Use the following data to calculate `Delta_("lattice") H^(@)` for NaBr. `Delta_("sub")H^(@)` 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^(@)` for `NaBr(s) - 360 kJ mol^(-1)`.

Explanation

Given that, `Delta_("sub") H^(Θ)` for `Na` metal `= 108.4 kJ mol^(-1)`
IE of `Na = 496 kJ mol^(-1) " of " Br = - 325 kJ mol^(-1), Delta_("diss") H^(Θ) " of " Br = 192 kJ mol^(-1), Delta_(f) H^(Θ) " for " NaBr = - 360.1 kJ mol^(-1)`
Born-Haber cycle for the fomation of `NaBr` is as

By applying Hess's law,
`Delta_(f) H^(Θ) + IE + Delta_("diss") H^(Θ) + Delta_("eg") H^(Θ) + U`
`-360.1 = 108.4 + 496 + 96 + (-325) - U`
`U = + 735.5 kJ mol^(-1)`