Adsorption Isotherms Applications - Explain, FAQs

Adsorption Isotherm

An adsorption isotherm is a graph or mathematical relationship that shows how the amount of a substance adsorbed on the surface of an adsorbent varies with the pressure of a gas (or concentration of a solution) at a constant temperature.

In other words, it expresses the relationship between the amount of adsorbate adsorbed per unit mass of adsorbent and the equilibrium pressure (or concentration) at a fixed temperature.

The most common adsorption isotherm is the Freundlich Adsorption Isotherm, represented by:

$\frac{x}{m}=k P^{1 / n}$

where:

• $x=$ mass of adsorbate adsorbed
• $m=$ mass of adsorbent
• $P=$ equilibrium pressure of the gas
• $k$ and $n=$ constants depending on the nature of the adsorbent and temperature

Logarithmic Form

$\log \left(\frac{x}{m}\right)=\log k+\frac{1}{n} \log P$
This equation gives a straight line when $\log (x / m)$ is plotted against $\log P$, with slope $1 / n$ and intercept $\log k$.

Factors Affecting Adsorption Isotherm

The amount of adsorption represented by an adsorption isotherm depends on several factors:

1. Nature and Surface Area of the Adsorbent

Adsorption increases with an increase in the surface area of the adsorbent. Porous and finely divided adsorbents such as activated charcoal, silica gel, and alumina provide more surface for adsorption and therefore show higher adsorption.

2. Nature of the Adsorbate

Easily liquefiable gases (having higher critical temperatures) are adsorbed more readily than gases that are difficult to liquefy. For example, NH₃ is adsorbed more strongly than H₂.

3. Pressure of the Gas

For the adsorption of gases on solids, the extent of adsorption generally increases with an increase in pressure. At very high pressures, the adsorption tends to reach a maximum value because the surface becomes saturated.

4. Temperature

Adsorption is generally an exothermic process. Therefore, an increase in temperature decreases the extent of adsorption, while lowering the temperature favours adsorption.

5. Activation of the Adsorbent

The adsorption capacity can be increased by activating the adsorbent. Activation is usually done by heating or treating the adsorbent to remove gases and moisture from its surface, thereby increasing the available surface area and active sites.

Types of Adsorption Isotherms

There are different types of adsorption isotherm models, all with their own characteristics. These include:

Langmuir Isotherm: Assumption of monolayer adsorption onto a surface comprising a finite number of adsorption sites. It is characterised by the equation:

The Langmuir adsorption isotherm equation is given by:

$$\theta=\frac{b P}{1+b P}$$

where:

$\theta$ is the fraction of the surface covered by the adsorbate

P is the pressure of the adsorbate.
b is the Langmuir constant

Freundlich Isotherm: Applied for heterogeneous surfaces, it can commonly be expressed as:

The Freundlich adsorption isotherm equation is given by:

$x / m=k P^{1 / n}$

where:
$x / m$ is the amount of adsorbate per unit mass of adsorbent.
$P$ is the pressure of the adsorbate.
$k$ and $n$ are empirical constants specific to the system.

BET Isotherm: Generalisation of the Langmuir model for multilayer adsorption, usually applied for porous materials.

Each model offers insights into different dimensions of the interaction of adsorbate with the surface, hence providing clues to choosing relevant models depending on the application.

Also Read:

• NCERT solutions for Class 12 Chemistry Chapter 5 Surface Chemistry
• NCERT Exemplar Class 12 Chemistry Solutions Chapter 5 Surface Chemistry
• NCERT notes Class 12 Chemistry Chapter 5 Surface Chemistry

Relevance and Applications

Adsorption isotherms find application in the following industrial and research sectors:

• Environmental Engineering: Adsorption isotherms turn out to be useful during water and air treatments in the selection and design of adsorbents for removing contaminants.
• Pharmaceuticals: The principles of adsorption are utilised in drug delivery systems by controlling the release of active ingredients.
• Chemical Engineering: Understanding adsorption isotherms helps in the design of catalysts and reactors.
• Agriculture: In soil science, adsorption isotherms are applied to the study of nutrient uptake and pesticide behaviour.

The applicability of adsorption isotherms in academic research falls under a very significant application. Characterisation of novel materials and understanding their surface properties are fundamentally important in the materials field. They provide insights into the adsorption mechanisms critical in developing innovative solutions in the domain of science.

Also Read

Some Solved Examples

Question 1: The curve showing the variation of adsorption with pressure at constant temperature is called?

1) An innovative
2) Adsorption isotherm
3) Adsorption isobar
4) All are incorrect

Solution:
The correct answer is option (2), Adsorption isotherm. An adsorption isotherm represents the relationship between the amount of gas adsorbed by an adsorbent and the pressure of the gas at a constant temperature.

Question 2:

At the equilibrium position in the process of adsorption $\qquad$
1) $\Delta H>0$

2) (correct) $\Delta H=T \Delta S$

3) $\Delta H>T \Delta S$

4) $\Delta H<T \Delta S$

Solution:

The answer is option (2).At equilibrium position during adsorption, $\Delta G=\Delta H-T \Delta S=0$ So that $\Delta H=T \Delta S$

Hence, the answer is option (2).

Question 3: For Freundlich adsorption isotherm, a plot of $\log \left(\frac{x}{m}\right)$ (y-axis) and $\log p$ ( $x$-axis) gives a straight line. The intercept and slope for the line are 0.4771 and 2 , respectively. The mass of gas, adsorbed per gram of adsorbent if the initial pressure is 0.04 atm , is $\times 10^{-4} g \cdot(\log 3=0.4771)$

Solution:

So, The mass of gas, adsorbed per gram of adsorbent means Need to find $\mathrm{x} / \mathrm{m}$ which is -

$
\begin{aligned}
& \frac{x}{m}=K P^{1 / n} \\
& \log \left(\frac{x}{m}\right)=\frac{1}{n} \log P+\log K \\
& \text { Slope }=\frac{1}{n}=2 \\
& \text { intercept }=\log K=0.4771 \\
& K=3
\end{aligned}
$

mass of gas adsorbed per gm of adsorbent $=\frac{x}{m}$

$\frac{\mathrm{x}}{\mathrm{~m}}=3 \times(0.04)^2=48 \times 10^{-4}$

Hence, the answer is (48)

Question 4: According to the Freundlich adsorption isotherm,

$\frac{x}{m}=k P^{1 / n}$
Which of the following graphs gives a straight line?
(A) $x / m$ vs $P$
(B) $\log (x / m)$ vs $\log P$
(C) $P$ vs $x / m$
(D) $\log P$ vs $x / m$

Solution:

Taking the logarithm of the Freundlich equation:

$\log \left(\frac{x}{m}\right)=\log k+\frac{1}{n} \log P$

This is the equation of a straight line of the form $y=c+m x$.
Hence, a plot of $\log (x / m)$ versus $\log P$ is a straight line.

Hence, the correct answer is option (B)

Question 5: For a gas adsorbed on a solid surface, the Freundlich adsorption isotherm is given by

$\frac{x}{m}=k P^{1 / n}$
If the pressure of the gas is increased 16 times and $n=2$, then the value of $x / m$ becomes:
(A) 2 times
(B) 4 times
(C) 8 times
(D) 16 times

Solution:

Given:

$\frac{x}{m} \propto P^{1 / n}$
Since $n=2$,

$\frac{x}{m} \propto P^{1 / 2}$
When pressure becomes 16 times,

$\begin{aligned}
\left(\frac{x}{m}\right)_{\text {new }} & =16^{1 / 2}\left(\frac{x}{m}\right)_{\text {old }} \\
= & 4\left(\frac{x}{m}\right)_{\text {old }}
\end{aligned}$
Therefore, adsorption per unit mass becomes 4 times.

Hence, the correct answer is option (B)

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