Question

Which one of the following is correct for the adsorption of a gas at a given temperature on a solid surface ?

• A. \(\Delta H>0, \Delta S>0\)
• B. \(\Delta H<0, \Delta S>0\)
• C. \(\Delta H>0, \Delta S<0\)
• D. \(\Delta H<0, \Delta S<0\)

Hint:

Adsorption is always exothermic and always results in a decrease in entropy. Think of it as gas molecules losing their "freedom" to move in 3D space.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Adsorption is a spontaneous process where molecules of a gas or liquid accumulate on the surface of a solid. Spontaneity and the physical nature of the process determine the signs of thermodynamic variables.
Step 2: Detailed Explanation:
1. Enthalpy (\(\Delta H\)): Adsorption involves the formation of new attractions (bonds or van der Waals forces) between the adsorbate and adsorbent. Bond formation is an exothermic process, so energy is released. Thus, \(\Delta H<0\).
2. Entropy (\(\Delta S\)): In the gas phase, molecules have high degrees of freedom and randomness. When they are adsorbed onto a solid surface, their movement becomes restricted, leading to a decrease in randomness. Thus, \(\Delta S<0\).
3. Gibbs Free Energy (\(\Delta G\)): For the process to be spontaneous, \(\Delta G\) must be negative. Since \(\Delta G = \Delta H - T\Delta S\), and \(\Delta S\) is negative (making \(-T\Delta S\) positive), \(\Delta H\) must be sufficiently negative to make \(\Delta G\) negative.
Step 3: Final Answer:
The correct thermodynamic parameters for adsorption are \(\Delta H<0\) and \(\Delta S<0\).