H-H, O=O and O-H bond enthalpy values are 436 kJ/mole, 498 kJ/mole and 486.5 kJ/mole respectively. Then $Δ H$ for the process $H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g)$ will be

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answer is B.

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Detailed Solution

$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g)$
$Δ_{r} H = H_{R} - H_{P}$
$\begin{array}{l} = [436 + \frac{1}{2} (498) - 2 (486.5)] kJ/mole \\ = - 288 kJ/mole \end{array}$

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