Using the Following Data to Calculate the Lattice Energy of NaBr(s)

Using the following data to calculate the lattice energy of NaBr(s) . ?IE and ?EA are enthalpy of ionization and electron attachment enthalpy, respectively. $$\begin{array} { l l } \mathrm { Na } ( \mathrm { s } ) \rightarrow \mathrm { Na } ( \mathrm { g } ) & \Delta _f H ^ { \circ } = + 107 \mathrm {~kJ} \\\mathrm { Na } ( \mathrm { g } ) \rightarrow \mathrm { Na } ^ { + } ( \mathrm { g } ) + \mathrm { e } ^ { - } & \Delta I E = + 496 \mathrm {~kJ} \\1 / 2 \mathrm { Br } _ { 2 } ( \mathrm {~g} ) \rightarrow \mathrm { Br } ( \mathrm { g } ) & \Delta_f H ^ { \circ } = + 112 \mathrm {~kJ} \\\mathrm { Br } ( \mathrm { g } ) + \mathrm { e } ^ { - } \rightarrow \mathrm { Br } ^ { - } ( \mathrm { g } ) & \Delta E A = - 325 \mathrm {~kJ} \\\mathrm { Na } ( \mathrm { s } ) + 1 / 2 \mathrm { Br } _ { 2 } ( \mathrm {~g} ) \rightarrow \mathrm { NaBr } ( \mathrm { s } ) & \Delta_f H ^ { \circ } = - 361 \mathrm {~kJ}\end{array}$$

A) -1401 kJ
B) -751 kJ
C) -241 kJ
D) +241 kJ
E) +751 kJ