question_answer 61)

  Use the following data to calculate \[{{\Delta }_{lattice}}{{H}^{O-}}\]for\[NaBr.\]\[{{\Delta }_{sub}}{{H}^{O-}}\]for sodium metal = \[108.4kJ\,mo{{l}^{-1}}\] lonization enthalpy of sodium = \[496kJ\,mo{{l}^{-1}}\]Electron gain enthalpy of bromine = \[-325\,kJ\,mo{{l}^{-1}}\] Bond dissociation enthalpy of bromine = \[192kJ\,mo{{l}^{-1}}\] \[{{\Delta }_{f}}{{H}^{O-}}\]for\[NaBr(s)=-360.1kJ\,mo{{l}^{-1}}\]

Answer:

      \[{{\Delta }_{f}}H_{NaBr}^{{}^\circ }={{\Delta }_{sub}}{{H}^{{}^\circ }}+\frac{1}{2}{{\Delta }_{d}}{{H}^{{}^\circ }}+{{\Delta }_{ie}}{{H}^{{}^\circ }}+{{\Delta }_{eg}}{{H}^{{}^\circ }}+{{\Delta }_{lattice}}H{}^\circ \] \[-360.1=+108.4+96+496-325+{{\Delta }_{lattice}}{{H}^{{}^\circ }}\] \[{{\Delta }_{lattice}}{{H}^{{}^\circ }}=-360.1-108.4-96-496+325\]                                 \[=-735.5kJ\,mo{{l}^{-1}}\]