why is the bond enthalpy of 2 O-H bonds in water different

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Solution

$W a t e r m o l e c u l e h a s H - O H b o n d t o b e c l e a v e d a n d O - H b o n d t o b e c l e a v e d . T h e s e t w o b o n d s h a v e d i f f e r e n t b o n d e n e r g y a s t h e b o n d s a r e d i f f e r e n t . O - H i n w a t e r, e n e r g y n e e d e d i s 428 k J / m o l . H - O H i n w a t e r, e n e r g y n e e d e d i s 497.8 k J / m o l . T h u s t h e a v e r a g e b o n d e n e r g y i s \frac{497.8 + 428}{2} = 464 k J / m o l$

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why is the bond enthalpy of 2 O-H bonds in water different

Water molecule has H-OH bond to be cleaved and O-H bond to be cleaved. These two bonds have different bond energy as the bonds are different. O-H in water, energy needed is 428 kJ/mol.H-OH in water, energy needed is 497.8 kJ/ mol.Thus the average bond energy is 497.8 + 4282 = 464 kJ/mol