The lattice energy of solid NaCl is 180 kcal/mol. The dissolution of the solid in water in the form of ions is endothermic to the extent of 1 kcal/mol. If the hydration energies of `Na^(+)` and `Cl^(-)` are in the ratio 6 : 5, what is the enthalpy of hydration of `Na^(+)` ion?

Options

A. `-85 kcal mol^(-1)`
B. `-98 kcal mol^(-1)`
C. `+82 kcal mol^(-1)`
D. `+100 kcal mol^(-1)`

Explanation

We know that
Ethalpy of solution =Lattice enegry + Hydration enthalpy
or `Delta_(sol)H^(Theta) = Delta_("lattice") H^(Theta) +Delta_("hyd")H^(Theta)`
`:. Delta_("hyd")H^(Theta) = Delta_("sol")H^(Theta) - Delta_("lattice")H^(Theta) = 1.0 - 180`
`=- 179.0 kcal mol^(-1)`
The ratio of hydration energies of `Na^(o+)` and `CI^(Theta)` is `6:5`.
`:.` Hydration enegry of `Na^(o+) = (6)/(11) xx (-179.0)`
`=- 97.63 kcal ~~ 98.0 kcal`