Q6(e). Sulfur melts but iodine sublimes on heating in an open vessel
Solid sulphur on heating in an open container melts, but solid iodine sublimes. Explain why.
Answer in brief
Iodine has a high vapour pressure below its melting point; in an open vessel its partial pressure stays below the triple-point pressure, so it passes straight to vapour. Sulfur has a very small vapour pressure up to its melting point, so it melts.
Key points
- a solid can melt only if the pressure of its own vapour above it reaches the triple-point pressure. If the partial pressure of the substance stays below it, heating moves the solid straight across the solid–vapour curve:…
- I₂ molecules are held only by van der Waals forces, so the solid has a high vapour pressure even at room temperature (violet vapour, smell). Its triple point lies just below its melting point (114°C), at an appreciable…
- In an open vessel the violet vapour diffuses away and mixes with air. The partial pressure of I₂ above the solid stays below the triple-point pressure, so the crystals vaporise from the surface as fast as they are heated and…
- the large S₈ rings have more intermolecular contact in the crystal and a much lower vapour pressure; at its melting point (113°C orthorhombic, 119°C monoclinic) the vapour pressure, and so the triple-point pressure, is tiny:…
The full model answer
Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.
- Phase-diagram rule: a solid can melt only if the pressure of its own vapour above it reaches the triple-point pressure. If the partial pressure of the substance stays below it, heating moves the solid straight across the solid–vapour curve: sublimation. (Solid CO₂, whose triple point lies above 1 atm, sublimes for the same reason.)
- Iodine: I₂ molecules are held only by van der Waals forces, so the solid has a high vapour pressure even at room temperature (violet vapour, smell). Its triple point lies just below its melting point (114°C), at an appreciable pressure that is still well below 1 atm.
- In an open vessel the violet vapour diffuses away and mixes with air. The partial pressure of I₂ above the solid stays below the triple-point pressure, so the crystals vaporise from the surface as fast as they are heated and sublime before melting. (In a closed tube, where the vapour builds up, iodine does melt to a dark liquid.)
- Sulfur: the large S₈ rings have more intermolecular contact in the crystal and a much lower vapour pressure; at its melting point (113°C orthorhombic, 119°C monoclinic) the vapour pressure, and so the triple-point pressure, is tiny: the sulfur vapour over the solid reaches it even in an open vessel. Heating therefore reaches the solid–liquid line first: sulfur melts to a yellow liquid.
- The deciding factor is not the melting point (the two are almost the same) but the vapour pressure at the melting point.
Application. sublimation is used to purify iodine, naphthalene and camphor from non-volatile impurities; sulfur is purified instead by melting and distillation (Frasch process gives molten sulfur).
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All parts of Question 6
- 6(a) cₘₚ, cₐᵥ, cᵣₘₛ on the Maxwell distribution curve10 marks
- 6(b) Coordination number of the cations in NaBr and KBr from radius ratios10 marks
- 6(c) Excess pressure inside a soap bubble10 marks
- 6(d) Maxwell relation (∂ S/∂ P)_T=-(∂ V/∂ T)_P from dG10 marks
- 6(e) Sulfur melts but iodine sublimes on heating in an open vessel10 marks