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The enthalpy changes at 298K in success...

The enthalpy changes at 298K in successive breaking of `O-H` bonds of water are
`H_(2)O(g) to H(g)+OH(g) , DeltaH=498kJ " mol"^(-1)`
`OH(g) to H(g)+O(g) , DeltaH=428kJ" mol"^(-1)`
The bond enthalpy of `O-H` bond is:

Text Solution

Verified by Experts

The correct Answer is:
463
Bond energy is the average energy required to break each bond of the same type in compund.
`Delta H_(av) = (498 + 428)/(2) = 462" kJ mol"^(-1)`
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The enthalpy changes at 298 K in successive breaking of O-H bonds of water, are H_(2)O(g) rarr H(g)+OH(g),DeltaH=498kJ mol^(-1) OH(g) rarr H(g)+O(g), DeltaH =428 kJ mol^(-1) The bond energy of the O-H bond is

H_2(g) + 1/2 O_2(g) to H_2O (l), DeltaH = - 286 kJ 2H_2(g) + O_2(g) to 2H_2O (l), DeltaH = …kJ

Knowledge Check

  • Dissociation of water takes place in two steps : H_(2)OrarrH+OH," "DeltaH=+497.8kJ OH rarr H+O," "DeltaH=428.5kJ Water is the bond energy of O-H bond ?

    A
    `"463.15 kJ mol"^(-1)`
    B
    `"428.5 kJ mol"^(-1)`
    C
    `"69.3 kJ mol"^(-1)`
    D
    `"926.3 kJ mol"^(-1)`

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