Question

Which of the following statements is correct for the spontaneous adsorption of a gas?

• A. \(\Delta S\) is negative and, therefore \(\Delta H\) should be highly positive
• B. \(\Delta S\) is negative and therefore, \(\Delta H\) should be highly negative.
• C. \(\Delta S\) is positive and therefore, \(\Delta H\) should be negative.
• D. \(\Delta S\) is positive and therefore, \(\Delta H\) should also be highly positive.

Hint:

Adsorption reduces the freedom of gas molecules, so entropy falls and the enthalpy change must be strongly negative.

Answer 1

The Correct Option is B

Step 1: Predict the sign of entropy.
A free gas has a lot of disorder. Once it is adsorbed it is held on the surface, so disorder drops: $\Delta S < 0$.

Step 2: Predict the sign of enthalpy.
Adsorption forms attraction between gas and surface and releases heat, so $\Delta H < 0$.

Step 3: Combine.
For $\Delta G = \Delta H - T\Delta S < 0$ with a negative $\Delta S$, the negative $\Delta H$ must be large enough to beat $T\Delta S$ in magnitude. This leaves option (B).

Final Answer:
Option (B). \[ \boxed{\Delta S < 0,\ \Delta H \ll 0} \]