Question

Determine the lattice energy for the formation of MgO(s) from Mg(s) and O_2​2​​(g) using the following equations:

Mg(s) \rightarrow→ Mg(g) , + 147 kJ/mol

Mg(g) \rightarrow→ Mg^{2+}​2+​​(g) + 2e^-​−​​ , + 2189 kJ/mol

O_2​2​​(g) \rightarrow→ 2O(g) , + 498 kJ/mol

O(g) + 2e^-​−​​ \rightarrow→ O^{2-}​2−​​(g) , +702 kJ/mol

Mg(s) + 1/2O_2​2​​(g) \rightarrow→ MgO(s) , -602 kJ/mol

Answer 1

Mg(s) → Mg(g) ,      H1=         + 147 kJ/mol

Mg(g) → Mg^{2+}​2+​​(g) + 2e^-​−​​ H2 = +2189 kJ/mol

O2​​(g) → 2O(g) ,              H3 = + 498 kJ/mol

O(g) + 2e^-​−​​ → O^2-       H4 = +702 kJ/mol

Mg ^2+ (g) + O2^- (g) ------------> MgO(s) H5 ( lattice energy)=

Mg(s) + 1/2O2​(g) → MgO(s) H= -602 kJ/mol

according to hess law

H   = H1 + H2 + H3 + H4 + H5

-602   = 147 +2189+498/2 +702 + H5

-602     = 147+2189+249 + 702+ H5

H5   = -3889KJ/mole lattice energy of MgO