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I am not able to understand why we use enthalpy of sublimation rather than enthalpy of fusion and enthalpy of vapourisation of solid sodium metal in a Born-Haber cycle for the formation of NaCl from $sodium_{(s)}$ and $chlorine_{(g)}$ in their standard states.

Also I have rarely heard of sublimation of such metals.Is it like sublimation of camphor.

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  • $\begingroup$ Thermodynamically, how would you relate the enthalpy of fusion plus enthalpy of vaporization with the enthalpy of sublimation? Think about that. $\endgroup$ – Jon Custer Jul 15 '16 at 13:48
  • $\begingroup$ @Jon Custer The sum may be equal to the enthalpy of sublimation.But I dont think this is the standard way of writing in textbooks.If it is written there, then it may mean that we are avoiding the liquid phase of the metal. $\endgroup$ – Sikander Jul 15 '16 at 13:50
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Sublimation of these metals is much like sublimation of dry ice. the only difference is that standard pressure is higher than the triple point of Na and lower than the triple point of dry ice.

You have a path: take Na and bring it from solid to gas phase. take Na gas and ionize it. Ionize fluorine. Burn Sodium in fluorine. Energies of each step are known. Now if you sum them up you get a net reaction of Na+ + F- = NaF

if by energy of fusion you mean "energy to melt material" then it brings you nowhere. You cannot efficiently (to 100%) ionize liquid Na. But you can easily measure the energy required to ionize Na vapor.

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  • $\begingroup$ Can you explain "...higher than the triple point of sodium and lower than the triple point of dry ice."Moreover I still don't get in your answer why didn't we use both enthalpy of fusion and enthalpy of vaporisation rather than enthalpy of sublimation as I mentioned in my question. $\endgroup$ – Sikander Jul 15 '16 at 15:47

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