3. Calculate the second ionization energy for calcium using the following information:
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
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.
To calculate the second ionization energy for calcium, we need to consider the steps involved in the process. The second ionization energy is the energy required to remove a second electron from a singly ionized species.
The overall equation for the second ionization energy is:
Ca^+ -> Ca^2+ + 1e^-
Step 1: Calculate the energy required to remove the first electron from calcium:
Ca -> Ca^+ + 1e^-
First ionization energy = 589.8 kJ/mol
Step 2: Calculate the energy released when fluorine accepts the electron:
F(g) + 1e^- -> F^-(g)
Electron affinity for fluorine = -328 kJ/mol
Step 3: Calculate the energy released when the Ca^+ ion and F^- ion combine to form CaF2(s):
Ca^+(g) + F^-(g) -> CaF2(s)
Lattice energy for CaF2(s) = -2630 kJ/mol
Step 4: Calculate the overall energy change for the process:
First ionization energy + Electron affinity + Lattice energy
= 589.8 kJ/mol + (-328 kJ/mol) + (-2630 kJ/mol)
= -2368.2 kJ/mol
Step 5: Calculate the energy required to remove the second electron:
Ca^+ -> Ca^2+ + 1e^-
Second ionization energy = First ionization energy + energy change
= 589.8 kJ/mol + (-2368.2 kJ/mol)
= -1778.4 kJ/mol
Therefore, the second ionization energy for calcium is approximately -1778.4 kJ/mol.