Answer
$\Delta H_{l}=-702.0\ kJ/mol$
Work Step by Step
Dissociation enthalpy: $1/2\ I_2(g)\rightarrow I(g),\ \Delta H_{1A}=+106.838\ kJ/mol$
Enthalpy of formation, gaseous sodium: $Na(s)\rightarrow Na(g),\ \Delta H_{2A}=+107.3\ kJ/mol$
Electron attachment energy, iodine: $I(g)+e^-\rightarrow I^-(g),\ \Delta H_{1B}=-295.16\ kJ/mol$
Ionization energy, sodium: $Na(g)\rightarrow Na^+(g)+e^-,\ \Delta H_{2B}=+496\ kJ/mol$
Lattice energy: $Na^+(g)+I^-(g)\rightarrow NaI(s),\ \Delta H_l$
Enthalpy of formation, sodium iodide: $Na(s)+1/2\ I_2(g)\rightarrow NaI(s),\ \Delta H_{f}=-287\ kJ/mol$
Combining equations we get:
$\Delta H_{1A}+\Delta H_{2A}+\Delta H_{1B}+\Delta H_{2B}+\Delta H_{l}=\Delta H_{f}$, so:
$\Delta H_{l}=-702.0\ kJ/mol$
