Question

During the adsorption of gas on a solid

• A. ΔG < O, ΔH < O, ΔS > O
• B. ΔG < O, ΔH < O, ΔS < O
• C. ΔG > O, ΔH < O, ΔS < O
• D. ΔG < O, ΔH > O, ΔS < O

Answer 1

The Correct Option is B

During adsorption, the gas molecules are attracted to the surface of the solid, which results in the formation of a bond between the gas and the solid. Adsorption is an exothermic process (releases heat), which implies that the change in enthalpy \( \Delta H \) is negative. The entropy \( \Delta S \) of the system decreases because the gas molecules become more ordered as they are adsorbed onto the solid surface, resulting in a decrease in randomness. The Gibbs free energy change \( \Delta G \) for the adsorption process is negative, indicating that the process is spontaneous. This can be explained by the relation: \[ \Delta G = \Delta H - T\Delta S \] Since both \( \Delta H < 0 \) and \( \Delta S < 0 \), the process can still be spontaneous if the enthalpy change is sufficiently negative to overcome the negative entropy change.

The correct option is (B): ΔG < O, ΔH < O, ΔS < O