Calculate the enthalpy change of 1 mole of reaction: \[ \begin{array}{l} \mathrm{Na}(\mathrm{s})...
0Calculate the enthalpy change of 1 mole of reaction:
\[
\begin{array}{l}
\mathrm{Na}(\mathrm{s})+\frac{1}{2} \mathrm{Br}_{2}(\mathrm{~g}) \longrightarrow \mathrm{NaBr}(\mathrm{s}) \text { in kcal } \\
\text { Given: } \Delta \mathrm{H}_{\text {sub }}(\mathrm{Na})=137 \mathrm{~kJ} \mathrm{~mole}^{-1} \text {; } \\
\Delta \mathrm{H}_{\text {bond dissociation }}\left(\mathrm{Br}_{2}(\mathrm{~g})\right)=144 \mathrm{~kJ} \text { mole }{ }^{-1} \\
\Delta \mathrm{H}_{1} \text { st ionisation }\left(\mathrm{Na}(\mathrm{g})=496 \mathrm{~kJ}^{\mathrm{mole}} \mathrm{e}^{-1}\right. \text {; } \\
\Delta \mathrm{H}_{1} \text { st electron affinity }\left(\mathrm{Br}(\mathrm{g})=-325 \mathrm{~kJ} \text { mole }^{-1}\right. \\
\Delta \mathrm{H}_{\text {Lattice energy }}(\mathrm{NaBr})=+742 \\
\end{array}
\]
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