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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 · 2018 · Paper I

Q4(d)(i). Slopes of the fusion and sublimation curves of water

  • 5 marks
  • Answered
Question

Explain why fusion curve of ice has negative slope whereas sublimation curve has positive slope in the phase diagram.

Answer in brief

Clapeyron: dp/dT=Δ H/TΔ V. Both Δ_(fus)H and Δ_(sub)H are positive. Ice is less dense than water, so Δ_(fus)V<0 and the fusion slope is negative (-135 bar K⁻¹); on sublimation the volume increases enormously, so Δ_(sub)V>0 and the slope is positive.

Key points

  • Along any two-phase line μ_α=μ_β, which gives the Clapeyron equation (dp)/(dT)=ΔₜᵣₛHT ΔₜᵣₛV. Melting and sublimation both absorb heat, so the sign of the slope is the sign of Δ V.
  • ice has an open, hydrogen-bonded structure and is less dense than water (0.917 against 1.000 g cm⁻³ at 0°C). Melting shrinks the volume: Δ_(fus)V=18.02-19.65=-1.63 cm³ mol⁻¹. 2pt] (dp)/(dT)=6008273.15×(-1.63×10⁻⁶)= -1.35×10⁷…
  • higher pressure lowers the melting point (by about 0.74 K per 100 bar). Pressure favours the denser phase, water (Le Chatelier).
  • ice → vapour; Δ_(sub)H>0 and the vapour occupies far more volume than the solid, Δ_(sub)V≈ Vₘ(g)=RT/p>0. The slope is positive: (dln p)/(dT)=Δ_(sub)HRT² (Clausius–Clapeyron), and the vapour pressure of ice rises steeply with T.

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. Along any two-phase line μ_α=μ_β, which gives the Clapeyron equation (dp)/(dT)=ΔₜᵣₛHT ΔₜᵣₛV. Melting and sublimation both absorb heat, so the sign of the slope is the sign of Δ V.
  2. Fusion: ice has an open, hydrogen-bonded structure and is less dense than water (0.917 against 1.000 g cm⁻³ at 0°C). Melting shrinks the volume: Δ_(fus)V=18.02-19.65=-1.63 cm³ mol⁻¹. 2pt] (dp)/(dT)=6008273.15×(-1.63×10⁻⁶)= -1.35×10⁷ Pa K⁻¹ (-135 bar K⁻¹)
  3. The line leans to the left: higher pressure lowers the melting point (by about 0.74 K per 100 bar). Pressure favours the denser phase, water (Le Chatelier).
  4. Sublimation: ice → vapour; Δ_(sub)H>0 and the vapour occupies far more volume than the solid, Δ_(sub)V≈ Vₘ(g)=RT/p>0. The slope is positive: (dln p)/(dT)=Δ_(sub)HRT² (Clausius–Clapeyron), and the vapour pressure of ice rises steeply with T.
  5. Water is unusual; for most substances (CO₂) the solid is denser and the fusion line leans to the right.

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Model answer to Q4(d)(i), 2018 Paper I: solution page 28
Solution page 28 of 55
Read the whole 2018 solution2018 question paper

All parts of Question 4