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^(o+)` and `CI^(Theta)` ions are in the ratio of `2:1` then the enthalpy of hydration of `K^(o+)` is `-20xK cal mol^(-1)`. Find the value of `x`.

Explanation

`KCI(s) rarr K^(o+)(g) +CI^(Theta) (g), DeltaH_(1) = 181 kcal mol^(-1)`
`KCI(s)+aq rarr K^(o+) (aq) +CI^(Theta) (aQ), DeltaH_(2) = 1.0 kcal mol^(-1)`
Let the enthalpy of hydration of`K^(o+)` is `2CI kcal mol^(-1)`
`K^(o+) (g) +aq rarr K^(o+) (aq), DeltaH_(3) = 2a`
`CI^(Theta) (g)+aq rarr CI^(Theta) (aq), DeltaH_(4) = a`
`:. DeltaH_(3) =- DeltaH_(1) +DeltaH_(2) - DeltaH_(4)`
`2a =- 181 +1 -a`
`3a =- 180, a =- 60`
`:. Delta_(hyd)H^(Theta) of K^(o+) = 2a =- 60 xx2 =- 120`
`:. -20x =-120`
`x =6`