Given that:
`DeltaG_(F)^(@)(CuO) =-30.4 "Kcal"//"mole"`
`DeltaG_(f)^(@)(Cu_(2)O)=-34.98 Kcal//"mole" " "T=298K`
Now on the basis of above data which of the following predications will be most appropriate under the standard conditons and reversible reaction.

Options

A. Finely divided form fof CuO Kept in excess `O_(2)` would be completely converted to `Cu_(2)O`
B. Finely divided form of `Cu_(2)O` kept in excess `O_(2)` would be Completely converted to CuO
C. Finely divided form of CuO kept in excess `O_(2)` would be converted to a mixture of CuO and `Cu_(2)O` (having more of CuO)
D. Finely divided form of CuO kept in excess `O_(2)` would be converted to a mixture of CuO and `Cu_(2)O` (having more of `Cu_(2)O`)

Explanation

To solve the problem, we need to analyze the given Gibbs free energy values for the reactions involving CuO and Cu2O. We will calculate the Gibbs free energy change for the reaction and then determine the equilibrium constant (K) to make predictions about the reaction's favorability.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction we are considering is:
\[ \text{Cu}_2\text{O (s)} + \text{O}_2 (g) \rightarrow 2 \text{CuO (s)} \]

2. **Write the Gibbs Free Energy Change Equation**:
The Gibbs free energy change (\( \Delta G^\circ \)) for the reaction can be calculated using the standard Gibbs free energies of formation (\( \Delta G_f^\circ \)):
\[ \Delta G^\circ = \Delta G_f^\circ (\text{products}) - \Delta G_f^\circ (\text{reactants}) \]

3. **Substitute the Given Values**:
From the problem, we have:
- \( \Delta G_f^\circ (\text{CuO}) = -30.4 \, \text{kcal/mol} \)
- \( \Delta G_f^\circ (\text{Cu}_2\text{O}) = -34.98 \, \text{kcal/mol} \)

Therefore, substituting these values into the equation:
\[ \Delta G^\circ = [2 \times (-30.4)] - [-34.98] \]
\[ \Delta G^\circ = -60.8 + 34.98 = -25.82 \, \text{kcal/mol} \]

4. **Convert to Joules**:
To convert kcal to Joules (1 kcal = 4184 J):
\[ \Delta G^\circ = -25.82 \, \text{kcal/mol} \times 4184 \, \text{J/kcal} = -107,000 \, \text{J/mol} \approx -1.07 \times 10^5 \, \text{J/mol} \]

5. **Calculate the Equilibrium Constant (K)**:
Using the relationship between Gibbs free energy and the equilibrium constant:
\[ \Delta G^\circ = -RT \ln K \]
Rearranging gives:
\[ K = e^{-\Delta G^\circ / RT} \]
Where \( R = 8.314 \, \text{J/(mol K)} \) and \( T = 298 \, \text{K} \):
\[ K = e^{1.07 \times 10^5 / (8.314 \times 298)} \]
\[ K = e^{42.6} \approx 10^{19} \]

6. **Conclusion**:
Since \( K \) is a very large number (approximately \( 10^{19} \)), this indicates that the reaction favors the formation of products (2 CuO) almost completely from Cu2O and O2 under standard conditions.

### Final Answer:
The most appropriate prediction under standard conditions and reversible reaction is that Cu2O will almost completely convert to 2CuO.