The Henry adsorption constant is the constant appearing in the linear adsorption isotherm, which formally resembles Henry's law; therefore, it is also called Henry's adsorption isotherm. It is named after British chemist William Henry. This is the simplest adsorption isotherm in that the amount of the surface adsorbate is represented to be proportional to the partial pressure of the adsorptive gas:[1]
X=KHP
where:
For solutions, concentrations, or activities, are used instead of the partial pressures.
The linear isotherm can be used to describe the initial part of many practical isotherms. It is typically taken as valid for low surface coverages, and the adsorption energy being independent of the coverage (lack of inhomogeneities on the surface).
The Henry adsorption constant can be defined as:[2]
KH=\lim\varrho →
\varrhos | |
\varrho(z) |
,
where:
\varrho(z)
\varrhos
Source:[2]
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
KH=
x' | |
\int\limits | |
-infty |
[\exp(-\betau)-\exp(-\betau0) ]dx-
infty | |
\int\limits | |
x' |
[1-\exp(-\betau) ]dx.
Here
x'
u=u(x)
u0
\beta=1/kBT
kB
T
Introducing "the surface of zero adsorption"
x0=-
0 | |
\int\limits | |
-infty |
\widetilde{\theta}(x)dx+
infty | |
\int\limits | |
0 |
\widetilde{\varphi}(x)dx,
where
\widetilde{\theta}=
\exp{(-\betau) | |
- |
\exp{(-\betau0)}}{1-\exp{(-\betau0)}}
and
\widetilde{\varphi}=
1-\exp{(-\betau) | |
}{1 |
-\exp{(-\betau0)}},
we get
KH(x')=[x'-x0(T)][1-\exp(-\betau0)]
and the problem of
KH
x0
Taking into account that for Henry absorption constant we have
kH=\lim\varrho →
\varrho(z') | |
\varrho(z) |
=\exp(-\betau0),
where
\varrho(z')
KH=
x' | |
\int\limits | |
-infty |
[
\widetilde{u | |
k | |
H |
(x)}-kH ]dx-
infty | |
\int\limits | |
x' |
[1-
\widetilde{u | |
k | |
H |
(x)} ]dx
where
\widetilde{u}(x)=
u(x) | |
u0 |
.
Source:[2]
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
x=\pminfty
KH=
infty | |
\int\limits | |
-infty |
[\exp(-\betau)-1 ]dx.
We see that in this case the
KH
u
Source:[3]
If a solid body is modeled by a constant hard-core field, then
KH=
x' | |
\int\limits | |
-infty |
\exp(-\betau)dx-
infty | |
\int\limits | |
x' |
[1-\exp(-\betau) ]dx,
or
KH(x')=x'-x0(T),
where
x0=-
0 | |
\int\limits | |
-infty |
\theta(x)dx+
infty | |
\int\limits | |
0 |
\varphi(x)dx.
Here
\theta=\exp{(-\betau)}
\varphi=1-\exp{(-\betau)}.
For the hard solid potential
x0=xstep,
where
xstep
KH(x')=x'-xstep.
The choice of the dividing surface, strictly speaking, is arbitrary, however, it is very desirable to take into account the type of external potential
u(x)
First,
x'
Second. In the case of weak adsorption, for example, when the potential is close to the stepwise, it is logical to choose
x'
x0
x0\pmR
R
In the case of pronounced adsorption it is advisable to choose
x'
KH
x'=x0
x'
Conversely, if
u0<0
x'
KH
Thus, except in the case of static membrane, we can always avoid the "negative adsorption" for one-component systems.