Calculate the magnitude of enthalpy change for \( 0.1 \) mole of \( \mathrm{NaBr} \) in kcal, ro...
0Calculate the magnitude of enthalpy change for \( 0.1 \) mole of \( \mathrm{NaBr} \) in kcal, rounding it off to nearest whole number.
\[
\mathrm{Na}(\mathrm{s})+\frac{1}{2} \mathrm{Br}_{2}(\mathrm{~g}) \longrightarrow \mathrm{NaBr}(\mathrm{s})
\]
Given : \( \Delta \mathrm{H}_{\text {sub }}(\mathrm{Na})=137 \mathrm{~kJ} \mathrm{~mole}^{-1} ; \quad \Delta \mathrm{H}_{\text {bond dissociation }}\left(\mathrm{Br}_{2}(\mathrm{~g})\right)=144 \mathrm{~kJ} \mathrm{~mole}^{-1} \)
\( \Delta \mathrm{H}_{\text {ist ionisation }}\left(\mathrm{Na}(\mathrm{g})=496 \mathrm{~kJ} \mathrm{~mole}^{-1} ; \quad \Delta \mathrm{H}_{\text {ist electron gain enthalpy }}\left(\mathrm{Br}(\mathrm{g})=-325 \mathrm{~kJ} \mathrm{~mole}^{-1}\right.\right. \)
\( \Delta \mathrm{H}_{\text {Lattice energy }}(\mathrm{NaBr})=+742 \mathrm{~kJ} \mathrm{~mole}^{-1} \)
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