The enthalpy of combustion of $H_{2} (g)$ 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

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$- 464 kJ {mol}^{- 1}$

$- 232 kJ {mol}^{- 1}$

$232 kJ {mol}^{- 1}$

$464 kJ {mol}^{- 1}$

answer is A.

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Detailed Solution

For the equation

$\begin{array}{l} H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g) Δ H = - 249 kJ {mol}^{- 1} \\ ε (H - H) + \frac{1}{2} eO = O) - 2 ε (O - H) = - 249 kJ {mol}^{- 1} \\ [433 + \frac{1}{2} (492)] kJ {mol}^{- 1} - 2 ε (O - H) = - 249 kJ {mol}^{- 1} \\ ε (O - H) = \frac{1}{2} [+ 249 + 433 + \frac{1}{2} (492)] kJ {mol}^{- 1} = 464 kJ {mol}^{- 1} \end{array}$