Assinale a alternativa que mais se aproxima da entalpia de rede do cloreto de cálcio.

Dados

  • ΔHf∘(CaClX2,  s)=−796 kJmol\Delta H_\mathrm{f}^{\circ}(\ce{CaCl2,\,\text{s}}) = \pu{-796 kJ//mol}
  • ΔHion∘(Ca)=590 kJmol\Delta H_\mathrm{ion}^{\circ}(\ce{Ca}) = \pu{590 kJ//mol}
  • ΔHion∘(CaX+)=1146 kJmol\Delta H_\mathrm{ion}^{\circ}(\ce{Ca^+}) = \pu{1146 kJ//mol}
  • ΔHge∘(Cl)=−349 kJmol\Delta H_\mathrm{ge}^{\circ}(\ce{Cl}) = \pu{-349 kJ//mol}
  • ΔHL(ClX2)=242 kJmol\Delta H_\mathrm{L}(\ce{Cl2}) = \pu{242 kJ//mol}
  • ΔHsub∘(Ca)=190 kJmol\Delta H_\mathrm{sub}^{\circ}(\ce{Ca}) = \pu{190 kJ//mol}
Gabarito 3X.02

ΔH°f(CaClX2X (s))=ΔH°sub(Ca)+ΔH°ion(Ca)+ΔH°ion(CaX+)+ΔHL(ClX2)+2ΔH°ge(Cl)+ΔH°rede(CaClX2)\Delta H\degree_f(\ce{CaCl2_{(s)}})=\Delta H\degree_{sub}(\ce{Ca})+\Delta H\degree_{ion}(\ce{Ca})+\Delta H\degree_{ion}(\ce{Ca^+})+\Delta H_{L}(\ce{Cl2})+2\Delta H\degree_{ge}(\ce{Cl})+\Delta H\degree_{rede}(\ce{CaCl2}) −796 kJmol=190 kJmol+590 kJmol+1150 kJmol+242 kJmol+(−698 kJmol)+ΔH°rede(CaClX2)\pu{-796 kJ//mol}=\pu{190 kJ//mol}+\pu{590 kJ//mol}+\pu{1150 kJ//mol}+\pu{242 kJ//mol}+(\pu{-698 kJ//mol})+\Delta H\degree_{rede}(\ce{CaCl2}) ΔH°rede(CaClX2)=−2270 kJmol\Delta H\degree_{rede}(\ce{CaCl2})=\pu{-2270 kJ//mol}