Surface Chemistry: Numerical Problems Set (1)

Question

According to the Freundlich adsorption isotherm, the mass of a gas adsorbed per gram of adsorbent ($x/m$) at a pressure of $0.5$ atm is $0.4$ g and at $1.0$ atm is $0.8$ g. Find the values of $k$ and $n$ in the Freundlich equation.

Solution — Step by Step

Why This Works

The Freundlich isotherm is an empirical relation that describes adsorption at moderate pressures. When $n = 1$, the relationship is linear — adsorption is directly proportional to pressure, like Henry’s law for dissolved gases. For $n > 1$, adsorption increases more slowly than pressure (saturating).

Taking logs converts the power law into a linear form, which is convenient when given multiple data points (you’d plot $\log(x/m)$ vs. $\log P$ and the slope gives $1/n$, intercept gives $\log k$).

Alternative Method

Take the ratio of the two data points to eliminate $k$:

$\frac{(x/m)_1}{(x/m)_2} = \left(\frac{P_1}{P_2}\right)^{1/n}$

$\frac{0.4}{0.8} = \left(\frac{0.5}{1.0}\right)^{1/n}$

$0.5 = (0.5)^{1/n} \implies 1/n = 1 \implies n = 1$

Then back-substitute to find $k$. Same answer, fewer algebraic steps.

Common Mistake

Students sometimes confuse the Freundlich isotherm with the Langmuir isotherm:

$\frac{x}{m} = \frac{abP}{1 + aP}$

The two have different functional forms and different applicability. Freundlich is empirical (works at moderate pressure); Langmuir assumes monolayer adsorption (valid up to higher pressures, saturates as $P \to \infty$).

Another trap: writing $1/n$ vs $n$ in the formula. The Freundlich isotherm uses $P^{1/n}$, not $P^n$. Read it as “P to the (1 over n)”.