The hydrogen bond between and is very strong and its strength may be analysed by setting up a Born–H — Thermodynamics and Thermochemistry Chemistry Question
Question
The hydrogen bond between $\text{F}^-$ and $\text{CH}_3\text{COOH}$ is very strong and its strength may be analysed by setting up a Born–Haber cycle with the following data (in kJ/mol): Lattice energy of $\text{KF.CH}_3\text{COOH}$, 734; enthalpy of vaporization of $\text{CH}_3\text{COOH}$, 20; enthalpy of solution of KF, 35; solvation energy of $\text{K}^+\text{(g)}$, –325; solvation energy of $\text{F}^-$, –389; enthalpy of formation of $\text{KF.CH}_3\text{COOH(s)}$ from KF(s) and $\text{CH}_3\text{COOH(l)}$, –25. Find the energy of the hydrogen bond between $\text{F}^-$ and $\text{CH}_3\text{COOH}$ in the gas phase (in kJ/mol).
💡 Solution & Explanation
Find lattice energy of KF(s): $\text{LE}(\text{KF}) = \Delta H_{\text{sol}} - \text{Solv}(\text{K}^+) - \text{Solv}(\text{F}^-) = 35 - (-325) - (-389) = 749\text{ kJ/mol}$. Construct cycle for $\text{KF(s)} + \text{CH}_3\text{COOH(l)} \to \text{KF} \cdot \text{CH}_3\text{COOH(s)}$ ($\Delta H = -25$): Break KF (+749), vaporize acetic acid (+20), form H-bond ($-E_{\text{HB}}$), form complex lattice ($-734$). Sum: $749 + 20 - E_{\text{HB}} - 734 = -25 \implies 35 - E_{\text{HB}} = -25 \implies E_{\text{HB}} = 60\text{ kJ/mol}$. Format as 0060.