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The lattice energy of solid `NaCI` is `180 kcal mol^(-1)`. The dissolution of the solid in `H_(2)O` is endothermic to the extent of `1.0 kcal mol^(-1)`. If the hydration energies of `Na^(o+)` and `CI^(Theta)` ions are in the ratio of `6:5` what is the enthalpy of hydration of sodium ion?A. `-85 kcal mol^(-1)`B. `-98 kcal mol^(-1)`C. `+82 kcal mol^(-1)`D. `+100 kcal mol^(-1)` |
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Answer» 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` |
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