The standard molar enthalpies of formation of and are –470 kJ and –847 kJ, respectively. Valence she — Thermodynamics and Thermochemistry Chemistry Question
Question
The standard molar enthalpies of formation of $\text{IF}_3\text{(g)}$ and $\text{IF}_5\text{(g)}$ are –470 kJ and –847 kJ, respectively. Valence shell electron-pair repulsion theory predicts that $\text{IF}_5\text{(g)}$ is square pyramidal in shape in which all I – F bonds are equivalent while $\text{IF}_3\text{(g)}$ is T-shaped (based on trigonal–bipyramidal geometry) in which I – F bonds are of different lengths. It is observed that the axial I – F bonds in $\text{IF}_3$ are equivalent to the I – F bonds in $\text{IF}_5$. Calculate the equatorial I – F bond strength (in kJ/mol) in $\text{IF}_3$. Some other details given are:<br>$\text{I}_2\text{(s)} \to \text{I}_2\text{(g)}; \Delta H = 62\text{ kJ}$<br>$\text{F}_2\text{(g)} \to 2\text{F(g)}; \Delta H = 155\text{ kJ}$<br>$\text{I}_2\text{(g)} \to 2\text{I(g)}; \Delta H = 149\text{ kJ}$
💡 Solution & Explanation
$\Delta H_f(\text{I,g}) = (62 + 149)/2 = 105.5\text{ kJ/mol}$. $\Delta H_f(\text{F,g}) = 155/2 = 77.5\text{ kJ/mol}$. Atomization of $\text{IF}_5 = 105.5 + 5(77.5) - (-847) = 1340\text{ kJ/mol}$. Energy per identical I-F bond in $\text{IF}_5$ (and thus axial in $\text{IF}_3$) $= 1340 / 5 = 268\text{ kJ/mol}$. Atomization of $\text{IF}_3 = 105.5 + 3(77.5) - (-470) = 808\text{ kJ/mol}$. $\text{IF}_3$ has 2 axial and 1 equatorial bond: $2(268) + E_{\text{eq}} = 808 \implies 536 + E_{\text{eq}} = 808 \implies E_{\text{eq}} = 272\text{ kJ/mol}$. Format as 0272.