Adsorption isotherm

An adsorption isotherm relates equilibrium uptake to surrounding pressure or concentration at fixed temperature. For a gas, specify the partial pressure of the adsorbing chemical species. An uptake trace during incomplete equilibration need not be an adsorption isotherm.

For the simple one-solute interface convention, surface excess per molecule is related to the derivative of surface tension $\gamma$ with respect to the logarithm of solute activity $a$ by the displayed equation. For ideal dilute behavior one may use concentration ratios in place of activity ratios. In molar units, ${\Gamma }_{\mathrm{mol}}=-(RT{)}^{-1}d\gamma /d\ln a$. A local molecular area estimate is $\Sigma =-k_{B}T/(d\gamma /d\ln a)$. Ionic surfactants can require counterion and activity factors; the simple single-component expression must not be transferred to them without specifying those constraints.

Consider an ideal gas in chemical equilibrium with homogeneous, identical adsorption sites at fixed temperature. Each intact molecule occupies one site; only one monolayer is allowed. Occupation of distinct sites is independent: neighboring bound molecules do not change binding or release rates. The simplest Langmuir adsorption model has one adsorbing chemical species, no splitting of molecules, and no competitors. Irving Langmuir's 1918 paper treats simple adsorption alongside more complicated cases; its scope exceeds this particular model.^[ref]

The derivation is a balance between arrivals finding vacancies and departures from occupied adsorption sites. Let $k_{\mathrm{a}}P$ be the arrival-and-binding rate per vacant site and $k_{\mathrm{d}}$ the departure rate per occupied site. Thus $k_{\mathrm{a}}$ has units of inverse pressure per unit time, whereas $k_{\mathrm{d}}$ has units of inverse time. The mean surface coverage obeysAt chemical equilibrium, setting the two terms equal givesThe product $bP$ is dimensionless. Here $P$ is the adsorbing gas's partial pressure, expressed in modern bar, with $1\mathrm{bar}=10^5\mathrm{Pa}$, and $b$ is expressed in ${\mathrm{bar}}^{-1}$. The pascal is the SI unit of pressure. This dimensional $b$ is not itself a dimensionless thermodynamic equilibrium constant. A larger $b$ gives greater surface coverage at the same $P$ and temperature, but does not specify how quickly chemical equilibrium is reached: multiplying both rate coefficients by the same factor leaves $b$ unchanged.

Choose $b=0.5$, $2$, and $8{\mathrm{bar}}^{-1}$ as illustrative parameters at one fixed temperature, without assigning a particular material. At $P=0.5\mathrm{bar}$, substitution gives, respectively,Solving $\theta =1/2$ gives $P_{1/2}=1/b$: the three half-occupation pressures are $2$, $0.5$, and $0.125\mathrm{bar}$. For $b=2{\mathrm{bar}}^{-1}$, reaching $\theta =0.9$ instead requires $bP=9$, or $P=4.5\mathrm{bar}$. If the complete-monolayer capacity is stipulated to be $q_{\mathrm{m}}=10\mathrm{mmol}{\mathrm{g}}^{-1}$, then $q=q_{\mathrm{m}}\theta$ predicts $q=5\mathrm{mmol}{\mathrm{g}}^{-1}$ at $P=0.5\mathrm{bar}$. This last number is an original hypothetical example, not a measured uptake.

The shape follows directly from the finite number of adsorption sites. For $bP\ll 1$, $\theta \approx bP$: nearly every arriving molecule finds a vacancy. For $bP\gg 1$, $1-\theta =1/(1+bP)\approx 1/(bP)$: vacancies become scarce. More formally,Thus surface coverage rises with diminishing slope and approaches one only as $P\to \infty$. This saturation is a site-capacity limit, not a claim that the gas has reached its condensation pressure. A second layer, different site binding strengths, interactions between bound molecules, or a changing surface invalidate parts of the setup. A good fit alone does not establish identical adsorption sites: a restricted pressure range can hide such differences. The Langmuir adsorption model is useful because it makes the capacity and the pressure scale of filling separate, testable quantities.

Competitive adsorption occurs when different chemical species seek the same adsorption sites. For intact molecules $A$ and $B$, retain identical adsorption sites and no interactions, but replace the vacant fraction by $1-{\theta }_A-{\theta }_B$. The same equilibrium balance givesHere $P_A$ and $P_B$ are partial pressures, not total pressure. Increasing $P_B$ can reduce ${\theta }_A$ at unchanged $P_A$: replacing $P_A$ by total pressure in a single-species Langmuir isotherm loses this competition.

In dissociative adsorption, an arriving molecule splits into bound fragments. For the idealized process $A_2+2\ast \rightleftharpoons 2A\ast$, let $\ast$ denote a vacant adsorption site and let $\theta$ count sites occupied by $A$. A mean-field approximation with independent vacancies takes the forward rate proportional to $P(1-\theta {)}^2$ and the reverse rate proportional to ${\theta }^2$. Equating them yieldswith a positive coefficient $K$ of inverse-pressure units that includes the chosen pair-counting convention. This square-root dependence illustrates why the intact-molecule Langmuir isotherm cannot simply be assumed for splitting molecules. Requirements for neighboring vacancies or interactions can further change the mean-field approximation.