Question

The ionization enthalpy of $Na^{+}$ formation from $\left(Na\right)_{\left(\right. g \left.\right)}$ is $495.8\,kJ\,mol^{- 1}$ , while the electron gain enthalpy of $Br$ is $-325.0\,kJ\,mol^{- 1}$ . Given the lattice enthalpy of $NaBr$ is $-728.4\,kJ\,mol^{- 1}$ . The energy for the formation of $NaBr$ ionic solid is $\left(-\times 10^{-1} \,kJ \,mol ^{-1}\right.$

Answer 1

Ionisation energy:
$Na ( s ) \longrightarrow Na ^{+}( g ) ; \Delta H =495.8\, kJ / mol$
Electron gain enthalpy:
$\frac{1}{2} Br _{2}(\ell)+ e ^{-} \longrightarrow Br ^{-}( g ) ; \Delta H =325 \, kJ / mol$
Lattice energy:
$Na ^{+}( g )+ Br ^{-}( g ) \longrightarrow NaBr ( s ) ; \Delta H =-728.4\, kJ / mol$
Formation enthalpy:
$Na ( s )+\frac{1}{2} Br _{2}(\ell) \longrightarrow NaBr ( s ) ; \Delta H =? $
$\Delta H =495.8-325-728.4$
$\Delta H =-557.6 \, kJ =-5576 \times 10^{-1}$ kJ$