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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

Q2(c)(ii). Δ G° of ammonia formation at 400 K

  • 10 marks
  • Answered
Question

Calculate Δ G° at 400 K temperature for the following reaction:
1/2N₂(g) + 3/2H₂(g) → NH₃(g)
Assume that Δ S° is independent of temperature.
(Given: Δ G°_(298 K)=-16.496 kJ mol⁻¹; S°_(NH₃)=192.45 J K⁻¹ mol⁻¹; S°_(N₂)=191.61 J K⁻¹ mol⁻¹; S°_(H₂)=130.68 J K⁻¹ mol⁻¹)

Answer in brief

Δ G°₄₀₀=-6.36 kJ mol⁻¹ (Δ S°=-99.38 J K⁻¹ mol⁻¹, Δ H°=-46.11 kJ mol⁻¹)

Key points

  • Δ S°=S°(NH₃)-12S°(N₂)-32S°(H₂)=192.45-95.805-196.02=-99.38 J K⁻¹ mol⁻¹
  • Δ H°=Δ G°₂₉₈+TΔ S°=-16.496+298(-0.09938)=-16.496-29.61=-46.11 kJ mol⁻¹
  • Δ G°₄₀₀=Δ H°-400 Δ S°=-46.11-400(-0.09938)=-46.11+39.75
  • Δ G°₄₀₀=-6.36 kJ mol⁻¹

The full model answer

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

Method, and why: Δ S° follows from the absolute entropies. Since (∂Δ G°/∂ T)ₚ=-Δ S°, a constant Δ S° makes Δ G° a straight line in T. A constant Δ S° also means Δ Cₚ=0, so Δ H° is constant too.

  1. Δ S°=S°(NH₃)-12S°(N₂)-32S°(H₂)=192.45-95.805-196.02=-99.38 J K⁻¹ mol⁻¹
  2. Δ H°=Δ G°₂₉₈+TΔ S°=-16.496+298(-0.09938)=-16.496-29.61=-46.11 kJ mol⁻¹
  3. Δ G°₄₀₀=Δ H°-400 Δ S°=-46.11-400(-0.09938)=-46.11+39.75
  4. Δ G°₄₀₀=-6.36 kJ mol⁻¹
  5. Directly from the slope: Δ G°₄₀₀=Δ G°₂₉₈-Δ S°(400-298)=-16.496+10.14=-6.36 kJ mol⁻¹

Reading the answer: Δ S° is negative because two moles of gas become one, so Δ G° rises with temperature. K=e^(-Δ G°/RT) falls from 779 at 298 K to 6.8 at 400 K, and Δ G° reaches zero at T=Δ H°/Δ S°=464 K. The derived Δ H°=-46.11 kJ mol⁻¹ is the tabulated enthalpy of formation of NH₃.

Application. the Haber process runs at about 450°C and 100 atm over a promoted iron catalyst. A low temperature favours the yield, but the rate is then too slow; the high pressure pushes the equilibrium towards NH₃, since the number of gas molecules falls.

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Model answer to Q2(c)(ii), 2026 Paper I: solution page 14
Solution page 14 of 55
Read the whole 2026 solution2026 question paper

All parts of Question 2