6
$\begingroup$

Why does Freundlich adsorption isotherm not have a term for surface area?

The formula given by Freundlich adsorption isotherm:

$$ \frac{x}{m}= k P ^{1/n}$$ Where $x$ is the mass of gas adsorbed on mass $m$ of the adsorbent and pressure $P$. $k$ and $n$ are constants which depend on the nature of the absorbent and the gas at a particular temperature.

There is no term for the surface area involved for adsorption and I had read that adsorption increases with the increase in surface area so if I take a highly porous substance, it will adsorb more gas on it. What am I missing here?

Edit: Actually, what I mean is that, let's say I have a single piece of adsorbent, it has a certain value of $x$ for a given $P$. Then I break it into 10 pieces, The value given by the formula is still the same, while there will for sure be an increase in $x$, due to increased surface area.

$\endgroup$
  • $\begingroup$ The equation is empirical where $x/m$ is equivalent to the fractional surface coverage often labelled $\theta$ as in the Langmuir isotherm. $k$ and $n >1$ are found by fitting the data. The Freundlich isotherm is unrealistic as there is no limiting value for $\theta$ but data is often followed by this expression for $\theta $ in the range $0.2\to 0.8$ $\endgroup$ – porphyrin Jun 28 at 15:53
5
$\begingroup$

The ratio

$$\frac{x}{m}= k P ^{1/n}$$

can be re-expressed in terms of the area $A$, surface-to-volume ratio $\lambda$, and density $\rho$ (note that $m=A\rho/\lambda)$:

$$\frac{x}{A}= k' P ^{1/n}$$

where

$$k'=k\rho/\lambda$$

The advantage of using $k$ as opposed to $k'$ is that you don't need to determine $\lambda$ to know how much (as in, what mass) adsorbent to use in a particular application (see e.g. Table 4 in ref 1).

There are two types of adsorbent material you are usually interested in:

  1. Porous material; in this case, breaking up the particles will not significantly increase $A$, and therefore not change k (since $\lambda$ will remain approx constant).
  2. Small particles with high $A/V$ ratio and adsorption on an outer surface of the particle. Now breaking up particles into smaller ones matters a lot, but this would be reflected in a different value of $k$, since $\lambda$ increases and $k\propto\lambda$.

Parameter $k'$ is more fundamental than $k$, as it reflects how much surface area the adsorbed molecules require, keeping in mind the range of applicability of the equation, as it does not predict e.g. monolayer saturation. Note that in neither scenario would a reduction in the size of the particles be expected to change $k'$ by much - certainly not in the first case, and, in the second case, depending on the importance of changes in adsorbent curvature or edge/corner effects.

Reference

  1. G. P. Jeppu, T. P. Clement, "A modified Langmuir-Freundlich isotherm model for simulating pH-dependent adsorption effects," Journal of Contaminant Hydrology 2012, 129–130, 46–53 (DOI: 10.1016/j.jconhyd.2011.12.001).
$\endgroup$
  • $\begingroup$ Actually, what I mean is that ,let's say I have a single piece of Adsorbent, it has a certain value of x for a given P. Then I break it into 10 pieces , The value given by the formula is still the same, while there will for sure be an increase is x, dur to increased surface area. $\endgroup$ – RandomAspirant Jun 28 at 17:54
  • $\begingroup$ @user232243 there are two types of adsorbent material you are usually interested in. (1) Porous material; in this case, breaking up the particles will not significantly increase A, and therefore not change k (since $\lambda$ will remain approx constant). (2) Small particles with high A/V ratio. Now breaking up particles into smaller ones matters a lot, but this would be reflected in a different value of k, since $\lambda$ increases and $k\propto\lambda$. $\endgroup$ – Buck Thorn Jun 28 at 21:36
  • 2
    $\begingroup$ @user232243, echoing Buck thorn here, you say in the question that $k$ depends on the "nature of the absorbent", and what this answer is saying, I think, is that part of what "nature of the absorbent" is, is, well, its specific surface area. So if you break up the absorbent into ten pieces, then its "nature" has changed and thus so would the value of $k$. $\endgroup$ – Curt F. Jun 29 at 0:55
  • $\begingroup$ Oh, ok Thanks a lot :) $\endgroup$ – RandomAspirant Jun 29 at 3:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.