The enthalpy changes at 298 K in successive breaking of $O - H$ bonds of water are
H₂O(g) $⟶$ H(g) + OH(g) ; $Δ H = 498 k J m o l^{- 1}$
OH(g) $⟶$ H(g) + O(g) ; $Δ H = 428 k J m o l^{- 1}$
The bond enthalpy of the O–H bond is:

Question

The enthalpy changes at 298 K in successive breaking of $O - H$ bonds of water are

H₂O(g) $⟶$ H(g) + OH(g) ; $Δ H = 498 k J m o l^{- 1}$

OH(g) $⟶$ H(g) + O(g) ; $Δ H = 428 k J m o l^{- 1}$

The bond enthalpy of the O–H bond is:

TestHub · Accepted Answer

Bond energy is the average energy required to break each bond of the same type in a compound. Δ H a v ​ = 2 498 + 428 ​ = 463 k J m o l − 1 .

Chemistry - Thermochemistry Question with Solution | TestHub