Chemistry · Lesson

Surfaces, adsorption and interfacial thermodynamics

Describe an interface through area work, surface excess and a stated adsorption model. Keep fluid tension, dividing-surface conventions, coverage kinetics and probe-dependent area distinct.

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By the end, you can

Calculate a stipulated fluid-area work and spherical pressure jump.
Use a nonionic Gibbs activity slope as a surface excess.
Separate equilibrium site occupancy, relaxation and conditional monolayer area.

Creating fluid area has a conditional reversible work cost

Fluid interfacial tension γ has units N m⁻¹=J m⁻². Under the compatible fixed-state conditions of a simple constant-γ model, reversible area work is γdA. For a spherical single interface with negligible deformation, the inside-minus-outside pressure jump is 2γ/r. A thin spherical film with two interfaces of nearly equal radius and the same γ instead approximates 4γ/r.

Temperature, composition and the surrounding phases matter for a real fluid tension. A pure-water equilibrium reference formulation has a specified temperature range and water/vapour conditions; it does not supply a universal tension for every water sample. Solid surface stress and surface free energy can differ despite related units. An interface is not literally a stretched rubber membrane.

An excess compares an interface with a bulk reference

A surface excess is Γ_i=(n_i−chosen bulk-reference amounts)/A and depends on the dividing-surface convention. It can be positive, zero or negative. It is not automatically the absolute count of all interfacial molecules. Adsorption enriches material near an interface; absorption instead describes uptake into the bulk.

For an equilibrium nonionic binary interface at fixed temperature T and pressure p with a solvent-zero dividing convention, Γ=−(1/RT)dγ/d ln a, where a is dimensionless solute activity. A decreasing tension with increasing activity gives positive excess under this convention. An ionic system needs its independently varying chemical potentials and activity assumptions; there is no universal extra factor 2 to apply to every surfactant slope.

A site isotherm does not fix a time scale by itself

In an invented equivalent-site, noninteracting monolayer model, dθ/dt=k_a a(1−θ)−k_dθ with clamped dimensionless reservoir activity a. Both k_a and k_d here have units s⁻¹. Equilibrium is θ_eq=Ka/(1+Ka), with dimensionless K=k_a/k_d. A concentration form K_c c/(1+K_c c) instead uses K_c in inverse concentration units.

At the stipulated fixed conditions, k_a=0.400 s⁻¹, k_d=0.100 s⁻¹ and a=0.500 give K=4.00 and θ_eq=2/3. Relaxation time is 1/(k_a a+k_d)=3.33 s. Multiplying both kinetic coefficients by 10 retains the isotherm but makes relaxation 10 times faster. The model omits site heterogeneity, lateral interactions and transport limits.

The equilibrium isotherm fixes coverage rather than a relaxation time

Equilibrium coverage θ

Dimensionless activity a

Equilibrium site fraction
— θ_eq = 4a / (1+4a)

Underlying data

Dimensionless activity a	Equilibrium coverage θ
0	0 · Equilibrium site fraction
0.125	0.3333333333333333 · Equilibrium site fraction
0.25	0.5 · Equilibrium site fraction
0.5	0.6666666666666666 · Equilibrium site fraction
1	0.8 · Equilibrium site fraction
2	0.8888888888888888 · Equilibrium site fraction

Original equivalent noninteracting monolayer site model at stipulated fixed conditions, using dimensionless reservoir activity a and K=k_a/k_d=4.00. Markers and dashed curve show equilibrium θ=Ka/(1+Ka), not a time trace. Both coefficients can be scaled together without changing this curve. Heterogeneous sites, lateral interactions and transport limitations are excluded.

An adsorption-derived area is conditional on the probe and model

If a monolayer occupancy is independently stipulated, the model area is n_mono N_A times a supplied area per adsorbed molecule. Specific area divides by adsorbent mass. This arithmetic is not a full BET analysis and does not prove that every pore or site is accessible.

Real results depend on the adsorptive probe, temperature, pretreatment, isotherm model and uncertainty. Micropore filling is not ordinary monolayer coverage, and an apparently straight interval alone does not validate a BET range. An inferred surface excess, a site coverage and a probe-dependent area are different quantities with different reference conditions.

Worked examples

Use one interface in a stipulated spherical model

For an invented fluid with constant γ=5.00 × 10⁻² N m⁻¹, a spherical single-interface radius is 1.00 × 10⁻⁴ m. Find its pressure jump and the area work for a separate area increase 2.00 × 10⁻⁴ m².

Single-interface Δp=2γ/r=1000 Pa.
The radius, positive tension and inside-minus-outside convention give a positive jump.
For the separate reversible area change, work=γΔA.
Multiply 5.00 × 10⁻² by 2.00 × 10⁻⁴ to obtain 1.00 × 10⁻⁵ J.

Δp=1.00 kPa and the stipulated area work is 10.0 μJ. A two-interface thin-film bubble has a different approximate pressure factor.

State a function before treating a slope as local

For an invented nonionic binary interface at fixed temperature T=300 K and pressure, assume γ is linear in ln a between a=0.0100 and 0.0200. The stipulated tensions are 0.0600 and 0.0570 N m⁻¹. Find the surface excess under the solvent-zero convention.

The ln a change is ln 2.
The slope is(0.0570−0.0600)/ln 2≈−0.00432809 N m⁻¹.
Γ=−slope/(RT)≈ 1.73516 × 10⁻⁶ mol m⁻².
Without the linear-in-ln a assumption, two points give an interval average rather than a proven local derivative.

Γ≈+1.74 × 10⁻⁶ mol m⁻² under the stated convention. It is an excess relative to the bulk reference, not automatically absolute molecular coverage.

Calculate equilibrium and relaxation from different information

Use the stipulated activity-model coefficients k_a=0.400 and k_d=0.100 s⁻¹ at a=0.500. Find coverage and relaxation time.

K=k_a/k_d=4.00 and Ka=2.00.
θ_eq=2/(1+2)=2/3.
The decay coefficient for a coverage perturbation is k_a a+k_d=0.300 s⁻¹.
Its reciprocal is approximately 3.33 s.

θ_eq≈ 0.667 and τ≈ 3.33 s. The ratio K alone would not have supplied τ.

Use a stipulated monolayer capacity without calling it a measured area

A model monolayer contains 1.00 × 10⁻⁴ mol on 1.00 g of solid, with supplied area 0.200 nm² per adsorbed molecule. Find model area and specific area.

Convert molecular area to 2.00 × 10⁻¹⁹ m².
The stipulated count is n_mono N_A≈ 6.02214 × 10¹⁹ molecules.
Multiply count by molecular area to obtain 12.0443 m².
Divide by 1.00 g for the mass-specific result.

Model area≈ 12.0 m² and specific area≈ 12.0 m² g⁻¹. Probe accessibility and monolayer-model validity remain assumptions.

Try it yourself

For K=4.00, which dimensionless activity gives θ_eq=1/2?
Show the answer and reasoning
a=0.250.
Half coverage requires Ka=1.
Does Γ always count the absolute number of molecules per interface area?
Show the answer and reasoning
No.
It is an excess under a selected dividing-surface convention.
If both k_a and k_d scale by 10, how do θ_eq and τ change at the same clamped activity?
Show the answer and reasoning
θ_eq stays fixed;τ becomes one tenth as large.
The coefficient ratio stays fixed while their sum in the relaxation rate scales.
Is micropore filling automatically a monolayer suitable for the simple area arithmetic?
Show the answer and reasoning
No.
The adsorption mechanism and probe-dependent model must be checked.

What to carry forward

Use the number and curvature of the fluid interfaces.
Surface excess depends on its bulk-reference convention.
An equilibrium isotherm and a kinetic relaxation time need different inputs.
Probe area calculations are conditional model inferences.

References

Teaching resources and reference material supporting this lesson’s explanations and examples.