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  4. Change in enthalpy for reaction 2H2O2(l) -> 2H2O(l) + O2(g) If heat of formation of H2O2(l) and H2O(l) are - 188 - 286 KJ/mol respectively: -

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Q. Change in enthalpy for reaction $\ce{2H_2O_2(l) -> 2H_2O(l) + O_2(g) }$
If heat of formation of $\ce{H_2O_2(l)}$ and $\ce{H_2O(l)}$ are - 188 & - 286 KJ/mol respectively : -

AIPMTAIPMT 2001Thermodynamics

Solution:

$2 H _2 O _2(l) \rightarrow 2 H _2 O _2(l)+ O _2(g) \Delta H=?$
$\Delta H=\left[2 \times \Delta H _f\right.$ of $H _2 O (l)+\left(\Delta H _f\right.$ of $\left.\left.O _2\right)\right]$
$-\left(2 \times \Delta H _f\right.$ of $\left.\left.H _2 O _2(l)\right)\right]$
$=[(2 \times-286)+(0)-(2 \times-188)]$
$=[-572+376]$
$=-196\, kJ / mol$