Question

At T(K), a gas is adsorbed on the surface of a solid. The signs of $\Delta H$ and $\Delta S$ are respectively:

• A. Negative, Negative
• B. Positive, Positive
• C. Positive, Negative
• D. Negative, Positive

Hint:

Adsorption is exothermic (releases heat) and decreases entropy (surface ordering).

Answer 1

The Correct Option is A

Step 1: Picture what happens during adsorption.
Gas molecules that were moving freely get held onto a solid surface. We must decide the signs of the enthalpy change $\Delta H$ and the entropy change $\Delta S$.

Step 2: Think about heat (enthalpy).
When gas molecules attach to the surface, new attractive bonds form between the gas and the solid. Forming bonds releases energy as heat, so $\Delta H$ is negative. Adsorption is an exothermic process.

Step 3: Think about disorder (entropy).
A free gas is very disordered because its molecules move everywhere. Once stuck on the surface, the molecules lose their freedom and become more ordered. Less freedom means lower entropy, so $\Delta S$ is negative.

Step 4: Check with spontaneity.
Adsorption happens on its own, so $\Delta G$ must be negative. Using $\Delta G = \Delta H - T\Delta S$, a negative $\Delta H$ helps make $\Delta G$ negative even though $-T\Delta S$ is positive.

Step 5: This is why it slows at high temperature.
Because $\Delta S$ is negative, raising the temperature makes the $-T\Delta S$ term work against adsorption, which fits the known fact that physical adsorption falls at higher temperature.

Step 6: State both signs.
Both $\Delta H$ and $\Delta S$ are negative.
\[ \boxed{\text{Negative, Negative}} \]