The ionization enthalpy of ${\mathrm{Na}}^{+}$ formation from ${\mathrm{Na}}_{\left(\mathrm{g}\right)}$ is $495.8\mathrm{kJ}{\mathrm{mol}}^{-1}$ , while the electron gain enthalpy of Br is -325.0 kJ ${\mathrm{mol}}^{-1}$. Given the lattice enthalpy of NaBr is $-728.4\mathrm{kJ}{\mathrm{mol}}^{-1}$ . The energy for the formation of  NaBr ionic solid is $(-)\mathit{\_}\mathit{\_}\mathit{\_}\mathit{\_}\mathit{\_}\mathit{\_}\mathit{\_}\mathit{\times }{\mathit{10}}^{\mathit{-}\mathit{1}}kJmo{l}^{\mathit{-}\mathit{1}}$ .