1. Tardigrade
  2. Question
  3. Chemistry
  4. 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

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

Thermodynamics

Solution:

$\Delta H_{reaction}=\Delta H^{\circ}_{f}\left(H_{2}O\right)-\Delta H^{\circ}_{f}\left(H_{2}O_{2}\right)$
$-23.5=x-\left(-44.8\right)$
or $x=-23.5-44.8=-68.3$