Gibbs Adsorption Isotherm: Surface Excess Formula

By Aniket Bhardwaj · 2 November 2026 · Thermodynamics

Surface chemistry at the postgraduate level goes well beyond simple qualitative ideas of physical adsorption. In competitive national tests such as CSIR NET, GATE Chemistry, and IIT-JAM, examiners frequently test the thermodynamic relationship connecting surface tension, solute concentration, and surface concentration. This relationship is quantified by the Gibbs adsorption isotherm. Many students lose marks in numerical problems not because the calculus is difficult, but because of unit mismatches between surface tension, gas constant values, and concentration derivatives. Understanding how solute molecules concentrate at a liquid-vapour interface enables you to solve both conceptual and numerical questions with speed and confidence.

Thermodynamic Basis of Surface Excess

When a solute dissolves in a liquid solvent, the solute molecules are generally not distributed uniformly up to the very top monolayer. Depending on the intermolecular forces, the solute either accumulates preferentially at the surface or prefers the interior bulk liquid. Josiah Willard Gibbs treated the interface as an infinitesimally thin mathematical boundary dividing two bulk phases. Let the total Gibbs free energy of the system be partitioned into bulk contributions and a surface contribution. At constant temperature and pressure, the total change in surface free energy satisfies the Gibbs-Duhem equation applied to the interfacial region.

For a two-component system comprising a solvent (component 1) and a dissolved solute (component 2), we place the Gibbs dividing plane such that the surface excess of the solvent is zero, meaning Γ₁ = 0. Under this convention, the surface excess of the solute Γ₂ (commonly written simply as Γ) represents the excess moles of solute per unit surface area compared to a uniform solution of identical bulk concentration. The fundamental differential form at constant temperature is:

Gibbs Adsorption Isotherm:

dγ = -Γ dμ

Since chemical potential is μ = μ° + RT ln a (where a is activity, or concentration c in dilute ideal solutions, so dμ = RT d(ln c) = (RT / c) dc):

Γ = - (1 / RT) × (dγ / d ln c) = - (c / RT) × (dγ / dc)

Variables and Standard SI Units:

Γ = surface excess concentration of solute (mol·m^-2)
γ = surface tension of the solution (N·m^-1 or J·m^-2)
c = bulk solute concentration (mol·m^-3 or mol·L^-1, noting conversion)
R = universal gas constant = 8.314 J·mol^-1·K^-1 (or 8.314 N·m·mol^-1·K^-1)
T = absolute temperature (K)
dγ/dc = rate of change of surface tension with respect to concentration (N·m^-1·mol^-1·m³ or N·m²·mol^-1)

Physical Meaning: Positive and Negative Adsorption

The sign of the derivative dγ/dc directly governs the sign of surface excess Γ:

Positive Surface Excess (Γ > 0): When a substance lowers the surface tension of water (dγ/dc < 0), the negative sign in the Gibbs equation makes Γ positive. Solute molecules concentrate strongly at the surface rather than remaining in the bulk. Such substances are surface-active agents, or surfactants (such as soaps, long-chain alcohols, and synthetic detergents).
Negative Surface Excess (Γ < 0): When a solute increases the surface tension of water (dγ/dc > 0), Γ becomes negative. The surface region contains a lower concentration of solute than the bulk solution. Strong electrolytes such as sodium chloride and sucrose exhibit negative surface excess because polar water molecules interact more strongly with ions than with each other, pulling water into the bulk and leaving the surface layer slightly depleted of ions.

For an ionic surfactant that dissociates into ν ions (for example, sodium dodecyl sulfate, which dissociates into two ions: SDS → Na⁺ + DS^- without excess background salt), the chemical potential changes as dμ = ν RT d ln c. The isotherm then takes the modified form: Γ = - [1 / (ν RT)] × (dγ / d ln c). In entrance exams, always check whether the question specifies a non-ionic surfactant or an ionic surfactant, and whether inert supporting electrolyte is present in excess.

Exam details

Source: practice problem written for this article (not a past-paper question)
Relevant exams: CSIR NET, GATE Chemistry, IIT-JAM, CUET-PG
Topic: Gibbs Adsorption Isotherm and Surface Excess
Difficulty: Advanced (Postgraduate / National Entrance Level)

Question: An aqueous solution of a non-ionic surfactant at 298.15 K shows a linear decrease in surface tension with natural logarithm of molar concentration in the range 0.0010 mol·L^-1 to 0.0100 mol·L^-1. The slope of surface tension versus ln c is found to be -0.01248 N·m^-1. Calculate the surface excess concentration Γ of the surfactant at 298.15 K, and determine the average surface area occupied per surfactant molecule in square nanometers (nm²). Take the gas constant R = 8.314 J·mol^-1·K^-1 and Avogadro constant N_A = 6.022 × 10²³ mol^-1.

Step 1: Identify the given data and verify units.

T = 298.15 K
R = 8.314 J·mol^-1·K^-1 = 8.314 N·m·mol^-1·K^-1
dγ / d ln c = -0.01248 N·m^-1
N_A = 6.022 × 10²³ mol^-1

Step 2: Apply the Gibbs adsorption isotherm for a non-ionic solute.

Γ = - (1 / RT) × (dγ / d ln c)

Substitute the values:

Γ = - [1 / (8.314 × 298.15)] × (-0.01248)

Compute RT product: 8.314 × 298.15 = 2478.8191 J·mol^-1.

Γ = 0.01248 / 2478.8191 = 5.03465 × 10^-6 mol·m^-2.

Step 3: Calculate the area occupied per molecule (A_m).

The number of molecules per square meter is given by N = Γ × N_A.

N = (5.03465 × 10^-6 mol·m^-2) × (6.022 × 10²³ molecules·mol^-1) = 3.03187 × 10¹⁸ molecules·m^-2.

The area per molecule in square meters is A_m = 1 / N:

A_m = 1 / (3.03187 × 10¹⁸) = 3.2983 × 10^-19 m².

Convert square meters to square nanometers (1 nm = 10^-9 m, so 1 nm² = 10^-18 m²):

A_m = (3.2983 × 10^-19 m²) / (10^-18 m²·nm^-2) = 0.3298 nm² ≈ 0.330 nm² (or 33.0 Å²).

Double-check by direct single-expression calculation: A_m = RT / [ -N_A × (dγ/d ln c) ] = 2478.8191 / [ 6.022 × 10²³ × 0.01248 ] = 2478.8191 / (7.515456 × 10²¹) = 3.2983 × 10^-19 m² = 0.330 nm². Both calculation routes give identical results.

Answer: Γ = 5.03 × 10⁻⁶ mol·m⁻² and area per molecule = 0.330 nm²

Question: The surface tension of dilute aqueous phenol solutions at 293.15 K varies with phenol concentration according to the empirical relationship γ = γ₀ - b × ln(1 + K·c), where pure water surface tension γ₀ = 0.0728 N·m^-1, constant b = 0.0150 N·m^-1, and K = 40.0 L·mol^-1. Calculate the surface excess Γ of phenol at a bulk concentration of c = 0.050 mol·L^-1. Take R = 8.314 J·mol^-1·K^-1.

Step 1: Differentiate the surface tension equation with respect to concentration c.

γ = γ₀ - b × ln(1 + K·c)

dγ / dc = - b × [K / (1 + K·c)]

Step 2: Evaluate the numerical derivative at c = 0.050 mol·L^-1.

K·c = 40.0 L·mol^-1 × 0.050 mol·L^-1 = 2.00 (dimensionless).

1 + K·c = 1 + 2.00 = 3.00.

dγ / dc = - 0.0150 × (40.0 / 3.00) = - 0.0150 × 13.3333 = - 0.200 N·m^-1·L·mol^-1.

Step 3: Convert concentration units to SI units to ensure consistency.

Notice that c × (dγ/dc) = (0.050 mol·L^-1) × (-0.200 N·m^-1·L·mol^-1) = -0.0100 N·m^-1. The concentration units (mol·L^-1) cancel out directly in the product c × (dγ/dc), leaving N·m^-1 (which equals J·m^-2).

Step 4: Compute the surface excess Γ.

Γ = - (c / RT) × (dγ / dc) = - [1 / RT] × [c × (dγ / dc)]

Compute RT at 293.15 K: RT = 8.314 × 293.15 = 2437.2491 J·mol^-1.

Γ = - [ -0.0100 N·m^-1 / 2437.2491 J·mol^-1 ] = 0.0100 / 2437.2491 = 4.103 × 10^-6 mol·m^-2.

Double-check by alternative route using dγ/d ln c: Since d ln c = dc / c, dγ/d ln c = c × (dγ/dc) = -b × [K·c / (1 + K·c)] = -0.0150 × (2.00 / 3.00) = -0.0100 N·m^-1. Then Γ = - (-0.0100) / 2437.25 = 4.103 × 10^-6 mol·m^-2. The two routes verify the numerical computation exactly.

Answer: 4.10 × 10⁻⁶ mol·m⁻²

Question: A 1:1 ionic surfactant MX dissociates completely into M⁺ and X^- ions in water without any added supporting electrolyte. At 300.0 K, the slope of surface tension against log₁₀(c) is measured to be -0.02763 N·m^-1 per decade of concentration. Calculate the surface excess Γ of the surfactant ion. (Take R = 8.314 J·mol^-1·K^-1).

Step 1: Understand the ionic dissociation factor ν.

For MX → M⁺ + X^-, total number of ions ν = 2. In the absence of excess salt, the electrical neutrality condition couples both ions, giving dμ = 2 RT d ln c.

Therefore, Γ = - [1 / (2 RT)] × (dγ / d ln c).

Step 2: Convert base-10 logarithmic slope to natural logarithmic slope.

Recall that ln c = 2.302585 × log₁₀ c, so dγ / d ln c = (1 / 2.302585) × (dγ / d log₁₀ c).

Given dγ / d log₁₀ c = -0.02763 N·m^-1.

dγ / d ln c = -0.02763 / 2.302585 = -0.0119995 N·m^-1.

Step 3: Calculate the surface excess Γ.

Compute denominator 2 RT: 2 × 8.314 × 300.0 = 4988.4 J·mol^-1.

Γ = - [ -0.0119995 / 4988.4 ] = 0.0119995 / 4988.4 = 2.4055 × 10^-6 mol·m^-2.

Double-check using combined formula: Γ = - (dγ / d log₁₀ c) / (2.302585 × 2 × R × T) = 0.02763 / (2.302585 × 4988.4) = 0.02763 / 11486.22 = 2.4055 × 10^-6 mol·m^-2. The calculation is confirmed.

Answer: 2.41 × 10⁻⁶ mol·m⁻²

Common Student Mistakes

Forgetting the Dissociation Factor (ν): In ionic surfactants like sodium dodecyl sulfate without added electrolyte, ν = 2. Dividing by RT instead of 2RT leads to an answer that is exactly double the correct value. If excess inert electrolyte (like 0.1 M NaCl) is present, the counterion chemical potential is fixed, and ν reverts to 1.
Confusing Natural Logarithm and Base-10 Logarithm: Experimental data is often plotted against log₁₀ c. Students regularly omit the factor of 2.303 when applying Γ = - (1 / RT) (dγ / d ln c), introducing a 2.3-fold error.
Unit Inconsistency in Surface Tension: Surface tension may be quoted in mN·m^-1 or dyn·cm^-1 (1 mN·m^-1 = 10^-3 N·m^-1 = 1 dyn·cm^-1). Using milliNewtons directly with R in J·mol^-1·K^-1 distorts the result by three orders of magnitude. Always convert surface tension to N·m^-1 before calculating.
Ignoring the Negative Sign: When surface tension drops with increasing concentration, dγ/dc is negative. The negative sign in front of the formula cancels this out to yield a positive surface excess. Writing a negative Γ for a surfactant indicates an algebra error.

Exam Relevance and Question Patterns

Target Examination	Question Type	Core Skills Tested
CSIR NET Chemical Sciences	Multiple Choice (Part B and Part C)	Determining surface excess from graphs of γ versus ln c; distinguishing non-ionic vs ionic surfactants with and without supporting electrolytes; computing maximum packing area per molecule.
GATE Chemistry	Numerical Answer Type (NAT) & MCQ	Exact calculation of Γ in SI units (mol·m^-2); finding cross-sectional area per molecule in Å² or nm²; differentiating empirical Szyszkowski equations.
IIT-JAM Chemistry	MCQ, MSQ, and NAT	Qualitative trends of positive and negative adsorption; definition of surface active agents; straightforward single-step evaluation of surface excess from tabular or slope data.
CUET-PG Chemistry	Multiple Choice Questions	Formula identification; sign of Γ for organic solutes versus inorganic salts; basic units of surface tension and surface concentration.

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