The enthalpy changes at 298K in successive breaking of O-H bonds of HOH are- H2Og- Hg-OHg- - H-498 kJ mol-1 OHg- Hg-Og- - H-428 kJ mol-1The bond enthalpy of the O-H bond is-

The correct option is B 463 kJ mol^ 1 Given dissociation enthalpy of #160; H OH=498 KJ/molGiven dissociation enthalpy of O H=428 KJ/mol ...

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Standard XII

Chemistry

Bond Order and Enthalpy

The enthalpy ...

Question

The enthalpy changes at $298$ K in successive breaking of $O - H$ bonds of $H O H$ are:
$H_{2} O (g) \to H (g) + O H (g), Δ H = 498$ kJ $m o l^{- 1}$
$O H (g) \to H (g) + O (g), Δ H = 428$ kJ $m o l^{- 1}$
The bond enthalpy of the $O - H$ bond is:

498

m o l^{- 1}

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463

m o l^{- 1}

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428

m o l^{- 1}

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70

m o l^{- 1}

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Solution

The correct option is B $463$ kJ $m o l^{- 1}$
Given dissociation enthalpy of $H - O H = 498$ KJ/mol
Given dissociation enthalpy of $O - H = 428$ KJ/mol
$∴$ Bond enthalpy of $O - H = \frac{498 + 428}{2} = \frac{926}{2} = 463$ KJ/mol.

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The enthalpy changes at 298K in successive breaking of O−H bonds of HOH are:H2O(g)→H(g)+OH(g),ΔH=498 kJ mol−1OH(g)→H(g)+O(g),ΔH=428 kJ mol−1The bond enthalpy of the O−H bond is:

The correct option is B 463 kJ mol−1Given dissociation enthalpy of H−OH=498 KJ/molGiven dissociation enthalpy of O−H=428 KJ/mol∴ Bond enthalpy of O−H=498+4282=9262=463 KJ/mol.

The enthalpy changes at $298$ K in successive breaking of $O - H$ bonds of $H O H$ are:
$H_{2} O (g) \to H (g) + O H (g), Δ H = 498$ kJ $m o l^{- 1}$
$O H (g) \to H (g) + O (g), Δ H = 428$ kJ $m o l^{- 1}$
The bond enthalpy of the $O - H$ bond is:

The correct option is B $463$ kJ $m o l^{- 1}$
Given dissociation enthalpy of $H - O H = 498$ KJ/mol
Given dissociation enthalpy of $O - H = 428$ KJ/mol
$∴$ Bond enthalpy of $O - H = \frac{498 + 428}{2} = \frac{926}{2} = 463$ KJ/mol.