Considere a reação: CClX3CHClX2(g)+2 HF(g)→CClX3CHFX2(g)+2 HCl \ce{ CCl3CHCl2(g) + 2 HF(g) -> CCl3CHF2(g) + 2 HCl }

Assinale a alternativa que mais se aproxima da entalpia de reação.

Dados

  • ΔHL(HF)=565 kJmol\Delta H_\mathrm{L}(\ce{HF}) = \pu{565 kJ//mol}
  • ΔHL(HCl)=431 kJmol\Delta H_\mathrm{L}(\ce{HCl}) = \pu{431 kJ//mol}
  • ΔHL(C−Cl)=338 kJmol\Delta H_\mathrm{L}(\ce{C-Cl}) = \pu{338 kJ//mol}
  • ΔHL(C−F)=484 kJmol\Delta H_\mathrm{L}(\ce{C-F}) = \pu{484 kJ//mol}
  • ΔHL(C−H)=412 kJmol\Delta H_\mathrm{L}(\ce{C-H}) = \pu{412 kJ//mol}
Gabarito 3X.05

ΔH=5ΔHL(C−Cl)+ΔHL(C−C)+ΔHL(C−H)+2ΔHL(HF)−(3ΔHL(C−Cl)+ΔHL(C−C)+ΔHL(C−H)+2ΔHL(C−F)+2ΔHL(HCl))\Delta H=5\Delta H_L(\ce{C-Cl})+\Delta H_L(\ce{C-C})+\Delta H_L(\ce{C-H})+2\Delta H_L(\ce{HF})-(3\Delta H_L(\ce{C-Cl})+\Delta H_L(\ce{C-C})+\Delta H_L(\ce{C-H})+2\Delta H_L(\ce{C-F})+2\Delta H_L(\ce{HCl})) ΔH=2⋅338 kJmol+2⋅565 kJmol−2⋅484 kJmol−2⋅431 kJmol\Delta H=2\cdot\pu{338 kJ//mol}+2\cdot\pu{565 kJ//mol}-2\cdot\pu{484 kJ//mol}-2\cdot\pu{431 kJ//mol} ΔH=−24 kJmol\Delta H=\pu{-24 kJ//mol}