The enthalpy of formation of liquid water at is –286 kJ. Given for and for and that the molar enthal — Thermodynamics and Thermochemistry Chemistry Question
Question
The enthalpy of formation of liquid water at $25^\circ\text{C}$ is –286 kJ. Given $C_P = 75.4\text{ J/K-mol}$ for $\text{H}_2\text{O(l)}$ and $33.4\text{ J/K-mol}$ for $\text{H}_2\text{O(g)}$ and that the molar enthalpy of vaporization of liquid water at $125^\circ\text{C}$ is 40.8 kJ/mol. The enthalpy of dissociation (in kJ/mol) of $\text{H}_2\text{O(g)}$ into $\text{H}_2$ and $\text{O}_2$ gases at $25^\circ\text{C}$ is
💡 Solution & Explanation
$\Delta H_{\text{vap}}$ at 298 K $= \Delta H_{\text{vap,398}} + \Delta C_P \Delta T = 40800 + (33.4 - 75.4)(298 - 398) = 40800 + (-42.0)(-100) = 45000\text{ J/mol} = 45.0\text{ kJ/mol}$. $\Delta H_f(\text{H}_2\text{O, g})$ at 298 K $= -286 + 45 = -241\text{ kJ/mol}$. The dissociation reaction is the exact reverse of formation, so $\Delta H_{\text{dissociation}} = +241\text{ kJ/mol}$. Format as 0241.