The enthalpies for the following reaction `(DeltaH^(0))` at `25^(@)C` are given.
`(1)/(2)H_(2)(g)+(1)/(2)O_(2)(g) to OH(g),DeltaH^(0)=10.06 "kcal"`
`H_(2)(g) to 2H(g),DeltaH^(0)=104.18"kcal"`
`O_(2)(g) to 2O(g),DeltaH^(0)=118.32 "kcal"`
Calculate O-H bond energy in the hydroxyl radical.

Explanation

From the given data, we have
`H_(2)(g) +O_(2)(g) to 2OH(g)` , `DeltaH^(@)=20.12kcal`
For reactants
Bond energy for 1 mole of H-H bonds `=104.18` kcal
Bond energy for 1 mole of O-O bonds `=118.32` kcal
and for product
Energy of formation of 2 moles of O-H bond `=2 xx x`
(where x is the energy of formation of 1 mole of O-H bonds)
Adding up all the energy changes, we get heat of the above react (`20.12` kcal)
i.e, `104.18+118.32+2xx x=20.12`
`x=-101.19kcal`
Hence bond energy of O-H bond `=+101.19` kcal per mole.