The standard molar enthalpies of formations of and are –286 and –188 kJ/mol, respectively. Molar ent — Thermodynamics and Thermochemistry Chemistry Question
Question
The standard molar enthalpies of formations of $\text{H}_2\text{O(l)}$ and $\text{H}_2\text{O}_2\text{(l)}$ are –286 and –188 kJ/mol, respectively. Molar enthalpies of vaporization of $\text{H}_2\text{O}$ and $\text{H}_2\text{O}_2$ are 44 and 53 kJ, respectively. The bond dissociation enthalpy of $\text{O}_2\text{(g)}$ is 498 kJ/mol. Calculate the bond dissociation enthalpy (in kJ/mol) of O – O bond in $\text{H}_2\text{O}_2$, assuming that the bond dissociation enthalpy of O – H bond is same in both $\text{H}_2\text{O}$ and $\text{H}_2\text{O}_2$.
💡 Solution & Explanation
$\Delta H_f(\text{H}_2\text{O,g}) = -286 + 44 = -242\text{ kJ/mol}$. $\Delta H_f(\text{H}_2\text{O}_2\text{,g}) = -188 + 53 = -135\text{ kJ/mol}$. Atomization of $\text{H}_2\text{O} = 2E_{\text{OH}} = 2\Delta H_f(\text{H}) + \Delta H_f(\text{O}) + 242$. Atomization of $\text{H}_2\text{O}_2 = 2E_{\text{OH}} + E_{\text{OO}} = 2\Delta H_f(\text{H}) + 2\Delta H_f(\text{O}) + 135$. Subtracting the two equations yields: $E_{\text{OO}} = \Delta H_f(\text{O}) + 135 - 242 = \Delta H_f(\text{O}) - 107$. With $\Delta H_f(\text{O}) = 498 / 2 = 249\text{ kJ/mol}$, $E_{\text{OO}} = 249 - 107 = 142\text{ kJ/mol}$. Format as 0142.