If the bond energies of $\mathrm{H}-\mathrm{H}, \mathrm{Br}-\mathrm{Br}$, and $\mathrm{H}-\mathrm{Br}$ are…
If the bond energies of $\mathrm{H}-\mathrm{H}, \mathrm{Br}-\mathrm{Br}$, and $\mathrm{H}-\mathrm{Br}$ are 433,192 and $364 \mathrm{~kJ} /\mathrm{mol}^{-1}$ respectively, the $\Delta \mathrm{H}^{\circ}$ for the reaction $\mathrm{H}_{2(g)}+\mathrm{Br}_{2(g)} \rightarrow 2 \mathrm{HBr}_{(g)}$ is:
$-261 \mathrm{~kJ}$
$+103 \mathrm{~kJ}$
$+261 \mathrm{~kJ}$
$-103 \mathrm{~kJ}$
Solution
$\mathrm{H}-\mathrm{H}+\mathrm{Br}-\mathrm{Br} \rightarrow 2 \mathrm{H}-\mathrm{Br}$
$\begin{array}{rrr}
433+192 & 2 & \times 364 \\
=625 & =728
\end{array}$
Energy absorbed Energy released
Net energy released $=\Sigma \mathrm{H}_{\mathrm{R}}-\Sigma \mathrm{H}_{\mathrm{p}}$ $=625-728=-103 \mathrm{~kJ}$.
Related Theory
Fluorine has high bond energy because it is smallest among these molecules so it is tightly bonded to other atoms due to greater nuclear charge.