use the following data to calculate Delta_(" lattice")H^(@) for NaBr . Delta("sub")H^(-) fpr sodium metal = 108 .4 kJmol^(-1)lonisation enthaply of sodium = 496 kJ mol^(-1) ltbr gt Electron gain enthalpy of bromine =325 kJ mol ^(-1) bond dissociation enthalpy of bromine =192 kJ mol ^(-1) Delta_(t)H^(-)for NaBr (s) = -360 .1 kJ mol ^(-1)

use the following data to calculate Delta_(" lattice")H^(@) for NaBr .

1.	use the following data to calculate Delta_(" lattice")H^(@) for NaBr . Delta("sub")H^(-) fpr sodium metal = 108 .4 kJmol^(-1)lonisation enthaply of sodium = 496 kJ mol^(-1) ltbr gt Electron gain enthalpy of bromine =325 kJ mol ^(-1) bond dissociation enthalpy of bromine =192 kJ mol ^(-1) Delta_(t)H^(-)for NaBr (s) = -360 .1 kJ mol ^(-1)
Answer» `-735.5 kJ mol^(-1)` `+735.5kJ mol ^(-1)` `-789.89 JK^(-1)mol ^(-1)` `+735 . 5 J mol ^(-1)` Solution :By applying Hess's law `Delta_(F)H^(Theta)=Delta_("sub")H^(@)+lE+Delta_(diss)H^(@)+Delta_(EQ)H^(@) +U` `-360.1 = 108.4 + 496 + 96 +(-325)+U` ` U =+ 735.5 kJ mol^(-1)`