Surface Chemistry: Common Mistakes and Fixes (3)

Question

A solid catalyst adsorbs $1.2 \times 10^{-3}$ mol of nitrogen per gram of catalyst at $77$ K and $1$ atm pressure. The Freundlich isotherm $x/m = k P^{1/n}$ is found to give $k = 1.0 \times 10^{-3}$ and $n = 2$. Verify the data and predict the adsorption at $4$ atm.

Solution — Step by Step

Final answer: predicted adsorption at $4$ atm $\approx 2 \times 10^{-3}$ mol/g (pure Freundlich) or $\approx 2.4 \times 10^{-3}$ mol/g if scaling for the $1$ atm offset.

Why This Works

Freundlich’s isotherm is empirical: $x/m \propto P^{1/n}$. Useful at moderate pressures. At very low pressures, Henry’s law ($x/m \propto P$) is more accurate. At very high pressures, $x/m$ saturates (Langmuir limit), and Freundlich overestimates.

The "$1/n$" power $< 1$ encodes diminishing returns: doubling pressure does not double adsorption.

Alternative Method

Use Langmuir’s isotherm $x/m = abP/(1 + bP)$. For low $P$ this reduces to $abP$ (linear). For high $P$ it saturates at $a$. JEE Advanced sometimes asks you to choose between models — Freundlich for the linear log-log range, Langmuir when saturation is hinted.