Understanding the Gibbs adsorption and surface coverage calculator — concept and principle
Surfactants lower surface tension because they crowd into the air-water surface. The Gibbs adsorption isotherm lets you work out how many molecules sit there per square metre, and how much area each one occupies, from the slope of a surface-tension curve. This calculator is used with tensiometry data on detergents, emulsifiers and wetting agents, typically from the steep region just below the critical micelle concentration (CMC).
How it is calculated
Γ = −(1 / nRT) · dγ/d(ln C), and area per molecule a = 1 / (Γ · N_A)
- Slope dγ/d(ln C) in mN/m: divided by 1000 to give N/m. It is negative because γ falls as C rises.
- T in °C, converted to kelvin; R = 8.314462618 J/(mol·K).
- n = 1 for nonionic surfactants (or ionic ones in excess salt), 2 for a 1:1 ionic surfactant without added salt.
- N_A = 6.02214076 × 10²³ /mol. Γ is shown in µmol/m² and a in Ų (1 Ų = 10⁻²⁰ m²).
The principle
Gibbs showed from thermodynamics that a solute which lowers surface tension must be enriched at the surface. At constant temperature, dγ = −Γ·RT·d(ln C) for a dilute ideal solution, so the steeper γ falls with ln C, the more is adsorbed. For an ionic surfactant without salt, the counter-ions adsorb along with the surfactant ions and both contribute, which gives the factor n = 2. Just below the CMC the surface is nearly saturated and the γ–ln C plot is close to straight, so one slope gives the maximum surface excess.
Worked example
- Inputs: slope = −8 mN/m = −0.008 N/m, T = 25 °C, n = 1.
- nRT = 1 × 8.314462618 × 298.15 = 2478.96 J/mol.
- Γ = 0.008 / 2478.96 = 3.2272 × 10⁻⁶ mol/m² = 3.227 µmol/m².
- Molecules per m²: 3.2272 × 10⁻⁶ × 6.02214 × 10²³ = 1.9434 × 10¹⁸.
- a = 1 / 1.9434 × 10¹⁸ m² = 5.1455 × 10⁻¹⁹ m² = 51.455 Ų.
| n (1 nonionic, 2 ionic) | Surface excess Γ (µmol/m²) |
|---|---|
| 0.25 | 12.90865457 |
| 0.5 | 6.454327287 |
| 0.75 | 4.302884858 |
| 1 | 3.227163644 |
| 1.25 | 2.581730915 |
| 1.5 | 2.151442429 |
| 2 | 1.613581822 |
How to read the result
Saturated monolayers of common surfactants usually give Γ of a few µmol/m² and areas of several tens of Ų per molecule. A small area means tightly packed, upright chains; a large area suggests a bulky head group, charge repulsion or incomplete coverage. Comparing a with the head-group cross-section helps judge how close the layer is to full packing.
Common mistakes
- Using a slope in log₁₀ C; multiply by 1/2.303 (that is, divide by ln 10) to convert to ln C first.
- Choosing n = 1 for an ionic surfactant in pure water, which doubles Γ and halves the area.
- Taking the slope above the CMC, where γ is flat and the slope is near zero.
Frequently asked questions
Does this calculator find the CMC itself?
No. The CMC is read from the break point where the γ–ln C curve levels off; this tool uses the slope just before that point.
When should I use n = 2?
For a 1:1 ionic surfactant such as SDS in water without added electrolyte. With a large excess of salt, use n = 1.
Why give the area in Ų?
Molecular head groups are a few ångströms across, so Ų gives readable numbers (1 nm² = 100 Ų).
Sources: J. W. Gibbs, "On the Equilibrium of Heterogeneous Substances", Transactions of the Connecticut Academy, 1876–1878 · M. J. Rosen and J. T. Kunjappu, Surfactants and Interfacial Phenomena, Wiley