Langmuir and BET isotherms: how much gas a surface holds, and how big the surface is

An isotherm is a record of how full the surface gets as you add gas

Intuition

Think of a car park. At first, cars take empty bays easily. As it fills, finding a free bay gets harder. When every bay is taken, extra cars have to double-park or wait. An adsorption isotherm is that story for gas molecules and a solid: how many molecules are stuck on at each pressure, with the temperature held fixed.

An isotherm gives the amount of gas adsorbed on a solid as a function of the relative gas pressure $p/p^0$. $p^0$ is the vapour pressure of the gas at that temperature. It is useful for finding the surface area and surface chemistry of a porous sample. The lecture’s historical example is Brunauer and Emmett’s iron catalyst data. They found the monolayer volume by extrapolating the straight part of the curve back to zero pressure (the “Point B method”), which the BET method has since replaced.

The two simplest models are Langmuir (one layer) and BET (many layers). Isotherms start to look alike when you normalise by the saturation vapour pressure, $P = p/p^0$. Brunauer sorted real isotherms into five types. Types I, II and IV come from strong gas-solid interaction, and type I is pure chemisorption. Types III and V come from weak interaction and are rarer.

Langmuir assumes one layer, equal sites and sites that ignore each other

Intuition

Langmuir’s car park has a fixed number of identical bays. A car parks in any free bay, and it does not matter whether the bay next door is taken. Once every bay is full, nothing more can stick. That is the one-layer limit.

The three assumptions are:

1. The maximum coverage is one monolayer (chemisorption). It works well for Type I isotherms.
2. Every adsorption site is equivalent.
3. The ability of a particle to bind does not depend on whether nearby sites are occupied.

Adsorption is treated as one equilibrium between bare sites $S_0$ and occupied sites $S_1$:

$X + S_0 \rightleftharpoons S_1$

$n_{ads}$ is the amount adsorbed and $n_{mono}$ is the amount needed for a monolayer. The fractional coverage is $\theta = n_{ads}/n_{mono}$.

$\theta = \frac{n_{ads}}{n_{mono}} = \frac{K_1[X]}{1 + K_1[X]} \qquad \text{(solute, concentration } [X]\text{)}$

$\theta = \frac{n_{ads}}{n_{mono}} = \frac{K_1 p}{1 + K_1 p} \qquad \text{(gas, pressure } p\text{)}$

Read the gas version in plain words. At low pressure $K_1 p$ is small, so $\theta \approx K_1 p$ and coverage rises in a straight line. At high pressure $K_1 p$ is large, so $\theta \to 1$ and the surface is full.

Straight lines are easier to fit than curves. Rearranging gives the linearised Langmuir isotherm:

$\frac{[X]}{n_{ads}} = \frac{[X]}{n_{mono}} + \frac{1}{K_1 n_{mono}} \qquad\qquad \frac{p}{n_{ads}} = \frac{p}{n_{mono}} + \frac{1}{K_1 n_{mono}}$

You control $[X]$ or $p$ and measure $n_{ads}$. A plot of $p/n_{ads}$ against $p$ is a straight line. The slope is $1/n_{mono}$ and the intercept is $1/(K_1 n_{mono})$, so one line gives both $n_{mono}$ and $K_1$. The lecture’s example is benzene on graphite, where the second panel is a straight line.

A bigger $K_1$ gives a steeper rise at low pressure, meaning higher affinity between gas and solid. All the curves plateau at 1, one monolayer, and all are Type I. If you know the size of the gas molecule, $n_{mono}$ gives the surface area of the solid.

BET lets layers pile up, with c telling you how keen the first layer is

BET is a slightly more elaborate model. Its assumptions:

1. More than one monolayer can adsorb, up to thick layers but not too thick.
2. Adsorption into layers 2, 3, 4 and so on all have the same equilibrium constant $K$ (not necessarily the same as the first layer’s).
3. When the layer becomes very thick, it behaves like bulk liquid.

Sites are bare ($S_0$), or covered by one ($S_1$), two ($S_2$) or three ($S_3$) molecules. Assumption 2 matters. It is the same effective constant that describes condensation, and it can be written as a free energy of condensation:

$\Delta G^\circ = -RT\ln K = \Delta H^\circ - T\Delta S^\circ$

The BET equation is

$\theta = \frac{n_{ads}}{n_{mono}} = \frac{cP}{(1-P)\left(1 + (c-1)P\right)}$

$\theta$ is the fractional coverage and may exceed 1 once multilayers form. $P = p/p^0$ is the pressure relative to where the liquid condenses. $c = K_1/K$ is the BET parameter, the relative ease of adsorbing on the first layer against all other layers.

Changing $c$ changes the shape. As $c$ increases, the first layer forms more readily and the bend is more pronounced. Multilayers always form as $p/p^0$ approaches 1. Raising $n_{mono}$ (a larger surface area) changes the y-scale but not the shape. BET covers Type I to III isotherms by changing $c$: type I is chemisorption of one layer, with $c \to \infty$. High affinity is $c > 1$ and low affinity is $c < 1$.

The BET equation can also be linearised, giving a straight line from which you read $n_{mono}$ and $c$:

$\frac{p}{n_{ads}\left(p^0 - p\right)} = \frac{1}{c\,n_{mono}} + \frac{c-1}{c\,n_{mono}}\,\frac{p}{p^0}$

In the form $y = b + mx$ you have $y = p/[n_{ads}(p^0 - p)]$, $x = p/p^0$, intercept $b = 1/(c\,n_{mono})$ and slope $m = (c-1)/(c\,n_{mono})$. The same equation works with volumes in place of moles:

$\frac{p}{V_{ads}\left(p^0 - p\right)} = \frac{1}{c\,V_{mono}} + \frac{c-1}{c\,V_{mono}}\,\frac{p}{p^0}$

To first approximation, moles and volume swap through the ideal gas equation, $pV = nRT$.

Turning a monolayer amount into a surface area takes three conversions

Try it: worked example

Nitrogen adsorbs on a porous sample and monolayer adsorption occurs at 5.0 mg of adsorbed N2. An adsorbed N2 molecule occupies 0.162 nm^2. Find the surface area.

Show answer and steps

Answer17.4 m^2 (with the slide's rounded steps).

1. Moles of N₂ in the monolayer: $\dfrac{5.0\times10^{-3}\ \text{g}}{28\ \text{g mol}^{-1}} = 0.178\times10^{-3}\ \text{mol}$.
2. Number of molecules: $0.178\times10^{-3}\ \text{mol}\times6.022\times10^{23}\ \text{molecules mol}^{-1} = 1.08\times10^{20}\ \text{molecules}$.
3. Area: $1.08\times10^{20}\times0.162\times10^{-18}\ \text{m}^2\ \text{molecule}^{-1} = 17.4\ \text{m}^2$.
4. A note on rounding: $0.162\times10^{-18}\ \text{m}^2 = 0.162\ \text{nm}^2$. Using the rounded $1.08\times10^{20}$ gives $17.5\ \text{m}^2$, while carrying the unrounded value gives $17.4\ \text{m}^2$. The slide prints $17.4\ \text{m}^2$, so use that.

Try it: worked example

Nitrogen on silica gives a linear BET plot with slope 319 mol^-1 and intercept 3.35 mol^-1. One N2 covers 0.162 nm^2. Find c, n_mono and the surface area.

Show answer and steps

Answerc = 96.2, n_mono = 3.10 x 10^-3 mol, area = 302 m^2.

1. Slope: $\dfrac{c-1}{c\,n_{mono}} = 319\ \text{mol}^{-1}$. Intercept: $\dfrac{1}{c\,n_{mono}} = 3.35\ \text{mol}^{-1}$.
2. Divide the slope by the intercept. The $c\,n_{mono}$ cancels and leaves $c - 1 = 319/3.35$.
3. So $c = 319\times\dfrac{1}{3.35} + 1 = 96.2$.
4. Put $c$ back into the intercept: $n_{mono} = \dfrac{1}{96.2\times3.35} = 3.10\times10^{-3}\ \text{mol}$.
5. Area: $3.10\times10^{-3}\times0.162\times10^{-18}\times6.02\times10^{23} = 302\ \text{m}^2$ (moles, times area per molecule in m², times Avogadro’s number).

Pores break BET because gas condenses inside them early

Real porous solids deviate from BET because of capillary condensation into pores below the saturated vapour pressure. Slit pores form in layered minerals (clays, graphite, talc, MoS₂). Pores also form where small particles meet in a powder, especially if they are deformable or sintered. Porous ceramics often show Type IV or V isotherms, with pore condensation and even hysteresis. BET cannot describe highly porous solids near their condensation point.

Exam detail

Learn the two Langmuir lines and the BET line as $y = b + mx$ with the slope and intercept written out. Know the full chain of the two worked examples: moles, molecules, area, and slope and intercept to $c$, then $n_{mono}$, then area.

Pitfall

In the BET line the slope is $(c-1)/(c\,n_{mono})$ and the intercept is $1/(c\,n_{mono})$. Mixing them up gives the wrong $c$. Divide slope by intercept first: that gives $c - 1$.

Recall

Why is the straight-line form of Langmuir or BET easier to use than the raw curve?

You control the pressure and measure the amount adsorbed. Plotting the rearranged quantities gives a straight line, and a straight line’s slope and intercept give $n_{mono}$ and $K_1$ (or $c$) directly. Fitting a curve by eye is much harder.