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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(d). Kₚ and Δ G^° for N₂O₄ ⇌ 2NO₂ at 300 K

  • 5 marks
  • Answered
Question

At 300 K temperature and 1 atm pressure, dinitrogen tetraoxide is 20 percent dissociated into nitrogen dioxide. Calculate the equilibrium constant (Kₚ) and standard free energy change for the reaction at this temperature:
N₂O₄(g) ⇌ 2NO₂(g)

Answer in brief

Kₚ=0.167 atm (= 1/6); standard Gibbs energy change Δ G^°=+4.47 kJ mol⁻¹ (standard state 1 atm)

Key points

  • at equilibrium N₂O₄ 1-α=0.80, NO₂ 2α=0.40, total 1.20. So p_(N₂O₄)=(0.80)/(1.20)P=2/3 atm and p_(NO₂)=(0.40)/(1.20)P=1/3 atm.
  • Kₚ=p_(NO₂)²p_(N₂O₄)=4α²P1-α²=4(0.20)²(1)1-0.04, Kₚ=0.167 atm (=1/6)
  • With the standard state p^°=1 atm, K=0.167 (dimensionless): 4pt] Δ G^°=-RTln K=-(8.314)(300)ln(0.1667), Δ G^°=+4.47 kJ mol⁻¹
  • With the IUPAC standard state of 1 bar, K=0.169 and Δ G^°=+4.44 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.

  1. Per mol N₂O₄: at equilibrium N₂O₄ 1-α=0.80, NO₂ 2α=0.40, total 1.20. So p_(N₂O₄)=(0.80)/(1.20)P=2/3 atm and p_(NO₂)=(0.40)/(1.20)P=1/3 atm.
  2. Kₚ=p_(NO₂)²p_(N₂O₄)=4α²P1-α²=4(0.20)²(1)1-0.04, Kₚ=0.167 atm (=1/6)
  3. With the standard state p^°=1 atm, K=0.167 (dimensionless): 4pt] Δ G^°=-RTln K=-(8.314)(300)ln(0.1667), Δ G^°=+4.47 kJ mol⁻¹
  4. With the IUPAC standard state of 1 bar, K=0.169 and Δ G^°=+4.44 kJ mol⁻¹.

Reading the answer: Δ G^°>0 only means K<1. Pure N₂O₄ starts at Q=0<K, so G falls until Q=K at α=0.20 (figure).

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Model answer to Q1(d), 2026 Paper I: solution page 4
Solution page 4 of 55
Read the whole 2026 solution2026 question paper

All parts of Question 1