Assinale a alternativa que mais se aproxima da entalpia de formação do metanol.

Dados

  • ΔHsub∘(grafite)=716 kJmol\Delta H_\mathrm{sub}^{\circ}(\ce{grafite}) = \pu{716 kJ//mol}
  • ΔHL(HX2)=436 kJmol\Delta H_\mathrm{L}(\ce{H2}) = \pu{436 kJ//mol}
  • ΔHL(OX2)=496 kJmol\Delta H_\mathrm{L}(\ce{O2}) = \pu{496 kJ//mol}
  • ΔHL(C−H)=412 kJmol\Delta H_\mathrm{L}(\ce{C-H}) = \pu{412 kJ//mol}
  • ΔHL(C−O)=360 kJmol\Delta H_\mathrm{L}(\ce{C-O}) = \pu{360 kJ//mol}
  • ΔHL(O−H)=463 kJmol\Delta H_\mathrm{L}(\ce{O-H}) = \pu{463 kJ//mol}
Gabarito 3X.07

ΔH°f(CH3OH)=ΔH°sub(grafita)+2ΔHL(HX2)+12ΔHL(OX2)−(3ΔHL(C−H)+ΔHL(C−O)+ΔHL(O−H))\Delta H\degree_f(CH3OH)=\Delta H\degree_{sub}(grafita)+2\Delta H_L(\ce{H2})+\frac{1}{2}\Delta H_L(\ce{O2})-(3\Delta H_L(\ce{C-H})+\Delta H_L(\ce{C-O})+\Delta H_L(\ce{O-H})) ΔH°f(CH3OH)=716 kJmol+872 kJmol+248 kJmol−1236 kJmol−360 kJmol−463 kJmol\Delta H\degree_f(CH3OH)=\pu{716 kJ//mol}+\pu{872 kJ//mol}+\pu{248 kJ//mol}-\pu{1236 kJ//mol}-\pu{360 kJ//mol}-\pu{463 kJ//mol} ΔH°f(CH3OH)=−223 kJmol\Delta H\degree_f(CH3OH)=\pu{-223 kJ//mol}