Using the thermodynamic data below and a value of -717 kJ/mole for the lattice enthalpy for KCl, calculate the ionization energy of K. $$\begin{array} { l l }
\text { Enthalpy of atomization of } \mathrm { K } & + 89 \mathrm {~kJ} / \mathrm { mol } \
\text { Enthalpy of formation of } \mathrm { KCl } ( \mathrm { s } ) & - 437 \mathrm {~kJ} / \mathrm { mol } \
\text { Enthalpy of formation of } \mathrm { Cl } ( \mathrm { g } ) & + 121 \mathrm {~kJ} / \mathrm { mol } \
\text { Electron attachment enthalpy of } \mathrm { Cl } ( \mathrm { g } ) & - 349 \mathrm {~kJ} / \mathrm { mol }
\end{array}$$
A) -576 kJ/mol
B) +141 kJ/mol
C) +419 kJ/mol
D) +576 kJ/mol
E) +597 kJ/mol

Question

Using the thermodynamic data below and a value of -717 kJ/mole for the lattice enthalpy for KCl, calculate the ionization energy of K. $$\begin{array} { l l } 
\text { Enthalpy of atomization of } \mathrm { K } & + 89 \mathrm {~kJ} / \mathrm { mol } \
\text { Enthalpy of formation of } \mathrm { KCl } ( \mathrm { s } ) & - 437 \mathrm {~kJ} / \mathrm { mol } \
\text { Enthalpy of formation of } \mathrm { Cl } ( \mathrm { g } ) & + 121 \mathrm {~kJ} / \mathrm { mol } \
\text { Electron attachment enthalpy of } \mathrm { Cl } ( \mathrm { g } ) & - 349 \mathrm {~kJ} / \mathrm { mol }
\end{array}$$
A) -576 kJ/mol
B) +141 kJ/mol
C) +419 kJ/mol
D) +576 kJ/mol
E) +597 kJ/mol

Quizplus · Accepted Answer

The ionization energy of K can be calculated using the Born-Haber cycle:  Ionization energy of K = Enthalpy of formation of KCl + Lattice enthalpy - Enthalpy of atomization of K - Enthalpy of formation of Cl - Electron attachment enthalpy of Cl.  = (-437) + 717 - 89 - 121 + 349 = +419 kJ/mol.

Using the Thermodynamic Data Below and a Value of -717