Determining the heat of formation for a substance

  • Ex: Use the following data to estimate ∆HfO for sodium chloride
  • Na(s) + ½ Cl2(g) --> NaCl(s)
  • Lattice energy -786 kJ/mol
  • Electron affinity(Cl) -348 kJ/mol
  • Ionization energy (Na) 495 kJ/mol
  • Bond energy (Cl2) 239 kJ/mol
  • Heat sublimation: ∆Hsub Na 108 kJ/mol
  • Na(s) à Na(g) is sublimation 108kJ/mol
  • Na(g) --> Na+(g) is ionization 495 kJ/mol
  • ½ Cl2(g) --> Cl(g) is breaking a bond and only have half a mole so 239/2 kJ/mol
  • Cl(g) --> Cl-(g) is electron affinity -348 kJ/mol
  • Na+(g) + Cl-(g) --> NaCl(s) is lattice energy -786 kJ/mol
  • Sum the heats and reactions:
  • Na(s) + 1/2Cl2(g) --> NaCl(s)      -412kJ/mol