Multicomponent

We can treat adsorption from a mixture of gases (at low pressures) by extending the Langmuir isotherm:

$A(g) + S \rightleftharpoons A(ads)$

$$B(g) + S \rightleftharpoons B(ads)$$

These reactions are not independent because both species compete for the same surface sites. The two equilibria must be solved simultaneously:

$$\frac{[A(ads)]}{P_A\times[S]} = K_A$$

$\frac{[B(ads)]}{P_B\times[S]} = K_B$

As for the one-component case, we define surface coverages $\theta_A = [A(ads)]/[S]_0$ and $\theta_B = [B(ads)]/[S]_0$, and note that

[S] + [A(ads)] + [B(ads)] = [S]_0$, or $[S]/[S]_0 = 1 - \theta_A - \theta_B

Then,

$$K_A = \frac{\theta_A}{P_A\times(1-\theta_A - \theta_B)}$$

$K_B = \frac{\theta_B}{P_B\times(1-\theta_A - \theta_B)}$

and dividing, we get

$$\frac{\theta_A}{\theta_B} = \frac{P_A}{P_B}\frac{K_A}{K_B}$$

The ratio of A:B on the surface is therefore different than the ratio of A:B in the gas phase, by precisely ${K_A}/{K_B}$, so the more strongly adsorbed component is enriched on the surface. This is referred to as the thermodynamic selectivity, the preference of the surface for one species of a mixture, and is the basis for many techniques to separate gas mixtures.