Assinale a alternativa que mais se aproxima da entalpia de sublimação do lítio.

Dados

  • ΔHf∘(LiI, s)=−292 kJmol\Delta H_\mathrm{f}^{\circ}(\ce{LiI,s}) = \pu{-292 kJ//mol}
  • ΔHrede∘(LiI)=753 kJmol\Delta H_\mathrm{rede}^{\circ}(\ce{LiI}) = \pu{753 kJ//mol}
  • ΔHion∘(Li)=520 kJmol\Delta H_\mathrm{ion}^{\circ}(\ce{Li}) = \pu{520 kJ//mol}
  • ΔHge∘(I)=−295 kJmol\Delta H_\mathrm{ge}^{\circ}(\ce{I}) = \pu{-295 kJ//mol}
  • ΔHL(IX2)=151 kJmol\Delta H_\mathrm{L}(\ce{I2}) = \pu{151 kJ//mol}
Gabarito 3X.03

ΔH°f(LiIX (s))=ΔH°sub(Li)+ΔH°ion(Li)+12ΔHL(IX2)+ΔH°ge(I)+ΔH°rede(LiI)\Delta H\degree_f(\ce{LiI_{(s)}})=\Delta H\degree_{sub}(\ce{Li})+\Delta H\degree_{ion}(\ce{Li})+\frac{1}{2}\Delta H_{L}(\ce{I2})+\Delta H\degree_{ge}(\ce{I})+\Delta H\degree_{rede}(\ce{LiI}) −292 kJmol=ΔH°sub(Li)+520 kJmol+75,5 kJmol+(−295 kJmol)+(−753 kJmol)\pu{-292 kJ//mol}=\Delta H\degree_{sub}(\ce{Li})+\pu{520 kJ//mol}+\pu{75,5 kJ//mol}+(\pu{-295 kJ//mol})+(\pu{-753 kJ//mol}) ΔH°sub(Li)=160,5 kJmol\Delta H\degree_{sub}(\ce{Li})=\pu{160,5 kJ//mol}