The enthalpy at 298 K of the reactionH2O2(l)→H2O(l)+12O2(g), is -23.5 kcal mol-1 and the enthalpy of formation of H2O2(l) is -44.8 kcal mol-1. The enthalpy of formation of H2O(l) is

ΔHreaction =ΔHf∘H2O2−ΔHf∘H2O2−23.5=x−(−44.8) or x=−23.5−44.8=−68.3

The enthalpy at 298 K of the reactionH2O2(l)→H2O(l)+12O2(g), is -23.5 kcal mol-1 and the enthalpy of formation of H2O2(l) is -44.8 kcal mol-1. The enthalpy of formation of H2O(l) is

ΔHreaction =ΔHf∘H2O2−ΔHf∘H2O2−23.5=x−(−44.8) or x=−23.5−44.8=−68.3

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The enthalpy at 298 K of the reaction
$H_{2} O_{2} (l) \to H_{2} O (l) + \frac{1}{2} O_{2} (g)$ , is -23.5 kcal mol^-1 and the enthalpy of formation of H₂O₂(l) is -44.8 kcal mol^-1. The enthalpy of formation of H₂O(l) is

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-68.3 kcal mol^-1

68.3 kcal mol^-1

91.8 kcal mol^-1

-91.8 kcal mol^-1

answer is A.

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$\begin{array}{l} Δ H_{reaction} = Δ H_{f}^{\circ} (H_{2} O_{2}) - {ΔH}_{f}^{\circ} (H_{2} O_{2}) \\ - 23.5 \\ = x - (- 44.8) \\ or x = - 23.5 - 44.8 = - 68.3 \end{array}$

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The enthalpy at 298 K of the reactionH2O2(l)→H2O(l)+12O2(g), is -23.5 kcal mol-1 and the enthalpy of formation of H2O2(l) is -44.8 kcal mol-1. The enthalpy of formation of H2O(l) is

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The enthalpy at 298 K of the reaction
$H_{2} O_{2} (l) \to H_{2} O (l) + \frac{1}{2} O_{2} (g)$ , is -23.5 kcal mol^-1 and the enthalpy of formation of H₂O₂(l) is -44.8 kcal mol^-1. The enthalpy of formation of H₂O(l) is