The enthalpy change at 298 K for decomposition is given in following steps :
Step-1: $H_{2} O (g) \to H (g) + OH (g) Δ H = 498 kJ / {mol}^{- 1}$
Step-2: $OH (g) \to H (g) + O (g) Δ H = 428 kJ / {mol}^{- 1}$
Then, value of mean bond enthalpy of $O - H$ bond will be:

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498 kJ/mol

463 kJ/mol

428 kJ/mol

70 kJ/mol

answer is B.

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Detailed Solution

$\begin{array}{l} H_{2} O ⟶ H + OH; Δ H = 498 kJ / mol \\ OH ⟶ H + O; ΔH = 428 KJ / mol \end{array}$

The bond enthalpy of $O - H$ bond will be

The average of both enthalpy required in 1 and 2

$\begin{array}{l} Δ H_{O - H} = \frac{{ΔH}_{1, OH} + {ΔH}_{2, OH}}{2} \\ = \frac{498 + 428}{2} \Rightarrow \frac{463}{2} \Rightarrow 463 KJ / mol \end{array}$

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