Surface Chemistry: Step-by-Step Worked Examples (6)

Question

The Freundlich adsorption isotherm at $25°C$ for the adsorption of acetic acid on activated charcoal gives a straight line on a $\log(x/m)$ vs $\log p$ plot, with slope $0.4$ and y-intercept $0.7$. Find the values of $n$ and $k$ in the Freundlich equation $\dfrac{x}{m} = k p^{1/n}$. NEET 2024 pattern.

Solution — Step by Step

Why This Works

The Freundlich isotherm is empirical — it fits experimental adsorption data over a moderate pressure range better than Langmuir’s monolayer model in many cases. Plotting in log-log coordinates linearises the power law, so we can extract $n$ and $k$ from a simple linear fit.

The exponent $1/n$ is always less than 1 in practice ($n > 1$), because adsorption rate slows as the surface fills up. If $n = 1$, the isotherm reduces to a linear (Henry-law-like) form.

Alternative Method

If you have specific $(p, x/m)$ data points instead of slope and intercept, take ratios. From two points: $\dfrac{(x/m)_2}{(x/m)_1} = \left(\dfrac{p_2}{p_1}\right)^{1/n}$. Solve for $n$, then back-substitute to find $k$.