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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:

A

`498 "KJ mol"^(-1)`

B

`428 "KJ mol"^(-1)`

C

`70 "KJ mol"^(-1)`

D

`463 "KJ mol"^(-1)`

Text Solution

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The correct Answer is:
D
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The H - O - O bond angle in H_(2)O_(2) (g) is

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The average O-H bond energy in H_(2)O with the help of following data. (1) H_(2)O(l)rarrH_(2)O(g), Delta H= +40.6 kJ mol^(-1) (2) 2H(g)rarrH_(2)(g), DeltaH= -435.kJ mol^(-1) (3) O_(2)(g)rarr2O(g), Delta H= +489.6 kJ mol^(-1) (4) 2H_(2)(g)+O_(2)(g)rarr2H_(2)O(l), Delta H= -571.6 kJ mol^(-1)

The bond energies of H - H and I - I bonds are 435 kJ mol^-1 and 150 kJ mol^-1 respectively. If DeltaH_f^o for HI is 26.5 kJ mol^-1 then bond enthalpy of H-I bond is

The enthalpy of the reaction H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(g) is DeltaH_(1) and that of H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l) is DeltaH_(2) . Then

ALLEN-THERMODYNAMICS -EXERCISE -3
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