Based on the following data, what is the F-F bond energy?
${ }^{\frac{1}{2}} \mathrm{H}_{2}(\mathrm{~g})+{ }^{\frac{1}{2}} \mathrm{~F}_{2}(\mathrm{~g}) \rightarrow \mathrm{HF}(\mathrm{g}) ; \Delta_{r} H=-272.5 \mathrm{~kJ} / \mathrm{mol}-\mathrm{rxn}$

$\begin{array}{ll}
Bond &Bond~Energy(kJ/mol)\
\text { H-H } & 435 \
\text { H-F } & 565
\end{array}$
A) 500 kJ/mol
B) 150 kJ/mol
C) -150 kJ/mol
D) -695 kJ/mol
E) 695 kJ/mol

Question

Based on the following data, what is the F-F bond energy? 
${ }^{\frac{1}{2}} \mathrm{H}_{2}(\mathrm{~g})+{ }^{\frac{1}{2}} \mathrm{~F}_{2}(\mathrm{~g}) \rightarrow \mathrm{HF}(\mathrm{g}) ; \Delta_{r} H=-272.5 \mathrm{~kJ} / \mathrm{mol}-\mathrm{rxn}$

$\begin{array}{ll}
Bond &Bond~Energy(kJ/mol)\
\text { H-H } & 435 \
\text { H-F } & 565
\end{array}$
A) 500 kJ/mol
B) 150 kJ/mol
C) -150 kJ/mol
D) -695 kJ/mol
E) 695 kJ/mol

Quizplus · Accepted Answer

The reaction enthalpy is given by the bond energies: $\Delta_r H = \text{(bonds broken)} - \text{(bonds formed)}$. For the reaction, $\Delta_r H = \frac{1}{2} \times 435 + \frac{1}{2} \times \text{F-F} - 565 = -272.5$. Solving for F-F gives 150 kJ/mol.

Based on the Following Data, What Is the F-F Bond