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The standard enthalpies of formation of H₂O_2(l) and H₂O_(l) are -187.8K.J.mole^-1 and –285.8 K.J. mole^-1 respectively. The $\Delta$ H^o for the decomposition of one mole of H₂O_2(l) to H₂O_(l) and O_2(g) is

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-98.0 K.J. mole^-1

+187.8 K.J. mole^-1

-473.6 K.J. mole^-1

+473.6 K.J. mole

answer is B.

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Detailed Solution

${H_2}{O_2}_{(l)} \to {H_2}{O_{(l)}} + \frac{1}{2}\,{O_2}_{(g)},\,{\Delta _r}H = ?$

$\begin{array}{l} {\Delta _r}H = [{\Delta _f}H({H_2}O) + \frac{1}{2}{\Delta _f}H({O_2})] - [{\Delta _f}H({H_2}{O_2})]\\\\ \,\,\,\,\,\,\,\,\,\, = [ - 285.8 + 0] - [ - 187.8]\\\\ \,\,\,\,\,\,\,\,\,\, = \, - 98KJ/mole \end{array}$

Note: Standard enthalpy of formation of an element in its standard state (or) uncombined state is Zero

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