Adsorption isotherms — Langmuir vs Freundlich comparison with graphs

Question

Compare the Langmuir and Freundlich adsorption isotherms. When does each apply, and how do their graphs differ? For the Freundlich isotherm $\frac{x}{m} = kP^{1/n}$ , what does the value of $1/n$ indicate?

(CBSE Class 12 / JEE Main pattern)

Solution — Step by Step

An adsorption isotherm is a graph of the amount adsorbed ( $x/m$ ) versus pressure ( $P$ ) at constant temperature. It answers: “How much gas sticks to the surface as we increase pressure?”

\frac{x}{m} = kP^{1/n} \quad (n > 1)

Taking log on both sides: $\log\frac{x}{m} = \log k + \frac{1}{n}\log P$

This is a straight line when plotted as $\log(x/m)$ vs $\log P$ , with slope $= 1/n$ and intercept $= \log k$ .

The value of $1/n$ lies between 0 and 1. If $1/n = 0$ , adsorption is independent of pressure (saturation). If $1/n = 1$ , adsorption is directly proportional to pressure. Most real cases fall between these limits.

\frac{x}{m} = \frac{aP}{1 + bP}

At low pressure ( $bP \ll 1$ ): $x/m \approx aP$ — linear, like Freundlich with $1/n = 1$ .

At high pressure ( $bP \gg 1$ ): $x/m \approx a/b$ — constant (saturation). This is the key feature Langmuir adds that Freundlich misses.

flowchart TD
    A["Choose isotherm model"] --> B{"Pressure range?"}
    B -->|"Low to moderate P"| C["Freundlich works\nx/m = kP^(1/n)"]
    B -->|"Full range including\nhigh P saturation"| D["Langmuir needed\nx/m = aP/(1+bP)"]
    C --> E["Limitation: does not\npredict saturation"]
    D --> F["Assumes: monolayer,\nuniform surface,\nno interaction between\nadsorbed molecules"]

Feature	Freundlich	Langmuir
Formula	$x/m = kP^{1/n}$	$x/m = aP/(1+bP)$
Saturation	Does not predict	Predicts monolayer saturation
Surface assumption	Non-uniform (heterogeneous)	Uniform (homogeneous)
Molecular interaction	Allows	No interaction assumed
Validity	Low to moderate pressure	Full pressure range
Linear form	$\log(x/m)$ vs $\log P$	$P/(x/m)$ vs $P$

Why This Works

Freundlich’s isotherm is empirical — it fits experimental data well at moderate pressures but fails at high pressures because it has no built-in saturation limit. Langmuir’s model is derived from first principles: assuming each surface site can hold at most one molecule (monolayer), all sites are equivalent, and adsorbed molecules don’t interact. This gives the $aP/(1+bP)$ form that naturally saturates.

Real surfaces often have both types of behaviour — Freundlich-like at low coverage and Langmuir-like at high coverage. The BET isotherm extends Langmuir to multilayer adsorption, but that is beyond CBSE/JEE Main scope.

Alternative Method — Linearisation for Graph Interpretation

For Langmuir, rearrange as: $\frac{P}{x/m} = \frac{1}{a} + \frac{b}{a}P$

Plot $P/(x/m)$ vs $P$ — you get a straight line with slope $b/a$ and intercept $1/a$ .

In JEE Main MCQs, you will often see a graph and need to identify which isotherm it represents. Freundlich: curve that keeps rising (no plateau). Langmuir: curve that rises and then flattens to a plateau. If the graph is a straight line with log axes, it is the linearised Freundlich form.

Common Mistake

Students often write $1/n > 1$ for the Freundlich isotherm. The correct range is $0 < 1/n < 1$ (since $n > 1$ ). If $1/n$ were greater than 1, adsorption would increase faster than linearly with pressure, which contradicts the physical expectation that surface sites become harder to fill as coverage increases.

Question

Solution — Step by Step

Why This Works

Alternative Method — Linearisation for Graph Interpretation

Common Mistake

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