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