For a reaction,
`Ag_(2)O(s)hArr2Ag(s)+(1)/(2)O_(2)(g)`
`DeltaH,DeltaS and T ` are 40.63 kJ `mol^(-1)`, 108.8 J `K^(-1)mol^(-1) and 373.3K` respectively. Predict the feasibility of the reaction:

Options

A. feasible
B. non-feasible
C. remains at equilibrium
D. not predicted

Explanation

To determine the feasibility of the reaction \( \text{Ag}_2\text{O}(s) \rightleftharpoons 2\text{Ag}(s) + \frac{1}{2}\text{O}_2(g) \), we need to calculate the Gibbs free energy change (\( \Delta G \)) using the provided values for enthalpy change (\( \Delta H \)), entropy change (\( \Delta S \)), and temperature (T). The reaction is feasible if \( \Delta G < 0 \).

### Step-by-Step Solution:

1. **Identify the given values**:
- \( \Delta H = 40.63 \, \text{kJ/mol} \)
- \( \Delta S = 108.8 \, \text{J/K} \cdot \text{mol} \)
- \( T = 373.3 \, \text{K} \)

2. **Convert \( \Delta H \) to Joules**:
- Since \( \Delta S \) is in Joules, we need to convert \( \Delta H \) from kJ to J:
\[ \Delta H = 40.63 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 40630 \, \text{J/mol} \]

3. **Use the Gibbs free energy formula**:
- The formula for Gibbs free energy is:
\[ \Delta G = \Delta H - T \Delta S \]

4. **Substitute the values into the formula**:
\[ \Delta G = 40630 \, \text{J/mol} - (373.3 \, \text{K} \times 108.8 \, \text{J/K} \cdot \text{mol}) \]

5. **Calculate \( T \Delta S \)**:
\[ T \Delta S = 373.3 \, \text{K} \times 108.8 \, \text{J/K} \cdot \text{mol} = 40515.04 \, \text{J/mol} \]

6. **Calculate \( \Delta G \)**:
\[ \Delta G = 40630 \, \text{J/mol} - 40515.04 \, \text{J/mol} = 114.96 \, \text{J/mol} \]

7. **Interpret the result**:
- Since \( \Delta G = 114.96 \, \text{J/mol} > 0 \), the reaction is non-feasible at the given temperature.

### Conclusion:
The reaction \( \text{Ag}_2\text{O}(s) \rightleftharpoons 2\text{Ag}(s) + \frac{1}{2}\text{O}_2(g) \) is non-feasible at \( 373.3 \, \text{K} \) because \( \Delta G \) is positive.

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