Lesson 6 · 35 min

Surfaces and adsorption: counting the molecules that cover a solid

From Langmuir’s balance on sites to BET multilayers: measure the surface areas of a titania, a mesoporous silica and a zeolite from nitrogen isotherms at 77 K, and learn to tell whether a fitting range can be trusted.

In this lesson

Tell physisorption from chemisorption, and explain why adsorption is almost always exothermic
Derive the Langmuir isotherm from the balance of adsorption and desorption, and say what the BET model adds
Use a BET plot to get the monolayer capacity, the constant C and the specific surface area from a nitrogen isotherm
Check a BET range with the consistency criteria, and recognise the pore structures behind Type I, II and IV isotherms

A gram of zeolite 13X has about 830 m² of surface — two basketball courts; a gram of the titania nanopowder in this lesson has about 56 m². How well catalysts, adsorbents, pigments and drug powders work depends largely on these invisible surfaces: reactions happen on them, impurities stick to them, drugs dissolve from them.

But how do you measure the surface of a pinch of powder? A ruler is useless. The trick is to count molecules: spread a single layer of gas molecules over the surface, count how many it takes, and multiply by the area each one occupies. This lesson starts from the balance at a single adsorption site and ends with the “BET surface area” measured every day in analytical laboratories — and with when that measurement gives the wrong answer.

Physisorption and chemisorption

A gas molecule that meets a solid surface can stay in two ways. Physisorption relies on van der Waals forces alone — the same forces that condense a gas to a liquid — with an enthalpy typically near −20 kJ/mol, comparable to an enthalpy of vaporisation; it is reversible, undone by pumping or gentle warming, and molecules can pile up layer on layer. Chemisorption forms a chemical bond between molecule and surface atoms, with enthalpies around −200 kJ/mol; it stops at one layer, and the molecule may dissociate — N₂ breaking its triple bond on iron in the Haber process is chemisorption.

Adsorption is almost always exothermic. A molecule pinned to a surface loses the entropy of free motion in three dimensions, so ΔS < 0; for ΔG = ΔH − TΔS to be negative, ΔH must be negative. The colder, the more is adsorbed — which is why surface areas are measured in liquid nitrogen at 77 K.

Langmuir: a dynamic balance on sites

Langmuir (1918) pictured the surface as a grid of adsorption sites, each holding at most one molecule, with no interactions between molecules. Let θ be the coverage, the fraction of sites occupied. Molecules land on empty sites at a rate proportional to the pressure p and to the empty fraction 1 − θ, and leave at a rate proportional to θ. At equilibrium the two rates are equal, $k_a p(1 - \theta) = k_d \theta$ , which gives the isotherm below with K = k_a/k_d.

\theta = \frac{Kp}{1 + Kp}

The Langmuir isotherm: at low pressure θ ≈ Kp (proportional to pressure); at high pressure θ → 1 (the sites run out).

$\theta$: coverage: fraction of sites occupied
$K$: adsorption equilibrium constant k_a/k_d
$p$: gas pressure

Predict first

A surface is half covered (θ = 0.5) at Kp = 1. Double the pressure: what does the coverage become?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

1.0 0.75 0.67 0.5

BET: molecules stacking on molecules

At 77 K nitrogen sits right at its boiling point. Nitrogen molecules already adsorbed are themselves a new “surface”, and more can stack on top as if condensing to a liquid. Brunauer, Emmett and Teller (1938) extended Langmuir’s balance to every layer: the first layer is held to the solid with a larger energy E₁, all higher layers like liquid nitrogen with the energy of vaporisation E_L; as the pressure approaches the saturation vapour pressure p⁰, the number of layers grows without limit — the gas condenses.

The result is the BET equation. It describes the amount adsorbed n in terms of the relative pressure x = p/p⁰ and a constant C, and nₘ is the amount that covers exactly one layer — the quantity we are after. The larger C, the more strongly the first layer is held compared with the ones above, and the sharper the “knee” of the isotherm at low pressure.

\frac{n}{n_m} = \frac{Cx}{(1 - x)(1 - x + Cx)} \quad\Longleftrightarrow\quad \frac{x}{n(1 - x)} = \frac{1}{n_m C} + \frac{C - 1}{n_m C}\,x

The BET equation and its linear form: plotting x/[n(1 − x)] against x should give a straight line whose slope s and intercept i yield nₘ = 1/(s + i) and C = s/i + 1.

$x$: relative pressure p/p⁰
$n$: amount adsorbed, cm³(STP)/g
$n_m$: monolayer capacity
$C$: BET constant, roughly exp[(E₁ − E_L)/RT]

Molecules per site assigned from the model’s own distribution

One layer or many: Langmuir and BET

Relative pressure p/p⁰ = 0.10 BET constant C = 74

Twenty-four adsorption sites on a solid surface

first layer (touching the solid)
upper layers (like liquid nitrogen)

Adsorption isotherm

BET
Langmuir (Kp = Cx)

Amount n/nₘ = 0.99: bare sites 11 %, one molecule 80 %, two or more 8.9 %.

The BET curve holds exactly one layer’s worth at p/p⁰ = 1/(√C + 1) = 0.104 — yet 10 % of the sites are still bare there while others already carry two.

C ≈ exp[(E₁ − E_L)/RT]: C = 74 means that at 77.3 K the first layer is bound about 2.77 kJ/mol more strongly than liquid nitrogen (whose enthalpy of vaporisation is 5.57 kJ/mol).

With C = 74 the BET curve holds exactly “one layer’s worth” at p/p⁰ = 1/(√C + 1) ≈ 0.104 — yet about 10 % of the sites are still bare while others carry two or three molecules. nₘ is a statistical amount: the adsorbed molecules spread flat would just cover the surface once. Drag x towards 0.8 and watch the stacks grow; switch to Langmuir, and the same first-layer balance never exceeds one layer.

From a monolayer to an area

a_{\text{BET}} = \frac{n_m}{V_m}\,N_A\,\sigma

Specific surface area = molecules in the monolayer × area per molecule. For nitrogen at 77 K, σ = 0.162 nm²; with n in cm³(STP)/g, V_m = 22 414 cm³/mol, so each 1 cm³(STP)/g of monolayer means 4.35 m²/g.

$n_m$: monolayer capacity, cm³(STP)/g
$V_m$: molar volume of a gas at STP
$N_A$: Avogadro constant
$\sigma$: area occupied by one adsorbed molecule

Worked example

The surface area of mesoporous silica MCM-41

The nitrogen isotherm of MCM-41 has nine points between p/p⁰ = 0.05 and 0.30; the BET plot has slope 4.42 × 10⁻³ g/cm³ and intercept 5.59 × 10⁻⁵ g/cm³. Find C, nₘ and the specific surface area.

Nitrogen at 77 K · real isotherms and a certified reference material

BET analysis: from isotherm to surface area

Lower limit p/p⁰ = 0.050 Upper limit p/p⁰ = 0.30

Adsorption isotherm

Rouquerol plot: n(1 − p/p⁰) must rise

BET plot: the selected points should fall on a line

Consistency checks

At least four points (9 now) pass
n(1 − x) rises throughout the range pass
C is positive (positive intercept) pass
nₘ lies between the amounts of the selected points pass
Linear: r² ≥ 0.999 pass

Slope 4.42 × 10⁻³, intercept 5.59 × 10⁻⁵ → C = 80.1, nₘ = 223.37 cm³(STP)/g, r² = 0.99996

BET surface area 972.2 m²/g — every check passes, so the value stands.

BET puts the monolayer at p/p⁰ = 0.10.

These are the raw data sent to 61 laboratories in an international round robin published in 2022. Between 0.05 and 0.30 the BET plot is very straight and all five checks pass. Drag the upper limit to 0.32 and beyond: the pores begin to fill with condensed nitrogen, the amount shoots up, the BET plot bends, r² drops below 0.999 and the “area” climbs past 1200 m²/g.

Predict first

Zeolite 13X’s micropores are already full of nitrogen by p/p⁰ ≈ 0.01. What happens if you still make the BET plot over the standard 0.05–0.30?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

A slightly low but usable area A negative C, and nearly every consistency check fails The same result as MCM-41 A perfectly straight line

The shape of the isotherm tells the pore size

Three common IUPAC isotherm types (micropores < 2 nm, mesopores 2–50 nm, macropores > 50 nm)
Type	Shape	Solid	Example here
I	steep rise at very low pressure, then a plateau	microporous: pores only one or two molecules wide	zeolite 13X
II	a “knee” at low pressure (monolayer), then a slow multilayer rise	nonporous or macroporous	titania SRM 1898
IV	like Type II at first, then a steep step (capillary condensation), then a plateau	mesoporous	MCM-41, step near p/p⁰ ≈ 0.33

The step in mesopores comes from capillary condensation: the vapour pressure over a curved liquid surface is lower than over a flat one (the Kelvin equation), so narrow pores flood with liquid nitrogen while p is still below p⁰, and the narrower the pore, the earlier the step. Its position can therefore be turned into a pore width — the job of other methods (BJH, density functional theory). For BET, the lesson is that the fitting range must stop short of the step.

Single point and multipoint

Quality-control laboratories often measure a single point: assuming C is so large that the intercept can be ignored, nₘ ≈ n(1 − x). The price is a systematic low bias — a single point returns only Cx/(1 − x + Cx) of the true value. For the titania reference material SRM 1898, NIST certifies both: multipoint 55.55 ± 0.70 m²/g and single point 53.85 ± 0.78 m²/g, 3.1 % lower; with the single point at p/p⁰ = 0.30 that corresponds to C ≈ 74.

Nitrogen at 77 K · real isotherms and a certified reference material

BET analysis: from isotherm to surface area

Lower limit p/p⁰ = 0.050 Upper limit p/p⁰ = 0.30

Adsorption isotherm

Rouquerol plot: n(1 − p/p⁰) must rise

BET plot: the selected points should fall on a line

Consistency checks

At least four points (11 now) pass
n(1 − x) rises throughout the range pass
C is positive (positive intercept) pass
nₘ lies between the amounts of the selected points pass
Linear: r² ≥ 0.999 pass

Slope 0.0770, intercept 1.09 × 10⁻³ → C = 71.7, nₘ = 12.80 cm³(STP)/g, r² = 0.99994

BET surface area 55.7 m²/g — every check passes, so the value stands.

BET puts the monolayer at p/p⁰ = 0.11.

NIST certificate: multipoint 55.55 ± 0.70 m²/g, single point 53.85 ± 0.78 m²/g.

The model titania is generated from the BET equation with the area and C from the NIST certificate, plus 0.3 % measurement scatter. The standard range gives a multipoint value inside the certificate’s uncertainty. Try keeping only the points from 0.20 to 0.30: the BET plot is still straight and r² is high, but nₘ comes out smaller than every selected amount — the monolayer is not in that range at all, and the third check fails; the area drifts to 56.2 m²/g, by luck not far from the true value, but nothing in the fit could tell you so. A good-looking line is not a trustworthy result.

Try the calculation

A sample has C = 100 and is measured by a single point at p/p⁰ = 0.30. By what percentage does the single-point area fall short of the multipoint area?

Try the calculation

The BET plot of a titania powder has slope 0.0770 g/cm³ and intercept 0.00109 g/cm³ (n in cm³(STP)/g). What is the specific surface area (m²/g)?

Concept check

Which observation shows that nitrogen at 77 K is physisorbed, not chemisorbed, on titania?

Adsorption continues past one layer, and pumping at the same temperature removes it completely The enthalpy of adsorption is about −200 kJ/mol The N≡N triple bond breaks Only one layer ever forms

Concept check

The SRM 1898 protocol asks for outgassing under vacuum at 200 °C for 2 to 4 hours before the measurement. What is it mainly for?

To remove adsorbed water and other vapours, which would otherwise occupy part of the surface To convert anatase into rutile To sinter the particles together To make the powder denser and easier to weigh

Key takeaways

Physisorption relies on van der Waals forces (about −20 kJ/mol), is reversible and can build multilayers; chemisorption forms chemical bonds (about −200 kJ/mol) and stops at one layer. Adsorption loses entropy, so it releases heat.
Langmuir: θ = Kp/(1 + Kp), from the dynamic balance of adsorption and desorption; BET extends the same balance to many layers, n/nₘ = Cx/[(1 − x)(1 − x + Cx)].
The slope and intercept of the BET plot give nₘ = 1/(s + i) and C = s/i + 1; each 1 cm³(STP)/g of nitrogen monolayer is 4.35 m²/g.
The fitting range must pass the checks: n(1 − x) rising, C positive, nₘ inside the range. Micropores (Type I) need very low pressures; mesopores (Type IV) must stop before capillary condensation.

Progress is stored only in this browser and never uploaded.

Data and references

I. Langmuir, “The adsorption of gases on plane surfaces of glass, mica and platinum”, J. Am. Chem. Soc. 40, 1361–1403 (1918).
S. Brunauer, P. H. Emmett & E. Teller, “Adsorption of gases in multimolecular layers”, J. Am. Chem. Soc. 60, 309–319 (1938).
M. Thommes et al., “Physisorption of gases, with special reference to the evaluation of surface area and pore size distribution (IUPAC Technical Report)”, Pure Appl. Chem. 87, 1051–1069 (2015).
J. Rouquerol, P. Llewellyn & F. Rouquerol, “Is the BET equation applicable to microporous adsorbents?”, Stud. Surf. Sci. Catal. 160, 49–56 (2007).
J. W. M. Osterrieth et al., “How reproducible are surface areas calculated from the BET equation?”, Adv. Mater. 34, 2201502 (2022); MCM-41 and zeolite 13X isotherms from the BETSI repository (MIT licence).
NIST, Certificate of Analysis, Standard Reference Material 1898, Titanium Dioxide Nanomaterial (2020); ISO 9277:2010, Determination of the specific surface area of solids by gas adsorption — BET method.
P. Atkins, J. de Paula & J. Keeler, Atkins’ Physical Chemistry, 11th edn (Oxford University Press, 2018), Focus 19.