Question

At the equilibrium position in the process of adsorption __________
(a)- $\Delta H>0$
(b)- $\Delta H=T\Delta S$
(c)- $\Delta H>T\Delta S$
(d)- $\Delta H=0$

Answer 1

Adsorption is a surface phenomenon. For the process to be in equilibrium the Gibbs free energy should zero. Adsorption is an exothermic process i.e. the change in enthalpy is negative.

Complete answer:
Let us first understand the adsorption phenomenon.
When there is attracting and retaining the molecules of a substance on the surface of a liquid or a solid forming into a higher concentration of the molecules on the surface is called adsorption.
The substance that is adsorbed on the surface is called the adsorbate and the substance on which it is adsorbed is called adsorbent.
The reverse process of adsorption i.e. removal of particles from the surface is called desorption.
Occlusion is the term used to define the adsorption of gases on the surface of metals.

Adsorption- an exothermic process:
When adsorption takes place, the residual forces on the surface of the adsorbent decreases. In other words, surface energy decreases. This appears in the form of heat which is called heat of adsorption. Hence, adsorption is an exothermic process, i.e., $\Delta {{H}_{adsorption}}$ is always negative.
Entropy changes during adsorption and adsorption equilibrium.

As the molecules of the adsorbate are held on the surface of the solid adsorbent, entropy decreases, i.e., $\Delta S$ is also negative. As $\Delta G=\Delta H-T\Delta S$, hence for the process of adsorption to occur, $\Delta G$ must be negative. As $\Delta S$ is negative, $\Delta G$can be negative only if $\Delta H$ is negative and $\Delta H>T\Delta S$ in magnitude. This is true in the beginning. However, as the adsorption proceeds, $\Delta H$ keeps on decreasing and $T\Delta S$ keeps on increasing till ultimately $\Delta H$ becomes equal $T\Delta S$ so that, $\Delta G=0$. This state is called adsorption equilibrium.

So, the correct answer is “Option B”.

Note: To reach the equilibrium state the process takes time that’s why the $\Delta H$ is negative at the beginning of the process and then becomes equal to $T\Delta S$. Equilibrium can only be attained in any reaction when the Gibbs free energy becomes 0.