O + 2e^–  → O^–2; $∆$H = 639 kJ/mole
O + e^– → O^–; $∆$H = –141 kJ/mole
O^– + e^– → O^–2; $∆$H = x kJ/mole
What is the value of x ?

O + 2e^–  → O^–2; $∆$H₁ = 639 kJ/mole ---(1)
O + e^– → O^–; $∆$H₂ = –141 kJ/mole ---(2)
O^– + e^– → O^–2; $∆$H = x kJ/mole ---(3)

(1)-(2)=(3)

 $$\begin{gathered} \therefore ∆\mathrm{H}=∆{\mathrm{H}}_1-∆{\mathrm{H}}_2 \\ \mathrm{x}=639-(-141)=+780 \mathrm{kj}/\mathrm{mol}. \end{gathered}$$