The standard enthalpies of formation of $H_{2} O_{2} (l) and H_{2} O (l) are - 187.8 kj . {mole}^{- 1}$ and $- 285.8 kj . {mole}^{- 1}$ mole− − respectively. The $∆ H^{o}$ for the decomposition of one mole of $H_{2} O_{2} (l) TO H_{2} O (l) and O_{2} (g)$ is

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-473.6 $kJ . {mole}^{- 1}$

+473.6 $kJ . {mole}^{- 1}$

+187.8 $kJ . {mole}^{- 1}$

-98.0 $kJ . {mole}^{- 1}$

answer is B.

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Detailed Solution

Standard enthalpy formation of $H_{2} O_{2} ∆ H = - 187.8 kJ$

Standard enthalpy formation of $H_{2} O_{2} \to H_{2} O + \frac{1}{2} O_{2}$

$∆ H_{r} = \sum H_{f, products} - \sum H_{f, Reactants}$

$- 285.8 kJ + \frac{1}{2} (O) - (- 187.8 kJ) = 98 kj . {mole}^{- 1}$

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