You are given the following thermodynamic data: bond dissociation enthalpies of $ ext{F-F}$ and $

Question

You are given the following thermodynamic data: bond dissociation enthalpies of $	ext{F-F}$ and $	ext{H-H}$ are $158 	ext{ kJ mol}^{-1}$ and $436 	ext{ kJ mol}^{-1}$ respectively. Furthermore, the standard enthalpy of formation ($\Delta_f H^\circ$) of gaseous $	ext{HF}$ is $-271 	ext{ kJ mol}^{-1}$.\ Based on these figures, the bond dissociation enthalpy of the $	ext{H-F}$ bond must be

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(a) : The defining equation for the formation of hydrogen fluoride is: $\frac{1}{2}	ext{H}_{2(g)} + \frac{1}{2}	ext{F}_{2(g)} ightarrow 	ext{HF}_{(g)}$. The enthalpy change for this reaction is given by: $\Delta_f H^\circ = \sum (	ext{B.E.}_{	ext{reactants}}) - \sum (	ext{B.E.}_{	ext{products}})$. Applying the formula: $-271 = [\frac{1}{2}(	ext{B.E.}_{	ext{H-H}}) + \frac{1}{2}(	ext{B.E.}_{	ext{F-F}})] - 	ext{B.E.}_{	ext{H-F}}$. Substituting the given values yields: $-271 = [\frac{1}{2}(436) + \frac{1}{2}(158)] - 	ext{B.E.}_{	ext{H-F}} = [218 + 79] - 	ext{B.E.}_{	ext{H-F}} = 297 - 	ext{B.E.}_{	ext{H-F}}$. Isolating the unknown variable: $	ext{B.E.}_{	ext{H-F}} = 297 + 271 = 568 	ext{ kJ mol}^{-1}$.