Question

During the adsorption of krypton on activated charcoal at low temperature.

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

Hint:

For any spontaneous process, \(\Delta G = \Delta H - T\Delta S\) is negative. For adsorption, since \(\Delta S < 0\), \(\Delta H\) must be sufficiently negative at low temperatures.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Adsorption is an exothermic process that results in a decrease in entropy.

Step 2: Detailed Explanation:
Adsorption is spontaneous at low temperatures. It is an exothermic process because energy is released when gas molecules bind to the surface (\(\Delta H < 0\)). Additionally, when gas molecules become confined to a surface, their freedom of movement decreases, leading to a decrease in entropy (\(\Delta S < 0\)).

Step 3: Final Answer:
During adsorption, \(\Delta H < 0\) and \(\Delta S < 0\), which corresponds to option (C).