Calculate `K_(sp)` for `AgCI`. Given:
`Delta_(f)H^(Θ) Ag^(o+) = 25.3 kcal mol^(-1)`
`Delta_(f)H^(Θ) C1^(Θ) =- 40.0 kcal mol^(-1)`
`Delta_(f)H^(Θ) AgC1=- 30.36 kcal mol^(-1)`
`S^(Θ) Ag^(o+), S^(Θ) C1^(Θ)`, and `S^(Θ) AgC1`are `17.7, 13.2` and `23.0 cal mol^(-1)`

Explanation

The process is represented by :
`AgC1(S) hArr Ag^(o+) + C1^(Θ)`.
`DeltaH^(Θ) = Delta_(f)H^(Θ) (Ag^(o+)) +Delta_(f)H^(Θ) (CI^(Θ)) - Delta_(f)H^(Θ) (AgCI)`
`= 25.3 + (-40.0) -(-30.36) = 15.7 kcal mol^(-1)`.
`DeltaS^(Θ) = S^(Θ) (Ag^(o+)) +S^(Θ) (CI^(Θ)) - S^(Θ) (AgCI)`
`= 17.7 +13.2 - 23.0 = 7.9 cal mol^(-1)`
`DeltaG^(Θ) = DeltaH^(Θ) - T DeltaS^(Θ) = 15.7 xx 106(3) - (298)(7.9) = 13350 cal`
`DeltaG^(Θ) =- 2.30RT log K`.
`:. -log K = (DeltaG^(Θ))/(2.30RT) = (13350)/((2.30)(1.99)(298)) =9.79`
`:. K = "Antilog" (-9.79) = "Antilog" (-9-0.79 +1-1)`
`= "Antilog" (bar(10).21) = 1.6 xx 10^(-10)`
The tabulated value of `K_(sp)` of `AgCI = 1 xx 10^(-10)`