The equilibrium constant at `25^(@)C` for the process:
`CO^(3+) (aq) +6NH_(3)(aq) hArr[Co(NH_(3))_(6)]^(3+)(aq)` is `2 xx 10^(7)`.
Calculate the value of `DeltaG^(Theta)` at `25^(@)C at 25^(@)C[R = 8.314 J K^(-1)mol^(-1)]`.
In which direction the reaction is spontaneous when the recatants and proudcts are in standard state?

Explanation

`DeltaG^(Theta)` is related to equilibrium constant, `K` as
`DeltaG^(Theta) =- 2.303 RT logK`
`DeltaG^(Theta) = ? , K = 2.0 xx 10^(7), T = 25 +273 = 298K`
`DeltaG^(Theta) =- 2.303 xx 8.314 xx 298 log 2 xx 10^(7)`
`=- 12023.4 J =- 12.023 kJ`
Since `DeltaG^(Theta)` is negative, the reactio is spontaneous in the forward direction.