Calculate in kJ/mol from the following data:<br> in glacial acetic acid <br> in glacial acetic acid — Thermodynamics and Thermochemistry Chemistry Question
Question
Calculate $\Delta H_{\text{vap}} \text{ [CH}_3\text{COOH(l)]}$ in kJ/mol from the following data:<br>$\Delta H_{\text{Solution}} \text{ [KF.CH}_3\text{COOH(s)]}$ in glacial acetic acid $= -3\text{ kJ/mol}$<br>$\Delta H_{\text{Solution}} \text{ [KF(s)]}$ in glacial acetic acid $= +35\text{ kJ/mol}$<br>The strength of H-bond between $\text{F}^-\text{(g)}$ and $\text{CH}_3\text{COOH(g)} = +46\text{ kJ/mol}$<br>Lattice enthalpy of $\text{KF.CH}_3\text{COOH(s)} = +734\text{ kJ/mol}$<br>Lattice enthalpy of KF(s) = $+797\text{ kJ/mol}$
💡 Solution & Explanation
The formation reaction $\text{KF(s)} + \text{CH}_3\text{COOH(l)} \to \text{KF} \cdot \text{CH}_3\text{COOH(s)}$ has $\Delta H = \Delta H_{\text{sol}}(\text{KF}) - \Delta H_{\text{sol}}(\text{Complex}) = 35 - (-3) = 38\text{ kJ/mol}$. Equivalent Born-Haber cycle equates this to: $\text{LE}(\text{KF}) + \Delta H_{\text{vap}} + \text{H-bond energy} - \text{LE}(\text{Complex}) = 38 \implies 797 + \Delta H_{\text{vap}} - 46 - 734 = 38 \implies 17 + \Delta H_{\text{vap}} = 38 \implies \Delta H_{\text{vap}} = 21\text{ kJ/mol}$. Format as 0021.