Which of the following curves is in accordance with Freundlich Adsorption Isotherm?

The correct answer is option 3.

The variation of extent of adsorption \((x/m)\) with pressure \((P)\) at a particular temperature was given mathematically by Freundlich in 1909. From the adsorption isotherm (Fig. 1), the following observations can be easily made:

(i) At low pressure, the graph is almost straight line which indicates that \(x/m\) is directly proportional to pressure (Fig. 2). This may be expressed as:

\[\frac{x}{m} \propto P\, \ or\, \ \frac{x}{m}= kP \, \ ------(i)\]

where \(k\) is a constant.

(ii) At high pressure, the graph becomes almost constant which means that \(x/m\) becomes independent of pressure (Fig. 2). This may be expressed as:

\[\frac{x}{m} = \, \ constant\, \ or\, \ \frac{x}{m}\propto \, \ P^0 \, \ (\text{Because} P^0 = 1)\]

\[or\, \ \frac{x}{m} = kP^0\, \ ------(ii)\]

(iii) Thus, in the intermediate range of pressure, \(x/m\) will depend upon the power of pressure which lies between 0 to 1 i.e., fractional power of pressure (probable range 0.1 to 0.5). This may be expressed as

\[\frac{x}{m}\, \ \propto \, \ P^{1/n}\]

\[or, \, \ \frac{x}{m} = kP^{1/n}\, \ -------(iii)\]

where \(n\) can take any whole number value which depends upon the nature of adsorbate and adsorbent. The above relationship is also called Freundlich's adsorption isotherm and is shown in Fig. 2.

Calculation of k and n of adsorption isotherm

The constants \(k\) and \(n\) can be determined as explained below:

Taking logarithms on both sides of Eq. (iii), we get

\[log\frac{x}{m} = log k\, \ + \, \ \frac{1}{n}log P\]

Thus, if we plot a graph between log (x/m) on y-axis (ordinate) and log P, on x-axis (abscissa), straight line will be obtained. This also shows the validity of Freundlich isotherm. The slope of the line (Fig. 3) is equal to \(1/n\) and the intercept is equal to \(log k\).