Surface Chemistry • Topic 1 of 3

Adsorption

Adsorption is the accumulation of a substance at a surface rather than in the bulk. When a gas or a dissolved solute is brought into contact with a solid, its concentration at the surface becomes higher than in the interior. The substance that is concentrated on the surface is the adsorbate; the surface that holds it is the adsorbent. Finely divided metals, charcoal, silica gel, alumina gel and clay are good adsorbents because they expose a very large surface area per gram.

Adsorption versus absorption. In adsorption the substance stays only on the surface, so the concentration changes sharply at the boundary. In absorption the substance is taken up uniformly throughout the bulk of the material, like water soaked into a sponge or ammonia dissolving through water. When both happen together the process is called sorption. Adsorption is rapid at first and then slows as the surface fills; absorption proceeds at a uniform rate.

Adsorption lowers the surface energy of the adsorbent, so it is always exothermic: $\Delta H$ is negative. During adsorption gas molecules lose freedom of motion, so the entropy change $\Delta S$ is also negative. For the process to be spontaneous, $\Delta G = \Delta H - T\Delta S$ must be negative, which requires $\Delta H$ to be sufficiently negative. As adsorption proceeds $\Delta H$ becomes less negative, and equilibrium is reached when $\Delta G = 0$.

Physisorption versus chemisorption. In physical adsorption (physisorption) the adsorbate is held by weak van der Waals forces. It has a low enthalpy of adsorption ($20$–$40\ \text{kJ mol}^{-1}$), is reversible, not specific, favoured by low temperature, and can build up multiple molecular layers. In chemical adsorption (chemisorption) the adsorbate forms chemical bonds with the surface. It has a high enthalpy of adsorption ($80$–$240\ \text{kJ mol}^{-1}$), is usually irreversible, highly specific, and forms only a single (unimolecular) layer. Physisorption can change into chemisorption as the temperature rises.

Factors affecting adsorption of gases. Adsorption increases with the surface area of the adsorbent, so porous and finely divided solids adsorb best. Easily liquefiable gases with higher critical temperatures (for example $\text{NH}_3$, $\text{HCl}$, $\text{CO}_2$, $\text{SO}_2$) are adsorbed more than permanent gases such as $\text{H}_2$, $\text{N}_2$ and $\text{O}_2$. Physisorption decreases with rising temperature (it is exothermic), while chemisorption first rises and then falls. Adsorption increases with pressure but eventually levels off when the surface is saturated.

Freundlich adsorption isotherm. At constant temperature the variation of the mass of gas adsorbed per gram of adsorbent, $\frac{x}{m}$, with pressure $p$ is given by $\frac{x}{m}=k\,p^{1/n}$, where $k$ and $n$ are constants for a given adsorbent-adsorbate pair at a given temperature, and $n \ge 1$. Taking logarithms gives the straight-line form $\log\frac{x}{m}=\log k+\frac{1}{n}\log p$. A plot of $\log\frac{x}{m}$ against $\log p$ has slope $\frac{1}{n}$ and intercept $\log k$. The equation holds over a limited range: at low pressure $\frac{x}{m}\propto p$ ($1/n \to 1$) and at high pressure $\frac{x}{m}$ becomes independent of pressure ($1/n \to 0$).

Applications. Adsorption underlies the creation of high vacuum, gas masks (activated charcoal adsorbs toxic gases), the control of humidity by silica and alumina gels, the removal of colouring matter from solutions, heterogeneous catalysis, ion-exchange and chromatographic separation, and froth flotation in metallurgy.

Freundlich adsorption isotherm and its log-log linear formpressure px/msaturationlog plog (x/m)slope = 1/nintercept = log k
1
Worked Example
Distinguish between adsorption and absorption with one example of each.
Solution
  1. In adsorption a substance accumulates only at the surface of a solid, so its concentration is higher at the surface than in the bulk.
  2. In absorption the substance is taken up uniformly throughout the entire bulk of the material.
  3. Example of adsorption: a gas such as $\text{NH}_3$ collecting on the surface of charcoal.
  4. Example of absorption: water being soaked uniformly into a sponge.

Answer: Adsorption is a surface phenomenon; absorption is a bulk phenomenon.

2
Worked Example
Why is adsorption always exothermic?
Solution
  1. A free solid surface has unbalanced (residual) attractive forces and therefore extra surface energy.
  2. When adsorbate molecules attach, these residual forces are satisfied and the surface energy decreases.
  3. The energy released as the surface energy falls appears as heat, so $\Delta H$ is negative.

Answer: Because adsorption reduces the surface energy of the adsorbent, releasing heat, $\Delta H < 0$.

3
Worked Example
Give three points of difference between physisorption and chemisorption.
Solution
  1. Forces: physisorption involves weak van der Waals forces; chemisorption involves chemical (covalent or ionic) bonds.
  2. Enthalpy: physisorption has low $\Delta H$ ($20$–$40\ \text{kJ mol}^{-1}$); chemisorption has high $\Delta H$ ($80$–$240\ \text{kJ mol}^{-1}$).
  3. Specificity and layers: physisorption is non-specific and multilayer; chemisorption is highly specific and forms a single layer.

Answer: Physisorption is weak, low-energy, non-specific and multilayer; chemisorption is strong, high-energy, specific and unilayer.

4
Worked Example
Why are powdered or finely divided substances more effective adsorbents than lumps of the same material?
Solution
  1. Adsorption is a surface phenomenon, so the amount adsorbed depends on the surface area exposed.
  2. Breaking a solid into a fine powder greatly increases the total surface area per gram.
  3. More surface means more sites available for the adsorbate.

Answer: Finely divided solids expose a far larger surface area, so they adsorb much more.

5
Worked Example
The Freundlich isotherm is $\frac{x}{m}=k\,p^{1/n}$. What does a plot of $\log\frac{x}{m}$ against $\log p$ look like, and what do its slope and intercept give?
Solution
  1. Take logarithms: $\log\frac{x}{m}=\log k+\frac{1}{n}\log p$.
  2. This is a straight line of the form $y = c + mx$ with $y=\log\frac{x}{m}$ and $x=\log p$.
  3. The slope of the line equals $\frac{1}{n}$ and the intercept on the $\log\frac{x}{m}$ axis equals $\log k$.

Answer: A straight line; slope $=\frac{1}{n}$ and intercept $=\log k$.

6
Worked Example
In an adsorption experiment $\frac{x}{m}=0.5\ \text{g g}^{-1}$ at $p=1\ \text{atm}$ and $\frac{x}{m}=1.0\ \text{g g}^{-1}$ at $p=4\ \text{atm}$. Find $1/n$ and $k$ in the Freundlich equation.
Solution
  1. Use $\log\frac{x}{m}=\log k+\frac{1}{n}\log p$ at the two points.
  2. Subtract: $\log\frac{1.0}{0.5}=\frac{1}{n}\log\frac{4}{1}$, i.e. $\log 2 = \frac{1}{n}\log 4 = \frac{1}{n}(2\log 2)$.
  3. Hence $\frac{1}{n}=\frac{1}{2}=0.5$.
  4. At $p=1\ \text{atm}$, $\log p = 0$, so $\log\frac{x}{m}=\log k$, giving $k=\frac{x}{m}=0.5$.

Answer: $\frac{1}{n}=0.5$ and $k=0.5\ \text{g g}^{-1}$.

Key Points

  • Adsorption is a surface phenomenon (substance accumulates at the surface), while absorption is a bulk phenomenon; sorption is when both occur together.
  • Adsorption is exothermic ($\Delta H<0$) and accompanied by a decrease in entropy ($\Delta S<0$); it is spontaneous as long as $\Delta G=\Delta H-T\Delta S<0$.
  • Physisorption uses weak van der Waals forces (low $\Delta H$, reversible, non-specific, multilayer, favoured by low $T$); chemisorption uses chemical bonds (high $\Delta H$, specific, unilayer).
  • Adsorption of a gas increases with surface area, with ease of liquefaction (higher critical temperature) and with pressure, and physisorption decreases with rising temperature.
  • Freundlich isotherm $\frac{x}{m}=k\,p^{1/n}$ ($n\ge 1$); the log form $\log\frac{x}{m}=\log k+\frac{1}{n}\log p$ gives a straight line of slope $\frac{1}{n}$ and intercept $\log k$.
Tap an option to check your answer0 / 4
Q1.Adsorption differs from absorption in that adsorption is:
Explanation: In adsorption the substance accumulates only at the surface; absorption is uniform throughout the bulk.
Q2.Which of the following is true of physisorption but NOT of chemisorption?
Explanation: Physisorption is due to weak van der Waals forces; chemisorption is specific, unilayer and high in enthalpy.
Q3.In the Freundlich isotherm $\frac{x}{m}=k\,p^{1/n}$, a plot of $\log\frac{x}{m}$ versus $\log p$ gives a straight line whose slope is:
Explanation: From $\log\frac{x}{m}=\log k+\frac{1}{n}\log p$, the slope is $\frac{1}{n}$ and the intercept is $\log k$.
Q4.Which of these gases is adsorbed to the greatest extent on a given mass of charcoal?
Explanation: Easily liquefiable gases with a higher critical temperature, such as $\text{CO}_2$, are adsorbed more than permanent gases like $\text{H}_2$, $\text{N}_2$ and $\text{O}_2$.