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(*) $= - 360.1 kJ mol^(-1)$

Answer 1

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