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The enthalpy change at 298 K of the reaction $H_2O_2(l) \longrightarrow H_2O (l) + \frac{1}{2} O_2 (g)$ is -23.5 kcal $mol^{-1}$ and enthalpy of formation of $H_2O_2(l)$ is -44.8 kcal $mol^{-1}$. The enthalpy of formation of $H_2O(l)$ is

(a) -68.3 kcal $mol^{-1}$
(b) 68.3 kcal $mol^{-1}$
(c) -91.8 kcal $mol^{-1}$
(d) 91.8 kcal $mol^{-1}$
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Answer: kcal $mol^{-1}$
(i) $H_2O_2(l) \longrightarrow H_2O (l) + \frac{1}{2} O_2 (g)$; $\Delta H = -23.5$ kcal $mol^{-1}$
(ii) $H_2(g) + O_2(g) \longrightarrow H_2O_2(l)$; $\Delta H = -44.8$ kcal $mol^{-1}$
Write equation (ii) in reverse direction, we get
(iii) $H_2O_2(l)\longrightarrow H_2(g) + O_2(g) $; $\Delta H = +44.8$ kcal $mol^{-1}$
Substract equation (iii) from (i), we get
$H_2O (l) + \frac{1}{2} O_2 (g) \longrightarrow H_2O_2(l) $; $\Delta H = -68.4$ kcal $mol^{-1}$
answered Feb 24, 2014 by mosymeow_1

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