JEE Main 2021 · 26 August · morning · Single correct · +4 / −1
Which one of the following is correct for the adsorption of a gas at a given temperature on a solid surface?
- (a) ΔH > 0, ΔS > 0
- (b) ΔH > 0, ΔS < 0
- (c) ΔH < 0, ΔS < 0
- (d) ΔH < 0, ΔS > 0
Answer and solution
Answer (c) ΔH < 0, ΔS < 0
Solution
Chemistry ma’am · 4 steps
Gas molecules lose freedom when they stick to a surface, so ΔS drops and ΔH must be negative to pay for it.
- Entropy first. In the gas phase the molecules move freely in three dimensions. Once adsorbed, they are held on the surface and can barely move, so disorder decreases and Δ S < 0.
- Spontaneity condition. Adsorption happens on its own at a given temperature, so Δ G must be negative.
Δ G = Δ H - TΔ S < 0
- Sign of ΔH. With Δ S < 0, the term -TΔ S is positive and works against spontaneity. The only way Δ G can still be negative is if Δ H is negative and large enough. That is why adsorption is always exothermic: bonds or attractions form between the gas and the surface, releasing heat.
- So. Adsorption of a gas on a solid has Δ H < 0 and Δ S < 0, which is option (c).
The idea to keep. Adsorption always lowers entropy because the gas loses its freedom of movement. A process with Δ S < 0 can be spontaneous only if it is exothermic, so Δ H < 0.
Answer from the NTA final answer key. Solution by PrepWiser.