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

We know that `DeltaG^(@)=-2.303RT" log "K_(C)`
Given `K_(C)=2xx10^(7),T=298K,R=8.314JK^(-1)mol^(-1)`
Thus, from above equation,
`DeltaG^(@)=-2.303xx8.314xx298log2xx10^(7)`
`=-12023.4J mol^(-1)=-12.023kJ" "mol^(-1)`
Since, `DeltaG^(@)` is negative hence reaction is spontaneous in forward direction.