The lattice energy of solid KCI is 181 kcal `mol^(-1)` and the enthalpy of solution of KCI in `H_(2)O` is 1.0 kcal `mol^(-1).` If the hydration enthalpies of `K^(Ө) and Cl ^(Ө)` ions are in the ratio of 2:1 then the enthalpy of hydration of `K^(o+)` is-20x K cal `mol^(-1).` Find the value of x.

Explanation

`KC l (s) to K ^(o+) (g) + Cl^(Ө)(g) , Delta H _(1) = 18 181 kcal mol^(-1)`
`KCl (s) + aq to K ^(o+) (aq) + Cl ^(Ө) (aq), Delta H _(2) = 1. 0 kcal mol ^(-1)`
Let the enthalpy of hydration of `K ^(o+)` is 2a kcal `mol ^(-1)`
`K ^(o+) (g) + aq to K ^(o+) (aq)_(2) " " Delta H _(3) = 2 a `
`Cl^(Ө) (g) + aq to Cl ^(Ө) (aq), " " Delta H _(4) =a`
`therefore Delta H _(3) =- Delta H _(1) + Delta H _(2) - Delta H _(4)`
`2a =- 181 + 1 - a " " 3a =- 180 , a =- 60`
`2a =- 181 + 10a`
`therefore Delta _( hyd) H ^(Ө) of K ^(o+) = 2a =- 60 xx 2 =- 120 therefore - 2 0 x =- 120 x = 6`