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Concept 03 / 06Surface Chemistry

Adsorption Isotherms: Freundlich Equation and Applications of Adsorption

Adsorption isotherms, the Freundlich adsorption isotherm and its logarithmic form, behaviour at low and high pressure, adsorption from solution, limitations, and applications of adsorption from gas masks and decolourisation to chromatography, froth flotation and adsorption indicators.

Figure 1
An adsorption isotherm shows how the extent of adsorption xm\frac{x}{m} (mass of adsorbate xx per unit mass mm of adsorbent) varies with pressure at constant temperature (see figure). Freundlich adsorption isotherm (1909). An empirical relation: xm=kp1/n\frac{x}{m} = kp^{1/n}, with n>1n > 1, where kk and nn depend on the adsorbent, the gas and the temperature. At a fixed pressure, xm\frac{x}{m} is smaller at higher temperature (physisorption). Taking logarithms: log⁡xm=log⁡k+1nlog⁡p\log\frac{x}{m} = \log k + \frac{1}{n}\log p. A plot of log⁡xm\log\frac{x}{m} against log⁡p\log p is a straight line with slope 1n\frac{1}{n} (between 00 and 11) and intercept log⁡k\log k. Limiting behaviour: when 1n=0\frac{1}{n} = 0, xm\frac{x}{m} is constant, independent of pressure (saturation at high pressure); when 1n=1\frac{1}{n} = 1, xm=kp\frac{x}{m} = kp, proportional to pressure (at low pressure). Experimentally, xm\frac{x}{m} rises linearly at low pressure and levels off at high pressure (when the surface is saturated); the Freundlich isotherm fits well only over intermediate ranges, which is its limitation. Adsorption from solution. Solids also adsorb solutes from solution: charcoal adsorbs acetic acid from its aqueous solution, and the extent follows xm=kC1/n\frac{x}{m} = kC^{1/n} (with concentration CC replacing pressure). Adsorption decreases with rising temperature and increases with surface area; the nature of the solvent matters. Applications of adsorption. Production of high vacuum: charcoal cooled in liquid air removes the last traces of gas. Gas masks: activated charcoal adsorbs poisonous gases in mines and warfare. Control of humidity: silica gel and alumina. Removal of colouring matter: animal charcoal decolourises sugar solutions. Heterogeneous catalysis: reactants are adsorbed on catalyst surfaces (Fe in the Haber process, Pt in contact and Ostwald processes, Ni in hydrogenation of oils). Separation of inert gases: coconut charcoal adsorbs noble gases to different extents at different temperatures. Curing diseases: some drugs kill germs by being adsorbed on them. Froth flotation: sulfide ores are concentrated because pine oil and frothing agents adsorb preferentially on ore particles. Adsorption indicators: dyes such as eosin and fluorescein change colour at the end point of precipitation titrations (argentometry) when adsorbed on the precipitate. Chromatographic analysis: separation depends on selective adsorption. Qualitative analysis: the lake test for Al3+\mathrm{Al^{3+}}, in which Al(OH)3\mathrm{Al(OH)_{3}} adsorbs blue litmus to form a blue lake. Key Takeaways: - xm=kp1/n\frac{x}{m} = kp^{1/n} (n>1n > 1); log⁡xm=log⁡k+1nlog⁡p\log\frac{x}{m} = \log k + \frac{1}{n}\log p. - Low pressure: xm∝p\frac{x}{m} \propto p; high pressure: xm\frac{x}{m} nearly constant. - From solution: xm=kC1/n\frac{x}{m} = kC^{1/n}. - Uses: gas masks, vacuum, decolourisation, catalysis, chromatography, froth flotation, indicators. - Pitfall: the Freundlich isotherm fails at high pressures, where adsorption saturates. Brain-Teaser 1: A Freundlich plot of log⁡xm\log\frac{x}{m} against log⁡p\log p has slope 0.50.5 and intercept 0.300.30. Find xm\frac{x}{m} at p=4 atmp = 4\,\mathrm{atm}. Solution: log⁡k=0.30\log k = 0.30 gives k=2.0k = 2.0; 1n=0.5\frac{1}{n} = 0.5. xm=2.0×40.5=4.0\frac{x}{m} = 2.0 \times 4^{0.5} = 4.0 (in the units of the data, say g\mathrm{g} per g\mathrm{g} of adsorbent). Brain-Teaser 2: When pressure increases from 11 to 16 atm16\,\mathrm{atm}, the extent of adsorption doubles. Find nn in the Freundlich equation. Solution: (x/m)2(x/m)1=(161)1/n=2\frac{(x/m)_{2}}{(x/m)_{1}} = \left(\frac{16}{1}\right)^{1/n} = 2, so 161/n=216^{1/n} = 2, giving 4n=1\frac{4}{n} = 1 and n=4n = 4. Brain-Teaser 3: Why does a finely powdered adsorbent work much better than a lump of the same mass? Solution: Adsorption occurs only at the surface. Powdering greatly increases the surface area per unit mass (a 1 cm1\,\mathrm{cm} cube cut into 1 μm1\,\mu\mathrm{m} cubes has 10410^{4} times more surface), exposing far more adsorption sites. Activated charcoal, with its huge internal pore area (hundreds of square metres per gram), is therefore very effective.