Two-Film Theory

Why most of the “action” happens in thin stagnant films next to the interface.

FoundationsMass transferConceptual model
⏱️ About 18 min

When a gas dissolves into a liquid, why doesn’t the entire liquid instantly reach equilibrium with the gas?

💡
The big idea: Whitman’s two-film theory models interphase mass transfer as diffusion through two thin stagnant films—one in the gas, one in the liquid—with interfacial equilibrium and resistances in series.
🎯 By the end, you'll be able to
  • State the two-film assumptions
  • Sketch and interpret profiles in each film
  • Explain interfacial equilibrium vs bulk conditions
  • Compute resistance fractions from film coefficients
📎 Helpful to know first
  • Diffusion Applications

The picture: two thin films, two resistances

In many gas–liquid operations (absorption, stripping, aeration), the bulk gas and bulk liquid are well-mixed by turbulence. However, right next to the interface there are thin regions where mixing is weak and transport occurs mainly by molecular diffusion.

Whitman’s two-film theory models these regions as a gas film and a liquid film in series. The interface itself is not a “barrier”; instead, the interface is assumed to be at local equilibrium.

🔑 Two-film assumptions (what you are assuming when you use it)

(1) Bulk phases are well-mixed (uniform bulk concentrations). (2) Thin stagnant films adjacent to the interface control diffusion. (3) At the interface, the two phases are in equilibrium (e.g., Henry’s law). (4) Steady transport through each film.

Gas phase (bulk) Liquid phase (bulk) gas film liquid film interface driving force p_A,b p_A,i x_A,i x_A,b well-mixed bulk well-mixed bulk diffusion diffusion
Whitman two-film model: diffusion through gas and liquid films in series; a step at the interface reflects interfacial equilibrium (not a resistance).

Resistances in series (concept)

Mass transfer through two films is like electrical resistors in series: the flux is the same through each film, but the driving force drops across each resistance.

Depending on solubility (Henry’s law) and hydrodynamics, either the gas film or the liquid film may dominate the overall resistance.

\[ R_{\text{total}} = R_G + R_L \]
Total resistance is the sum of gas-film and liquid-film resistances (in a consistent basis).
✨ “Controlling film” means “largest resistance”

If one resistance is much larger than the other, it controls the flux. Improving mixing in the non-controlling phase will barely change the rate.

📝 Worked example: You want to compare gas-film vs liquid-film resistance for absorption of a dilute gas A into water at 25°C. Given: gas-film coefficient k_G = 0.020 mol/(m²·s·atm), liquid-film coefficient k_L = 2.0×10⁻⁵ m/s, Henry’s constant H = 1.64×10³ atm/(mole fraction). Use a liquid-phase basis where the comparable resistances are R_L = 1/k_L and R_G = 1/(H·k_G). Compute the fraction of total resistance in each film.
  1. Compute liquid-film resistance: R_L = 1/k_L = 1/(2.0×10⁻⁵) = 50,000 s/m.
  2. Compute H·k_G: H·k_G = (1.64×10³)·(0.020) = 32.8 mol/(m²·s·(mole fraction)).
  3. Compute gas-film resistance on liquid basis: R_G = 1/(H·k_G) = 1/32.8 = 0.03049 (in the same liquid-basis resistance units).
  4. Total resistance: R_total = R_L + R_G ≈ 50,000 + 0.03049 ≈ 50,000.03049.
  5. Resistance fractions: f_L = R_L/R_total ≈ 50,000/50,000.03049 = 0.99999939; f_G = R_G/R_total ≈ 0.03049/50,000.03049 = 6.10×10⁻⁷.
✓ Liquid film ≈ 99.99994% of resistance; gas film ≈ 6.10×10⁻⁵% (liquid-film controlled).
✏️ Practice: Using the same resistance comparison idea on a liquid-phase basis: k_G = 0.010 mol/(m²·s·atm), k_L = 4.0×10⁻⁵ m/s, H = 4.4×10⁴ atm/(mole fraction). Compute the fraction of total resistance in the liquid film.
fraction
Solution
  1. R_L = 1/k_L = 1/(4.0×10⁻⁵) = 25,000.
  2. H·k_G = (4.4×10⁴)·(0.010) = 440.
  3. R_G = 1/(H·k_G) = 1/440 = 0.0022727273.
  4. R_total = 25,000 + 0.0022727273 = 25,000.0022727273.
  5. f_L = R_L/R_total = 25,000/25,000.0022727273 = 0.9999999091.

Check your understanding

1. In Whitman’s two-film theory, what is assumed at the gas–liquid interface?
Two-film theory assumes no resistance at the interface itself; instead, the interface is at local equilibrium, so a “jump” can occur when switching variables (p to x).
2. If the liquid-film resistance is much larger than the gas-film resistance, which action most increases the transfer rate?
The controlling (largest) resistance dominates; reducing it (increasing k_L) has the biggest effect.
3. Why can a step-like change appear at the interface in a profile diagram?
Interfacial equilibrium can relate gas-phase partial pressure to liquid-phase mole fraction, so the plotted quantity changes when switching definitions.
🔎 Related lessons
Diffusion Applications · soon
✅ Key takeaways
  • Two-film theory places diffusion resistance in thin gas and liquid films adjacent to the interface.
  • The interface is assumed to be at local equilibrium (e.g., Henry’s law), not a resistive barrier.
  • Resistances add in series; the larger resistance controls the overall flux.
  • Comparing resistance fractions helps identify which side to “fix” to increase rate.
➡️ Next, we turn these ideas into practical coefficients: individual film coefficients (k) and overall coefficients (K) that combine the two resistances.
Want to test yourself on this? Try the Chemical Aptitude test →