Henry adsorption constant

If a solid body is modeled by a constant field and the structure of the field is such that it has a penetrable core, then

${\displaystyle K_{H}=\int \limits _{-\infty }^{x'}{\big [}\exp(-\beta u)-\exp(-\beta u_{0}){\big ]}dx-\int \limits _{x'}^{\infty }{\big [}1-\exp(-\beta u){\big ]}dx.}$

Here ${\displaystyle x'}$ is the position of the dividing surface, ${\displaystyle u=u(x)}$ is the external force field, simulating a solid, ${\displaystyle u_{0}}$ is the field value deep in the solid, ${\displaystyle \beta =1/k_{B}T}$, ${\displaystyle k_{B}}$ is the Boltzmann constant, and ${\displaystyle T}$ is the temperature.

Introducing "the surface of zero adsorption"

${\displaystyle x_{0}=-\int \limits _{-\infty }^{0}{\widetilde {\theta }}(x)dx+\int \limits _{0}^{\infty }{\widetilde {\varphi }}(x)dx,}$

where

${\displaystyle {\widetilde {\theta }}={\frac {\exp {(-\beta u)}-\exp {(-\beta u_{0})}}{1-\exp {(-\beta u_{0})}}}}$

and

${\displaystyle {\widetilde {\varphi }}={\frac {1-\exp {(-\beta u)}}{1-\exp {(-\beta u_{0})}}},}$

we get

${\displaystyle K_{H}(x')=[x'-x_{0}(T)][1-\exp(-\beta u_{0})]}$

and the problem of ${\displaystyle K_{H}}$ determination is reduced to the calculation of ${\displaystyle x_{0}}$.

Taking into account that for Henry absorption constant we have

${\displaystyle k_{H}=\lim _{\varrho \rightarrow 0}{\frac {\varrho (z')}{\varrho (z)}}=\exp(-\beta u_{0}),}$

where ${\displaystyle \varrho (z')}$ is the number density inside the solid, we arrive at the parametric dependence

${\displaystyle K_{H}=\int \limits _{-\infty }^{x'}{\big [}k_{H}^{{\widetilde {u}}(x)}-k_{H}{\big ]}dx-\int \limits _{x'}^{\infty }{\big [}1-k_{H}^{{\widetilde {u}}(x)}{\big ]}dx}$

where

${\displaystyle {\widetilde {u}}(x)={\frac {u(x)}{u_{0}}}.}$

If a static membrane is modeled by a constant field and the structure of the field is such that it has a penetrable core and vanishes when ${\displaystyle x=\pm \infty }$, then

${\displaystyle K_{H}=\int \limits _{-\infty }^{\infty }{\big [}\exp(-\beta u)-1{\big ]}dx.}$

We see that in this case the ${\displaystyle K_{H}}$ sign and value depend on the potential ${\displaystyle u}$ and temperature only.

If a solid body is modeled by a constant hard-core field, then

${\displaystyle K_{H}=\int \limits _{-\infty }^{x'}\exp(-\beta u)dx-\int \limits _{x'}^{\infty }{\big [}1-\exp(-\beta u){\big ]}dx,}$

or

${\displaystyle K_{H}(x')=x'-x_{0}(T),}$

where

${\displaystyle x_{0}=-\int \limits _{-\infty }^{0}\theta (x)dx+\int \limits _{0}^{\infty }\varphi (x)dx.}$

Here

${\displaystyle \theta =\exp {(-\beta u)}}$

${\displaystyle \varphi =1-\exp {(-\beta u)}.}$

For the hard solid potential

${\displaystyle x_{0}=x_{step},}$

where ${\displaystyle x_{step}}$ is the position of the potential discontinuity. So, in this case

${\displaystyle K_{H}(x')=x'-x_{step}.}$

The choice of the dividing surface, strictly speaking, is arbitrary, however, it is very desirable to take into account the type of external potential ${\displaystyle u(x)}$. Otherwise, these expressions are at odds with the generally accepted concepts and common sense.

First, ${\displaystyle x'}$ must lie close to the transition layer (i.e., the region where the number density varies), otherwise it would mean the attribution of the bulk properties of one of the phase to the surface.

Second. In the case of weak adsorption, for example, when the potential is close to the stepwise, it is logical to choose ${\displaystyle x'}$ close to ${\displaystyle x_{0}}$. (In some cases, choosing ${\displaystyle x_{0}\pm R}$, where ${\displaystyle R}$ is particle radius, excluding the "dead" volume.)

In the case of pronounced adsorption it is advisable to choose ${\displaystyle x'}$ close to the right border of the transition region. In this case all particles from the transition layer will be attributed to the solid, and ${\displaystyle K_{H}}$ is always positive. Trying to put ${\displaystyle x'=x_{0}}$ in this case will lead to a strong shift of ${\displaystyle x'}$ to the solid body domain, which is clearly unphysical.

Conversely, if ${\displaystyle u_{0}<0}$ (fluid on the left), it is advisable to choose ${\displaystyle x'}$ lying on the left side of the transition layer. In this case the surface particles once again refer to the solid and ${\displaystyle K_{H}}$ is back positive.

Thus, except in the case of static membrane, we can always avoid the "negative adsorption" for one-component systems.