Question:
At T(K), a gas is adsorbed on the surface of a solid. The signs of $\Delta H$ and $\Delta S$ are respectively:
At T(K), a gas is adsorbed on the surface of a solid. The signs of $\Delta H$ and $\Delta S$ are respectively:
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Adsorption is exothermic (releases heat) and decreases entropy (surface ordering).
Updated On: Jun 10, 2026
- Negative, Negative
- Positive, Positive
- Positive, Negative
- Negative, Positive
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The Correct Option is A
Solution and Explanation
Step 1: Concept
Adsorption is a spontaneous process, so $\Delta G < 0$. $\Delta G = \Delta H - T\Delta S$.
Step 2: Analysis
Adsorption releases heat, so $\Delta H < 0$. Gas molecules become ordered on the surface, reducing randomness, so $\Delta S < 0$.
Step 3: Conclusion
Both $\Delta H$ and $\Delta S$ are negative.
Final Answer: (A)
Adsorption is a spontaneous process, so $\Delta G < 0$. $\Delta G = \Delta H - T\Delta S$.
Step 2: Analysis
Adsorption releases heat, so $\Delta H < 0$. Gas molecules become ordered on the surface, reducing randomness, so $\Delta S < 0$.
Step 3: Conclusion
Both $\Delta H$ and $\Delta S$ are negative.
Final Answer: (A)
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