Question

Which of the following statement is true about the adsorption?

• 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 \( \Delta S < 0 \)
• E. \( \Delta H = 0 \) and \( \Delta S > 0 \)

Hint:

Think of adsorption as "forming a bond" between the surface and the molecule. Since bond formation releases energy, the process is exothermic (\( \Delta H < 0 \)).

Answer 1

The Correct Option is A

Concept: Adsorption is the process of accumulation of molecular species at the surface rather than in the bulk. We evaluate its thermodynamics using the Gibbs Free Energy equation: \( \Delta G = \Delta H - T\Delta S \).

Step 1: Analyze Entropy (\( \Delta S \)).
When a gas is adsorbed on a solid surface, its molecules lose freedom of movement. Their randomness decreases as they become "fixed" to the surface. Therefore, the entropy change is always negative: \[ \Delta S < 0 \]

Step 2: Analyze Enthalpy (\( \Delta H \)).
Adsorption is a spontaneous process (\( \Delta G < 0 \)). For \( \Delta G \) to be negative when \( \Delta S \) is already negative, the term \( \Delta H \) must be sufficiently negative to overcome the positive \( -T\Delta S \) term. \[ \Delta G = \Delta H - T(\text{negative value}) \] Thus, adsorption is almost always an exothermic process: \[ \Delta H < 0 \]