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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)(i). Enthalpy of formation of HI(g) from the gaseous elements at 225°C

  • 10 marks
  • Answered
Question

At 25°C, the latent heat of sublimation per mole of iodine is 62.3 kJ mol⁻¹ and the standard enthalpy of formation of HI(g) is 24.7 kJ mol⁻¹. Calculate the enthalpy change which occurs when HI(g) is formed from the gaseous elements at 225°C. The mean molar heat capacities over the temperature range 25°C to 225°C are:
H₂(g); C_(p,m)=29.08 J K⁻¹ mol⁻¹
I₂(g); C_(p,m)=33.56 J K⁻¹ mol⁻¹
HI(g); C_(p,m)=29.87 J K⁻¹ mol⁻¹

Answer in brief

Δ H(498 K)=-6.74 kJ mol⁻¹ per mole of HI (-13.48 kJ for H₂ + I₂ → 2HI)

Key points

  • 1/2H₂(g) + 1/2I₂(s) → HI(g), Δ H=+24.7 kJ mol⁻¹
  • 1/2I₂(s) → 1/2I₂(g), Δ H=12(62.3)=+31.15 kJ mol⁻¹
  • 1/2H₂(g) + 1/2I₂(g) → HI(g), Δ H₂₉₈=24.7-31.15=-6.45 kJ mol⁻¹
  • Δ Cₚ=C_(p,m)(HI)-12C_(p,m)(H₂)-12C_(p,m)(I₂)=29.87-12(29.08+33.56)=-1.45 J K⁻¹ 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.

Given: Δ_(sub)H(I₂)=62.3 kJ mol⁻¹; Δ_fH(HI, g)=24.7 kJ mol⁻¹ at 298 K; T₂=225°C=498 K.

Method, and why: the standard enthalpy of formation refers to iodine in its standard state at 25°C, the solid. The question asks for the reaction from gaseous iodine, so the sublimation step is first removed (Hess's law). The result is then carried from 298 K to 498 K with Kirchhoff's law.

  1. Standard formation: 1/2H₂(g) + 1/2I₂(s) → HI(g), Δ H=+24.7 kJ mol⁻¹
  2. Sublimation: 1/2I₂(s) → 1/2I₂(g), Δ H=12(62.3)=+31.15 kJ mol⁻¹
  3. Hess's law at 298 K: 1/2H₂(g) + 1/2I₂(g) → HI(g), Δ H₂₉₈=24.7-31.15=-6.45 kJ mol⁻¹
  4. Δ Cₚ=C_(p,m)(HI)-12C_(p,m)(H₂)-12C_(p,m)(I₂)=29.87-12(29.08+33.56)=-1.45 J K⁻¹ mol⁻¹
  5. Kirchhoff's law with constant (mean) Δ Cₚ: Δ H₄₉₈=Δ H₂₉₈+Δ Cₚ (T₂-T₁)=-6.45+(-1.45×10⁻³)(200)
  6. Δ H₄₉₈=-6.45-0.29=-6.74 kJ mol⁻¹ per mole of HI
  7. For H₂(g) + I₂(g) → 2HI(g) all terms double: Δ H₄₉₈=2(-6.74)=-13.48 kJ

Check round the cycle: cooling the reactants to 298 K, -31.32×200=-6.26 kJ; reaction at 298 K, -6.45 kJ; heating HI to 498 K, +29.87×200=+5.97 kJ. The sum is -6.74 kJ.

Reading the answer: from gaseous iodine the reaction is exothermic; the positive Δ_fH of HI comes entirely from subliming the iodine. Δ Cₚ is small, so 200 K of heating changes Δ H by only 0.29 kJ mol⁻¹.

Application. reaction enthalpies are tabulated at 298 K, while gas reactions such as H₂ + I₂ ⇌ 2HI are studied and run at higher temperatures. Kirchhoff's law gives Δ H at the working temperature from the heat capacities, for use in the van 't Hoff equation.

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Model answer to Q2(c)(i), 2026 Paper I: solution page 13
Solution page 13 of 55
Read the whole 2026 solution2026 question paper

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