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: Understanding the Concept:
Spontaneity needs \(\Delta G < 0\). When a gas sticks to a solid surface, its molecules lose freedom of movement, so \(\Delta S\) is negative.

Step 2: Key Formula or Approach:
\[ \Delta G = \Delta H - T\Delta S \]

Step 3: Detailed Explanation:
With \(\Delta S < 0\) the term \(-T\Delta S\) is positive. This works against spontaneity. To still get \(\Delta G < 0\), \(\Delta H\) must be negative with a large magnitude. Adsorption is exothermic.

Step 4: Check the options.
(A) says \(\Delta H\) should be highly positive, which would make \(\Delta G\) even more positive. (C) and (D) claim \(\Delta S\) is positive, which is wrong because the gas is confined to the surface. Only (B) fits.

Final Answer:
\(\Delta S\) is negative and \(\Delta H\) is highly negative, option (B). \[ \boxed{\Delta S < 0,\ \Delta H \ll 0} \]