Question

Calculate the lattice energy of potassium oxide from the following data:

Enthalpy of sublimation of potassium: +89.24 kJ/mol
Bond energy of oxygen: +498 kJ/mol
First ionization energy of potassium: +419 kJ/mol
1st electron affinity of oxygen: -141 kJ/mol
2nd electron affinity of oxygen: +744 kJ/mol
deltaHf potassium oxide: −363.17 kJ/mol


the answer is - 2232 kJ/mol. i do not know how to go about getting this answer. i know youre supposed to add everything and have it equal -363.17 but i don't know if there is something that i should exclude or something that i'm supposed to multiply by two. i just really need help please

Answer 1

2K(s) + 1/2O2(g)---->K2O deltaH=−363.17 kJ or better still,
K2O --->2K(s) + 1/2O2(g)deltaH=363.17
take note that the equation is reversed above.
2K(s)---->2K(g) deltaH=2(+89.24 kJ/mol )=178.48kJ
2K(g)---->2K+(g) deltaH=2(+419 kJ/mol )=838kJ
1/2O2(g)---->O(g) deltaH=1/2(498 kJ/mol )=249kJ
O(g)---->O-(g) deltaH=-141kJ
O-(g)---->O2-(g) deltaH=744
as such, from the energy cycle (Born Haber cycle for K2O, the LE=-(363.17+178.48+838+249-141+744)=2231.65~2232kJ/mol

Answer 2

2K(s) + 1/2O2(g)---->K2O deltaH=−363.17 kJ or better still,
K2O --->2K(s) + 1/2O2(g)deltaH=363.17
take note that the equation is reversed above.
2K(s)---->2K(g) deltaH=2(+89.24 kJ/mol )=178.48kJ
2K(g)---->2K+(g) deltaH=2(+419 kJ/mol )=838kJ
1/2O2(g)---->O(g) deltaH=1/2(498 kJ/mol )=249kJ
O(g)---->O-(g) deltaH=-141kJ
O-(g)---->O2-(g) deltaH=744
as such, from the energy cycle (Born Haber cycle for K2O, the LE=-(363.17+178.48+838+249-141+744)=-2231.65~-2232kJ/mol