Question

Using the thermochemical data and an estimated value of -2235.2 kJ/mol for the lattice energy for potassium oxide, calculate the value for the second electron affinity of oxygen [O⁻ + e- → O²⁻]. (The answer is 748.5 kJ/mol)

Quantity	Numerical Value (kJ/mol)
Enthalpy of atomization of K	89
Ionization energy of K	418.8
Enthalpy of formation of solid K2O	-363
Enthalpy of formation of O(g) from O2(g)	249.1
First electron attachment enthalpy of O   -141.0

Answer 1

since K2O we have 2K we multiply energies related to K by 2

Enthalphy of formation of K2O = 2 x Enthalphy of atomisation of K + 2 xIonization energy of K +

                          Enthalphy formation of O from O2 + 1st Electrona ffinity of O + second electron affinityO

                                                 + Lattice energy  

-363 = ( 2x89)+(2 x418.8)+249.1 -141 Electronaffinity 2 - 2235.2)

electron affinity of O 2nd = 748.5 KJ/mol