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Use the following data to calculate Delt...

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)`

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Given that, `Delta_("sub") H^(@)` for Na metal = `180.4 kJ mol^(-1)`
IE of `Na = 496 kJ mol^(-1) , Delta_(eg) H^@ "of" Br = -325 kJ mol^(-1) , Delta_("diss") H^@ " of" Br = 192 kJ mol^(-1) , Delta_(f) H^(@)`
For `NaBr = 360.1 kJ mol^(-1)`
Born-Haber cycle for the formation of NaBr is as

By applying Hess's law: `Delta_fH^@ = Delta_("sub")H^(@) + IE + Delta_("diss")H^(@) + Delta_(eg)H^(@) + U`
`-360.1 = 108.4 + 496 + 96 + (-325) - U`
`U = +735.5 kJ mol^(-1)`.
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