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).

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

1.	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).
Answer» SOLUTION :Given that, `Delta_("sub") H^( THETA )` for `Na` metal `= 108.4 "kJ mol"^(-1)" IE"` of `Na= 496 "kJ mol"^(-1) , Delta_("eg") H^( Theta )` of `BR = - 325 "kJ mol"^(-1), Delta_("DISS") H^( Theta )` of `Br = 192 "kJ mol"^(-1) , Delta H^( Theta )` for `NaBr = - 360.1 "kJ mol"^(-1)` Born-Haber cycle for the formation of NaBr is as By applying Hess.s law, `Delta_(f) H^( Theta )= Delta_("sub") H^( Theta ) + "IE" +Delta_("diss") H^( Theta ) + Delta_("eg") H^( Theta ) + U - 360.1 = 108.4 + 496 + 96 + (-325) - U` `U= + 735.5 "kJ mol"^(-1)`