If the enthalpy of combustion of ${\mathrm{CH}}_4$ and ${\mathrm{CH}}_3\mathrm{OH}$ are $\text{'}\mathrm{x}\text{'}$ and $\text{'}\mathrm{y}\text{'}$ respectively, then find the enthalpy of the reaction ${\mathrm{CH}}_4+\frac{1}{2}{\mathrm{O}}_2⟶{\mathrm{CH}}_3\mathrm{OH}$

The formation of an oxide ion from Oxygen atom takes place as shown below. This two-step process is unfavourable, even though ${\mathrm{O}}^{2-}$ is iso-electronic with Neon, due to the fact that

${\mathrm{O}}_{\left(\mathrm{g}\right)}+{\mathrm{e}}^{-}⟶{\mathrm{O}}_{\left(\mathrm{g}\right)}^{-}; {\mathrm{\Delta H}}_{\mathrm{f}}°=-141 \mathrm{kJ}.{\mathrm{mol}}^{-1}$

${\mathrm{O}}_{\left(\mathrm{g}\right)}^{-}+{\mathrm{e}}^{-}⟶{\mathrm{O}}_{\left(\mathrm{g}\right)}^{2-};{\mathrm{\Delta H}}_{\mathrm{f}}^0=+780 \mathrm{kJ}.{\mathrm{mol}}^{-1}$