Assinale a alternativa que mais se aproxima da segunda afinidade eletrônica do enxofre.

Dados

  • ΔHrede∘(KX2S)=2050 kJmol\Delta H_\mathrm{rede}^{\circ}(\ce{K2S}) = \pu{2050 kJ//mol}
  • ΔHf∘(KX2S,  s)=−381 kJmol\Delta H_\mathrm{f}^{\circ}(\ce{K2S,\,\text{s}}) = \pu{-381 kJ//mol}
  • ΔHsub∘(S)=277 kJmol\Delta H_\mathrm{sub}^{\circ}(\ce{S}) = \pu{277 kJ//mol}
  • ΔHsub∘(K)=90 kJmol\Delta H_\mathrm{sub}^{\circ}(\ce{K}) = \pu{90 kJ//mol}
Gabarito 3X.04

ΔH°f(KX2SX (s))=2ΔH°sub(K)+2ΔH°ion(K)+ΔH°sub(S)+ΔH°ge(S)+ΔH°ge(SX−)+ΔH°rede(KX2S)\Delta H\degree_f(\ce{K2S_{(s)}})=2\Delta H\degree_{sub}(\ce{K})+2\Delta H\degree_{ion}(\ce{K})+\Delta H\degree_{sub}(\ce{S})+\Delta H\degree_{ge}(\ce{S})+\Delta H\degree_{ge}(\ce{S^-})+\Delta H\degree_{rede}(\ce{K2S}) −381 kJmol=180 kJmol+8 eV 6,02⋅1023 mol−1 1,6⋅10−22kJeV+277 kJmol+(−2 eV 6,02⋅1023 mol−1 1,6⋅10−22kJeV)+ΔH°ge(SX−)+(−2050 kJmol)\pu{-381 kJ//mol}=\pu{180 kJ//mol}+\pu{8 eV 6,02\cdot 10^{23} mol^{-1} 1,6\cdot 10^{-22}\frac{kJ}{eV}}+\pu{277 kJ//mol}+(\pu{-2 eV 6,02\cdot 10^{23} mol^{-1} 1,6\cdot 10^{-22}\frac{kJ}{eV}})+\Delta H\degree_{ge}(\ce{S^-})+(\pu{-2050 kJ//mol}) Eae2(S) 6,02⋅1023 mol−1 1,6⋅10−22kJeV=634,08 kJmol  ⟹  Eae2(S)≈6,583 eV\pu{E_{ae2}(\ce{S}) 6,02\cdot 10^{23} mol^{-1} 1,6\cdot 10^{-22}\frac{kJ}{eV}}=\pu{634,08 kJ//mol}\implies E_{ae2}(\ce{S})\approx\pu{6,583 eV}