Use the following data to calculate $\Delta _{lattice}H^{\theta }$ for $NaBr$. $\Delta _{sub}H^{\theta }$ for sodium metal $= 108.4 \ kJ \ mol^{-1}$,ionization enthalpy of sodium $= 496 \ kJ \ mol^{-1}$,electron gain enthalpy of bromine $= -325 \ kJ \ mol^{-1}$,bond dissociation enthalpy of bromine $= 192 \ kJ \ mol^{-1}$,$\Delta _{f}H^{\theta }$ for $NaBr_{(s)}$ $= -360.1 \ kJ \ mol^{-1}$.

(D) According to the Born-Haber cycle,the standard enthalpy of formation $\Delta _{f}H^{\theta }$ is given by the sum of all energy changes involved in the process:
$\Delta _{f}H^{\theta } = \Delta _{sub}H^{\theta } + IE + \frac{1}{2}\Delta _{diss}H^{\theta } + \Delta _{eg}H^{\theta } + \Delta _{lattice}H^{\theta }$
Substituting the given values:
$-360.1 = 108.4 + 496 + \frac{1}{2}(192) + (-325) + \Delta _{lattice}H^{\theta }$
$-360.1 = 108.4 + 496 + 96 - 325 + \Delta _{lattice}H^{\theta }$
$-360.1 = 375.4 + \Delta _{lattice}H^{\theta }$
$\Delta _{lattice}H^{\theta } = -360.1 - 375.4 = -735.5 \ kJ \ mol^{-1}$