Question

The lattice enthalpy of solid $NaCl$ is $772kJmo{l}^{-1}$ and enthalpy of solution is $2kJmo{l}^{-1}$. If the hydration enthalpy of $N{a}^{+}\&C{l}^{-}$ions are in the ratio of $3:2.5$, what is the magnitude of enthalpy of hydration (in $kJ$ ) of chloride ion?

Answer 1

Answer: 350

Lattice enthalpy $=772KJ/Mol$
$\frac{(\Delta H.E{)}_{N{a}^{+}}}{(\Delta H.E{)}_{C{l}^{-}}}=\frac{30}{25}=\frac{6}{5}$
$\underset{(x)}{(\Delta HE{)}_{N{a}^{+}}}=\frac{6}{5}\underset{(y)}{(\Delta HE{)}_{C{l}^{-}}}$
$x=\frac{6}{5}y$
$(\Delta H{)}_{\text{solution }}=$ Lattice enthalpy $+$ H.E. of $N{a}^{+}+$H.E. of $C{l}^{-}$
$2=272+(y)+\frac{6}{5}(y)$
$-770=\frac{11}{5}y$;
$y=\frac{770\times 5}{11}=350KJ$