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Multicomponent

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

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

B(g)+SB(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:

[A(ads)]PA×[S]=KA\frac{[A(ads)]}{P_A\times[S]} = K_A

[B(ads)]PB×[S]=KB\frac{[B(ads)]}{P_B\times[S]} = K_B

As for the one-component case, we define surface coverages θA=[A(ads)]/[S]0\theta_A = [A(ads)]/[S]_0 and θB=[B(ads)]/[S]0\theta_B = [B(ads)]/[S]_0, and note that

[S]+[A(ads)]+[B(ads)]=[S]0[S] + [A(ads)] + [B(ads)] = [S]_0, or [S]/[S]0=1θAθB[S]/[S]_0 = 1 - \theta_A - \theta_B

Then,

KA=θAPA×(1θAθB)K_A = \frac{\theta_A}{P_A\times(1-\theta_A - \theta_B)}

KB=θBPB×(1θAθB)K_B = \frac{\theta_B}{P_B\times(1-\theta_A - \theta_B)}

and dividing, we get

θAθB=PAPBKAKB\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 KA/KB{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.