Adsorption arises due to the fact that the surface particles of the adsorbent are not in the same environment as the particles inside the bulk. The extent of adsorption increases with the increase of surface are per unit mass of the adsorbent at a given temperature and pressure. Another important factor featuring adsorption is the heat of adsorption. During adsorption, there is always a decrease in residual forces of the surface, therefore, it is invariably an exothermic process or \(\Delta H\) and \(\Delta S\) are always negative. There are mainly two types of adsorption of gases on solids. In physisorption, the attractive forces are mainly van der Waals forces while in cemisorption, actual bonding occurs between the particles of adsorbate and adsorbent. Generally, easily liquifying gases are adsorbed more easily on the surface of a solid as compared to the gases which are liquified with difficulty. Freundlich gave an emperical relationship between the quantity of gas adsorbed by unit mass of solid adsorbent and pressure, at a particular temperature.

Identify the correct order of ease of adsorption of the following gases on the surface of charcoal:

A. \(CH_4\)

B. \(H_2\)

C. \(NH_3\)

D. \(CO_2\)

Choose the correct answer from the options given below:

C > D > A > B

The correct answer is option 4. C > D > A > B.

Let us delve into the factors influencing the adsorption of gases onto charcoal:

Polarity of the Gas Molecule:

Polar molecules tend to have stronger interactions with the surface of the adsorbent (charcoal in this case) because of dipole-dipole interactions or hydrogen bonding. Non-polar molecules rely more on van der Waals forces for adsorption.

Molecular Size and van der Waals Forces:

Larger molecules generally have stronger van der Waals forces because they have more electrons that can contribute to these forces. This makes larger molecules more likely to be adsorbed compared to smaller ones. Smaller molecules with weaker van der Waals forces are less likely to be adsorbed.

Boiling Point:

Gases with higher boiling points are easier to adsorb because they condense more readily onto surfaces. Higher boiling points usually indicate stronger intermolecular forces.

Analyzing Each Gas:

A. \(CH_4\) (Methane)

Nature: Non-polar, small molecule.

Molecular Size: Slightly larger than hydrogen but still small.

Boiling Point: \(-161.5^\circ \text{C}\).

van der Waals Forces: Present, but weaker compared to larger molecules.

Adsorption Potential: Moderate. Since \(CH_4\) is non-polar and relatively small, its adsorption on charcoal is less significant than that of larger or more polar molecules.

B. \(H_2\) (Hydrogen)

Nature: Non-polar, very small molecule.

Molecular Size: Smallest among the given gases.

Boiling Point: \(-252.9^\circ \text{C}\), the lowest among the options.

van der Waals Forces: Weak due to its small size and low polarizability.

Adsorption Potential: Lowest. Due to its small size, low boiling point, and weak van der Waals forces, \(H_2\) is the least likely to be adsorbed onto charcoal.

C. \(NH_3\) (Ammonia)

Nature: Polar, moderate size.

Molecular Size: Larger than \(H_2\) and \(CH_4\), smaller than \(CO_2\).

Boiling Point: \(-33.3^\circ \text{C}\).

Polarity and Hydrogen Bonding: Strong dipole moment and ability to form hydrogen bonds make \(NH_3\) highly interactive with adsorbents.

Adsorption Potential: Highest. The strong polarity and ability to form hydrogen bonds with the surface of charcoal make \(NH_3\) the most readily adsorbed gas among the given options.

D. \(CO_2\) (Carbon dioxide)

Nature: Non-polar, but with a significant linear structure that allows for stronger van der Waals forces.

Molecular Size: Larger than \(NH_3\), \(CH_4\), and \(H_2\).

Boiling Point: \(-78.5^\circ \text{C}\).

van der Waals Forces: Stronger than \(CH_4\) and \(H_2\) due to its size.

Adsorption Potential: High. While \(CO_2\) is non-polar, its larger size and stronger van der Waals forces make it more likely to be adsorbed compared to \(CH_4\) and \(H_2\), but less than \(NH_3\) due to the lack of polarity.

Thus, the correct order of ease of adsorption is: \(C > D > A > B\)

This means that \(NH_3\) has the highest ease of adsorption, followed by \(CO_2\), then \(CH_4\), and finally \(H_2\) which has the lowest ease of adsorption.