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

Question

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

Quizplus · Accepted Answer

The electron attachment enthalpy of Cl(g) can be calculated using the Born-Haber cycle. The cycle involves the following steps: sublimation of Na(s), ionization of Na(g), dissociation of Cl2(g), formation of NaCl(s), and the lattice energy. By rearranging the cycle and solving for the electron attachment enthalpy, we find it to be -349 kJ/mol.

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

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