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?

The equilibrium constant at `25^(@)C` for the process: `CO^(3+) (aq) +

1.	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?
Answer» We know, `DeltaG^(Theta) =- 2.303 RT log_(10)K_(c)` Given, `K_(c) =2xx10^(5),T = 298 K,R = 8.314 J K^(-1)mol^(-1)` Thus, form the above equation, `DeltaG^(Theta) =- 2.303 xx 8.314 xx 298 log 2xx10^(5) = 8.588 kJ`

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