3. Calculate the second ionization energy for calcium using the following information:

Question

o Bond dissociation energy for gaseous molecular fluorine = 158 k/mol o First ionization energy for calcium = 589.8 kJ/mol
o Heat of sublimation for calcium = 178.2 kJ/mol
o Electron affinity for fluorine = -328 kJ/mol
o Lattice energy for CaF2(s) = -2630 kJ/mol
o Heat of formation for CaF2(s) = -1215 kJ/mol

Answer 1

The second ionization energy for calcium can be calculated using the equation:

Second ionization energy = first ionization energy + lattice energy + heat of sublimation - bond dissociation energy - heat of formation

Plugging in the given values:

Second ionization energy = 589.8 kJ/mol + (-2630 kJ/mol) + 178.2 kJ/mol - 158 kJ/mol - (-1215 kJ/mol)

Second ionization energy = 589.8 kJ/mol - 2630 kJ/mol + 178.2 kJ/mol - 158 kJ/mol + 1215 kJ/mol

Second ionization energy = -804.8 kJ/mol

Therefore, the second ionization energy for calcium is -804.8 kJ/mol.