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UPSC Civil Services (Mains)

Chemistry optional

  • 2 papers
  • 250 marks each
  • 3 h per paper
  • 5 of 8 questions, Q1 and Q5 compulsory

Paper I covers physical and inorganic chemistry.

UPSC Civil Services (Mains), Chemistry optional · 2026 · Paper I

Q1(j). Why low temperature favours adsorption of a gas

  • 5 marks
  • Answered
Question

Spontaneous adsorption of gas on solid surface favours low temperature. Explain, why.

Answer in brief

Adsorption lowers the entropy (Δ S<0), so it is spontaneous only if it is exothermic (Δ H<0), as almost all adsorptions are. In Δ G=Δ H-TΔ S the term -TΔ S is positive and grows with T, so Δ G is most negative at low temperature.

Key points

  • A gas molecule moves in three dimensions. On the surface it is held in a layer and loses most of its translational motion, so Δ S is negative.
  • For Δ G=Δ H-TΔ S<0 with Δ S<0, Δ H must be negative and |Δ H|>T|Δ S|. Adsorption is exothermic: about -20 kJ mol⁻¹ for physisorption and about -200 kJ mol⁻¹ for chemisorption.
  • As T rises, -TΔ S grows until Δ G=0 at T=Δ H/Δ S. Above this the gas desorbs.
  • gas + surface ⇌ adsorbed gas + heat. Cooling shifts the equilibrium to the right, so the amount adsorbed (x/m) at constant pressure falls with temperature (adsorption isobar).

The full model answer

Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.

  1. A gas molecule moves in three dimensions. On the surface it is held in a layer and loses most of its translational motion, so Δ S is negative.
  2. For Δ G=Δ H-TΔ S<0 with Δ S<0, Δ H must be negative and |Δ H|>T|Δ S|. Adsorption is exothermic: about -20 kJ mol⁻¹ for physisorption and about -200 kJ mol⁻¹ for chemisorption.
  3. As T rises, -TΔ S grows until Δ G=0 at T=Δ H/Δ S. Above this the gas desorbs.
  4. Le Chatelier: gas + surface ⇌ adsorbed gas + heat. Cooling shifts the equilibrium to the right, so the amount adsorbed (x/m) at constant pressure falls with temperature (adsorption isobar).

Chemisorption needs an activation energy, so its isobar first rises with T, then falls.

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Model answer to Q1(j), 2026 Paper I: solution page 8
Solution page 8 of 55
Read the whole 2026 solution2026 question paper

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