Use the following data to calculate Delta_("lattice") H^(Θ) " for " NaBr. Delta_("Sub")H^(Θ) for sodium metal = 108.4 kJ mol^(-1), ionisation 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^(Θ) " for " NaBr(s) = - 360.1 kJ mol^(-1)

Use the following data to calculate Delta_("lattice") H^(Θ) " for " N

1.	Use the following data to calculate Delta_("lattice") H^(Θ) " for " NaBr. Delta_("Sub")H^(Θ) for sodium metal = 108.4 kJ mol^(-1), ionisation 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^(Θ) " for " NaBr(s) = - 360.1 kJ mol^(-1)
Answer» Solution :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)`