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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 · 2017 · Paper I

Q3(a)(i). Signs of Δ H and Δ G when hot tea cools

  • 10 marks
  • Answered
Question

At constant pressure, the temperature of a fixed amount of hot tea in a cup decreases spontaneously to room temperature. What will be the sign of (I) Δ H and (II) Δ G of this process? Give reason(s) in support of your answer.

Answer in brief

(I) Δ H<0: heat leaves the tea at constant pressure. (II) Δ G of the tea (the system) is positive: G rises as T falls, (∂ G/∂ T)ₚ=-S. The test "Δ G<0 for a spontaneous change" holds only at constant T and p; here T falls, and the spontaneity shows as Δ Sₜₒₜₐₗ>0.

Key points

  • Sign of Δ H. At constant pressure, with no work other than expansion, Δ H=qₚ. The tea gives heat to the room, qₚ<0, so Δ H<0. In numbers: Δ H=∫_Tₕₒₜ^(T₀)Cₚ dT=Cₚ(T₀-Tₕₒₜ)<0.
  • Sign of Δ G. G is a state function, and at constant pressure dG=-S dT. The tea has S>0 and dT<0, so Δ G=-∫_Tₕₒₜ^(T₀)S dT>0Δ G>0 The Gibbs energy of the tea rises as it cools (G=H-TS; TS falls more than H).
  • No contradiction. Δ G≤0 is the test of spontaneity only for a change at constant T and p. The tea's temperature changes, so its Δ G says nothing about spontaneity.
  • The real criterion is the total entropy. Tea: Δ Sₜₑₐ=Cₚln(T₀/Tₕₒₜ)<0. Room (at T₀): Δ Sᵣₒₒₘ=-Δ H/T₀=Cₚ(Tₕₒₜ-T₀)/T₀>0. Since (x-1)>ln x for x=Tₕₒₜ/T₀>1, Δ Sₜₒₜₐₗ>0: the cooling is spontaneous.

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. Sign of Δ H. At constant pressure, with no work other than expansion, Δ H=qₚ. The tea gives heat to the room, qₚ<0, so Δ H<0. In numbers: Δ H=∫_Tₕₒₜ^(T₀)Cₚ dT=Cₚ(T₀-Tₕₒₜ)<0.
  2. Sign of Δ G. G is a state function, and at constant pressure dG=-S dT. The tea has S>0 and dT<0, so Δ G=-∫_Tₕₒₜ^(T₀)S dT>0Δ G>0 The Gibbs energy of the tea rises as it cools (G=H-TS; TS falls more than H).
  3. No contradiction. Δ G≤0 is the test of spontaneity only for a change at constant T and p. The tea's temperature changes, so its Δ G says nothing about spontaneity.
  4. The real criterion is the total entropy. Tea: Δ Sₜₑₐ=Cₚln(T₀/Tₕₒₜ)<0. Room (at T₀): Δ Sᵣₒₒₘ=-Δ H/T₀=Cₚ(Tₕₒₜ-T₀)/T₀>0. Since (x-1)>ln x for x=Tₕₒₜ/T₀>1, Δ Sₜₒₜₐₗ>0: the cooling is spontaneous.
  5. Numbers. One mole of tea taken as water, 80°C → 25°C (Cₚ=75.3 J K⁻¹ mol⁻¹, Sₘ=69.9 J K⁻¹ mol⁻¹ at 298 K): Δ H=-4.14 kJ; Δ Sₜₑₐ=-12.7 J K⁻¹; Δ Sᵣₒₒₘ=+13.9 J K⁻¹; Δ Sₜₒₜₐₗ=+1.1 J K⁻¹; Δ Gₜₑₐ=-∫ S dT=+4.21 kJ.
  6. Where "Δ G<0" comes from. The quantity that is negative is Δ H-T₀Δ Sₜₑₐ, the Gibbs energy change counted at the fixed room temperature T₀: Δ H-T₀Δ Sₜₑₐ=-T₀Δ Sₜₒₜₐₗ=-0.34 kJ. Answer keys that print Δ G<0 mean this; for the tea itself Δ G>0.

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Model answer to Q3(a)(i), 2017 Paper I: solution page 11
Solution page 11 of 38
Read the whole 2017 solution2017 question paper

All parts of Question 3