Calculate the average $ ext{S-F}$ bond dissociation enthalpy (in $ ext{kJ mol}^{-1}$) in sulfur he

Question

Calculate the average $	ext{S-F}$ bond dissociation enthalpy (in $	ext{kJ mol}^{-1}$) in sulfur hexafluoride ($	ext{SF}_6$). The standard enthalpies of formation at $298	ext{ K}$ for $	ext{SF}_6(g)$, $	ext{S}(g)$, and $	ext{F}(g)$ are $-1100	ext{ kJ mol}^{-1}$, $+275	ext{ kJ mol}^{-1}$, and $+80	ext{ kJ mol}^{-1}$, respectively. (Report your answer to the nearest integer)

Professor PC Thomas Classes · Answer

The atomization reaction of gaseous sulfur hexafluoride is: $	ext{SF}_6(g) ightarrow 	ext{S}(g) + 6	ext{F}(g)$. The standard enthalpy change for this atomization process represents the total energy required to break all six identical $	ext{S-F}$ bonds: $\Delta H^\circ_{	ext{atomization}} = \Delta H^\circ_f[	ext{S}(g)] + 6\Delta H^\circ_f[	ext{F}(g)] - \Delta H^\circ_f[	ext{SF}_6(g)]$. Substituting the given values: $\Delta H^\circ_{	ext{atomization}} = 275 + 6(80) - (-1100) = 275 + 480 + 1100 = 1855	ext{ kJ mol}^{-1}$. Since there are six equivalent $	ext{S-F}$ single bonds in a molecule of $	ext{SF}_6$, the average $	ext{S-F}$ bond enthalpy is $E_{	ext{S-F}} = 1855 / 6 = 309.17	ext{ kJ mol}^{-1} \approx 309	ext{ kJ mol}^{-1}$.