The enthalpy at 298 K of the reaction H2O2(l)→ H2O(l)+(1/2)O2(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

Question

The enthalpy at 298 K of the reaction H2O2(l)→ H2O(l)+(1/2)O2(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 (A) -68.3 kcal mol-1 (B) 68.3 kcal mol-1 (C) -91.8 kcal mol-1 (D) 91.8 kcal mol-1

Tardigrade Admin · Accepted Answer

Δ Hreaction=Δ H°f(H2O)-Δ H°f(H2O2) -23.5=x-(-44.8) or x=-23.5-44.8=-68.3

Q. 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 $- 23.5 k c a l m o l^{- 1}$ and the enthalpy of formation of $H_{2} O_{2} (l)$ is $- 44.8 k c a l m o l^{- 1}$ . The enthalpy of formation of $H_{2} O (l)$ is

$- 68.3 k c a l m o l^{- 1}$

$68.3 k c a l m o l^{- 1}$

$- 91.8 k c a l m o l^{- 1}$

$91.8 k c a l m o l^{- 1}$

$Δ H_{re a c t i o n} = Δ H_{f}^{\circ} (H_{2} O) - Δ H_{f}^{\circ} (H_{2} O_{2})$
$- 23.5 = x - (- 44.8)$
or $x = - 23.5 - 44.8 = - 68.3$

Q. The enthalpy at 298K of the reaction H2​O2​(l)→H2​O(l)+21​O2​(g) is −23.5kcalmol−1 and the enthalpy of formation of H2​O2​(l) is −44.8kcalmol−1. The enthalpy of formation of H2​O(l) is

−68.3kcalmol−1

68.3kcalmol−1

−91.8kcalmol−1

91.8kcalmol−1