Thelattice energy of NaCl is 180 kcal // mol . The dissociation of the solid in water in the form of ioins is endothermicto the extent of 1 kcal // mol.If the solvation energies ofNa^(+) and Cl^(-) ions are in theratesof6:5 , calculate the enthalpy of hydration of Na^(+) ions.

Thelattice energy of NaCl is 180 kcal // mol . The dissociation of the

1.	Thelattice energy of NaCl is 180 kcal // mol . The dissociation of the solid in water in the form of ioins is endothermicto the extent of 1 kcal // mol.If the solvation energies ofNa^(+) and Cl^(-) ions are in theratesof6:5 , calculate the enthalpy of hydration of Na^(+) ions.
Answer» Solution :(i) `NaCl(s) rarr Na^(+) (g) + Cl^(-) (g), DeltaH =180kcal mol^(-1)` (ii) `Na^(+) Cl^(-)(s) overset( + aq)(rarr) Na^(+)(aq), Cl^(-)(aq) , DELTA H = 1 kcalmol^(-1)` (iii) `Na^(+)(g)+aq rarr Na^(+)(aq) , Delta H = 6 x ( `Aim) (iv) `Cl^(-)(g) +aq rarrCl^(-) (aq), DeltaH = 5X` Eqn. (ii) - Eqn. (iv) GIVES Enq. (i) `:. 180 = 1-5x ` or ` - 11X = 179 `or ` x= ( -179)/( 11)` `:. ` Enthalpy of hydration of`Na^(+)`ion `= 6X = ( -179)/( 11) xx 6 =-97.6 kcal mol^(-1)`