Question

Use the following data to calculate `Delta_("lattice") H^(c-)` for NaBr.
`Delta_("sub") H^(c-)` for sodium metal `=108.4 kJ mol^(-1)`
Ioniztion enthalpy ofsodium `= 496kJ mol^(-1)`
Electron gain enthalpy of bromine `= - 325 kJ mol^(-1)`
Bond dissociation enthalpy of bromine `= 192kJ mol^(-1)`
`Delta_(f)H^(c-)` for NaBr(s) `= - 360.1 kJ mol^(-1)`

Answer 1

Proceed as in calculation of lattice enthalpy for NaCl on page .
We get `Delta_("lattice") H^(@) ( NaBr)= - 360 .1 kJ mol^(-1)`