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Heat of Formation

The enthalpy ...

Question

The enthalpy of combustion of $H_{2} (g)$ at 298 K to give $H_{2} O (g)$ is -249 kJ mol $^{- 1}$ and bond enthalpies of H-H and O-O are 433 kJ mol $^{- 1}$ and 492 kJ mol $^{- 1}$ respectively. The bond enthalpy of O-H is

A

464 k J m o l^{- 1}

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B

- 464 k J m o l^{- 1}

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C

232 k J m o l^{- 1}

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D

- 232 k J m o l^{- 1}

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Solution

The correct option is A $464 k J m o l^{- 1}$
We have
Given - $ϵ_{H - H} = 433 k J m o l^{- 1}$ & $ϵ_{O = O} = 492 k J m o l^{- 1}$
Hence, $ϵ_{H - H} + \frac{1}{2} ϵ_{O = O} - 2 ϵ_{O - H} = (433 + \frac{1}{2} \times 492) k J m o l^{-} - 2 ϵ_{O - H} = - 249 k J m o l^{- 1}$
Hence , $ϵ_{O - H} = 464 k J m o l^{- 1}$

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