Question

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

Answer 1

Answer: 5576

Ionisation energy:
$Na(s)⟶N{a}^{+}(g);\Delta H=495.8kJ/mol$
Electron gain enthalpy:
$\frac{1}{2}B{r}_2(\ell )+e^{-}⟶B{r}^{-}(g);\Delta H=325kJ/mol$
Lattice energy:
$N{a}^{+}(g)+B{r}^{-}(g)⟶NaBr(s);\Delta H=-728.4kJ/mol$
Formation enthalpy:
$Na(s)+\frac{1}{2}B{r}_2(\ell )⟶NaBr(s);\Delta H=?$
$\Delta H=495.8-325-728.4$
$\Delta H=-557.6kJ=-5576\times 10^{-1}$ kJ$