The triple point of Iodine occurs at 0.12 atm and at standard temperature it is a solid on the phase diagram.

enter image description here

I understand that therefore, when heating iodine at 1 atm, it (at least some of it) will pass through the liquid phase, and it is also obvious from observing heated Iodine that it yields liquid when heated gradually. Also, just as liquids can evaporate below the boiling point, I understand that some Iodine can escape from a sample in the vapor form when the vapor pressure is lower than the atmospheric pressure.

My question is 2 part:

1) Is it correct to call the change form solid to gas below equilibrium vapor pressure "sublimation". In other words, is sublimation equivalent to the term "boiling" or is it equivalent to the term evaporation in the change of phase from liquid to vapor, or both? (Is there a separate term, even if out of use, for the solid to vapor transition below equilibrium pressure?)

2) At 1 atm, when Iodine molecules in a solid sample escape into the vapor form, either at standard temperature, or as the sample is heated, do they all briefly pass through the liquid phase first, either at a vapor pressure below 1 atm or as vapor pressure is raised above 1 atm. . enter image description here

Iodine's crystal vapor pressure will of course be <0.12 atm below the triple point pressure as can be seen on both graphs. A converse but equivalent question would be whether vaporous Iodine at 1 atm can crystalize and deposit without liquifying.

  • $\begingroup$ Related: chemistry.stackexchange.com/questions/37413/… $\endgroup$ Dec 9 '16 at 20:36
  • $\begingroup$ So trying to distill out the essence of my question, is the heat of sublimation going to be equal to the heat of fusion+ heat of vaporization at 1 atm? Anyway, when heating Iodine in a flask, it is very easy to see liquid Iodine in the bottom of the flask. It can be swirled and will "freeze" to fill the shape of the container. When the vapor deposits on the sides of the flask, it looks to be atomized, not crystals formed from condensed droplets of liquid. $\endgroup$ Dec 9 '16 at 21:07
  • $\begingroup$ But it looks like what may be the case is that melting does not happen spontaneously below the melting point because the liquid is not free to escape contact with the solid, but vaporization can happen because the energy can be carried away. $\endgroup$ Dec 9 '16 at 21:09

To answer part 1 of your question:

It is correct to call the change form solid to gas below the equilibrium vapor pressure sublimation. And there is no separate term for vaporization of a solid above the equilibrium vapor pressure; it is still sublimation. Similarly, a boiling liquid is still perfectly correctly said to be evaporating.

To answer part 2 of your question:

At temperatures below the melting point of iodine, it will sublimate until the equilibrium vapor pressure is reached without going through a liquid phase in that process.

Once the iodine is heated to it's melting point, it will continue the same process of vaporizing to it's equilibrium vapor pressure. Because it is now a liquid, we just call that evaporation rather than sublimation.

If the iodine is further heated to it's boiling point at 1 atmosphere, then it will continue to rapidly evaporate as it boils away.

Regarding the thermodynamic relationships between these processes as mentioned in one of your comments (which I think you should just edit into the question), what you have said is correct. This is a path-independent thermodynamic process, so with respect to the ΔH of the phase-change processes, it doesn't matter how you get from the solid phase to the vapor phase. ΔH for the direct solid-vapor transition (sublimation) is equal to the ΔH of melting plus the ΔH of evaporation.

I hope I have addressed all your points, please ask for clarification in the comments if I've left anything unclear.

  • $\begingroup$ I'd quibble a tiny bit. At the melting point you could have both solid and liquid iodine. It all has to go to liquid before you could get above the melting point of course. $\endgroup$
    – MaxW
    Feb 19 '17 at 1:16
  • $\begingroup$ The "Once the iodine is heated to it's melting point..." statement could be interpreted as saying that the vapor pressure would continually increase with increased heat input, which as you say is not the case. Once the solid starts to melt, the additional heat will go into melting the solid and neither temperature nor vapor pressure will increase again until all of the solid is melted. Of course at the melting point both the liquid and solid have the same vapor pressure, and the rate of sublimation of the solid is equal to the rate of evaporation of the liquid. Hope I got the right quibble ;) $\endgroup$
    – airhuff
    Feb 19 '17 at 1:42
  • $\begingroup$ Paragraph starting "Once the iodine is heated to it's melting point, ..." then the second sentence starts off "Since it now all a liquid ..." // My quibble is that it doesn't all have to be liquid at the melting point. $\endgroup$
    – MaxW
    Feb 19 '17 at 1:46

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