Answer the question on the basis of passage given below:
Adsorption is the phenomenon of attracting and retaining the molecules of a substance on the surface of a solid resulting into a higher concentration on the surface than in the bulk. The variation in the amount of gas adsorbed by the adsorbent with pressure at constant temperature is expressed as isotherm. A catalyst is a substance, when adsorbed enhances the rate of a chemical reaction without itself getting used up in the reaction. Depending on the type of the particles of the dispersed phase, colloids are classified as multimolecular, macromolecular and associated colloids and soap is such an example of associated colloids. Colloids show various properties such as Tyndall effect, colour, Brownian movement and charge development. There are positively charged sols and negatively charged sols. Due to this, electrophoresis and electroosmosis are shown by colloids.
Following is not an equation for Freundlich adsorption isotherm-
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\log\frac{x}{m}=\log k-\frac{1}{n}\log P\)
The Freundlich adsorption isotherm is an empirical relation between the mass of gas adsorbed per unit mass of adsorbent (x/m) and the pressure at a fixed temperature:
\( \displaystyle \frac{x}{m} = k\,P^{1/n} \quad (n \gt 1) \)
Taking logarithms of both sides gives the straight-line form, with slope 1/n and intercept log k:
\( \displaystyle \log\frac{x}{m} = \log k + \frac{1}{n}\log P \)
The term (1/n) log P is added, because x/m increases with pressure. Option (4) subtracts it, which would mean adsorption falls as pressure rises, so it is not a form of the Freundlich isotherm.
Option (1) \(\frac{x}{m}=k.P^{\frac{1}{n}}\) — A form of the Freundlich isotherm. This is the isotherm itself for adsorption of a gas.
Option (2) \(\log\frac{x}{m}=\log k+\frac{1}{n}\log P\) — A form of the Freundlich isotherm. This is its logarithmic form; log x/m against log P is a straight line of slope 1/n.
Option (3) \(\frac{x}{m}=k.c^{\frac{1}{n}}\) — A form of the Freundlich isotherm. This is the same isotherm for adsorption from solution, with concentration in place of pressure.
Option (4) \(\log\frac{x}{m}=\log k-\frac{1}{n}\log P\) — Not a Freundlich equation. The (1/n) log P term must be added, not subtracted.
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic is not there in the latest NCERT book (2026-27), and it is also out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
Freundlich Adsorption Isotherm – “Freundlich, in 1909, gave an empirical relationship between the quantity of gas adsorbed by unit mass of solid adsorbent and pressure at a particular temperature. The relationship can be expressed by the following equation: \(\frac{x}{m} = k.P^{1/n}\) (n > 1) … Taking logarithm of eq. (5.1) \(\log\frac{x}{m} = \log k + \frac{1}{n}\log P\)”
Adsorption from Solution Phase – “… The precise mechanism of adsorption from solution is not known. Freundlich’s equation approximately describes the behaviour of adsorption from solution with a difference that instead of pressure, concentration of the solution is taken into account, i.e., \(\frac{x}{m} = k\,C^{1/n}\)”