Calculate `Delta_(r)G^(Theta)` for the following reactions using `Delta_(f)G^(Theta)` values and and predict which reactions are spontaneous.
a. `Ca(s) +CI_(2)(g) rarr CaCI_(2)(s)`
b. `H_(2)O(s) rarr Hg(l) +(1)/(2)O_(2)(g)`
c. `NH_(3)(g) +2O_(2)(g) rarr HNO_(3)(l) +H_(2)O(l)`
`Delta_(f)G^(Theta)` value `(kJ mol^(-1))` are:
`CaCI_(2)(s) =- 748.1, HgO(s) =- 58.84`
`NH_(3)(g) =- 16.45, HNO_(3)(l) = - 80.71`,
`H_(2)O(l) =- 237.13`

Explanation

`Delta_(r)G^(Theta)`can be calculated by using the relation.
`Delta_(r)G^(Theta) = sum Delta_(r)G^(Theta) ("products") -sum Delta_(r)G^(Theta) ("reactants")`
a. `Ca(s) +CI_(2) (g) rarr CaCI_(2)(s)`
`Delta_(r)G^(Theta) = Delta_(f)G^(Theta) [CaCI_(2)(s)]`
`-{Delta_(f)G^(Theta)[Ca(s)] +Delta_(f)G^(Theta) [CI_(2)(g)]}`
`=- 748.1 -(0+0) =- 748.1 kJ`
Reaction is spontaneous.
b. `HgO(s) rarr Hg(l) +(1)/(2)O_(2)(g)`
`Delta_(r)G^(Theta) =DeltaG^(Theta) [Hg(l) +(1)/(2)DeltaG^(@) [O_(2)(g)] -Delta_(f)G^(Theta) [HgO(s)]`
`=0+0 -(-58.54) = 58.84 kJ`
Reaction is not spontaneous
c. `NH_(3)(g) +2O_(2)(g) rarr HNO_(3)(l) +H_(2)O(l)`
`Delta_(r)G^(Theta) =Delta_(f)G^(Theta) [HNO_(3)(l)] +Delta_(f)G^(Theta) [H_(2)O(l)]`
`-{Delta_(f)G^(Theta) [NH_(3)(g)] +2Delta_(f)G^(Theta) [O_(2)(g)]}`
`=- 80.71 +(-237.13) -{-16.45+0}`
`=- 80.71 -237.13 +16.45 =- 301.39kJ`
Reaction is spontanoeus.
a. `DeltaG^(Theta) =- 2.303 RT logK`