The Lattice Energy of NaCl(s) Is -786 KJmol A) -1921 KJ/mol
B) -473 KJ/mol
C) -349 KJ/mol
D)

Multiple Choice

The lattice energy of NaCl(s) is -786 kJmol. Use this value and the following thermochemical data to determine the electron attachment enthalpy of Cl(g) . ?IE is the enthalpy of ionization. $$\begin{array} { l l } \mathrm { Na } ( \mathrm { s } ) \rightarrow \mathrm { Na } ( \mathrm { g } ) & \Delta _ { f } H ^ { \circ } = + 107 \mathrm {~kJ} / \mathrm { mol } \\\mathrm { Na } ( \mathrm { g } ) \rightarrow \mathrm { Na } ^ { + } ( \mathrm { g } ) + \mathrm { e } ^ { - } & \Delta I E = + 496 \mathrm {~kJ} / \mathrm { mol } \\1 / 2 \mathrm { Cl } _ { 2 } ( \mathrm {~g} ) \rightarrow \mathrm { Cl } ( \mathrm { g } ) & \Delta _ { f } H ^ { \circ } = + 121 \mathrm {~kJ} / \mathrm { mol } \\\mathrm { Na } ( \mathrm { s } ) + 1 / 2 \mathrm { Cl } _ { 2 } ( \mathrm {~g} ) \rightarrow \mathrm { NaCl } ( \mathrm { s } ) & \Delta _ { f } H ^ { \circ } = - 411 \mathrm {~kJ} / \mathrm { mol }\end{array}$$

A) -1921 kJ/mol
B) -473 kJ/mol
C) -349 kJ/mol
D) +349 kJ/mol
E) +473 kJ/mol