Bond enthalpies of ${\mathrm{H}}_2,{\mathrm{O}}_2$ are 436kJ/mole and 498kJ/mole respectively. Calculate O–H bond enthalpy if  ${\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{H}}^0$ of water is –286kJ/mole

${\mathrm{\Delta }}^0\mathrm{H}={\mathrm{H}}_{\mathrm{R}}^0-{\mathrm{H}}_{\mathrm{P}}^{\mathrm{o}}$

*** Bonds are broken in reactants, endothermic
        Bonds are formed in products, exothermic
$\begin{array}{l}\mathrm{H}-\mathrm{H}+\frac{1}{2} \mathrm{O}= \mathrm{O}\to \mathrm{H}-\mathrm{O}-\mathrm{H};\mathrm{\Delta H}=-286\mathrm{kJ}\\ -286=+436+\frac{1}{2}\left(498\right)-2\mathrm{x}\\ 2\mathrm{x}=436+249+286\\ \\ \mathrm{x}=\left(\frac{436+249+286}{2}\right)\mathrm{kJ}\\ \mathrm{X}=485.5\mathrm{kJ}\\ \mathrm{O}–\mathrm{H}\mathrm{bond}\mathrm{enthalpy}=485.5\mathrm{kJ}\end{array}$