrium partitioning can give different concentrations in different phases.

4. Keep the bulk concentrations and film coefficients, but supply m = 4 instead of 2. Find the common flux and interfacial concentrations. What would m = 8 imply?

   Show the answer and reasoning

   For m = 4, J = (4−4×0.5)/(1/0.002+4/0.004) = 2/1500 = 0.001333 mol/(m²·s). Then C2i = 0.5+J/0.004 = 0.8333 and C1i = 4 C2i = 3.3333 mol/m³. Both films give the same flux. At m = 8 the phase-1-basis driving difference is 4−8×0.5 = 0, so J = 0 even though the two bulk concentrations are unequal. Their equilibrium relationship, not raw equality, determines the zero-driving condition.

What to carry forward

A transfer estimate needs a defined flux, physical frame, unit-consistent gradient and maintained or evolving boundary conditions.