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

To solve the problem, we need to calculate the standard Gibbs free energy change (ΔG°) for the given reaction at 25°C and determine the direction of spontaneity based on the sign of ΔG°.

### Step-by-step Solution:

1. **Identify the Given Data:**
- Equilibrium constant (Kc) = \(2 \times 10^7\)
- Temperature (T) = \(25°C = 298 K\)
- Universal gas constant (R) = \(8.314 \, J \, K^{-1} \, mol^{-1}\)
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