For the reaction `Ag_(2)O(s)rarr 2Ag(s)+1//2O_(2)(g)`, which one of the following is true :

Options

A. `Delta H = Delta E`
B. `Delta H = 1//2 Delta E`
C. `Delta H lt Delta H`
D. `Delta H gt Delta E`

Answer

To solve the problem regarding the reaction \( \text{Ag}_2\text{O}(s) \rightarrow 2\text{Ag}(s) + \frac{1}{2}\text{O}_2(g) \), we need to analyze the relationship between the change in enthalpy (\( \Delta H \)) and the change in internal energy (\( \Delta E \)).

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction is given as:
\[ \text{Ag}_2\text{O}(s) \rightarrow 2\text{Ag}(s) + \frac{1}{2}\text{O}_2(g) \]

2. **Determine the Change in Moles of Gas (\( \Delta N_G \))**:
- **Products**: The gaseous product is \( \frac{1}{2} \text{O}_2 \), which contributes \( 0.5 \) moles of gas.
- **Reactants**: There are no gaseous reactants, so this contributes \( 0 \) moles of gas.
- Therefore, the change in moles of gas (\( \Delta N_G \)) is:
\[ \Delta N_G = \text{Moles of gaseous products} - \text{Moles of gaseous reactants} = 0.5 - 0 = 0.5 \]

3. **Use the Relationship Between \( \Delta H \) and \( \Delta E \)**:
The relationship between the change in enthalpy and the change in internal energy is given by:
\[ \Delta H = \Delta E + \Delta N_G RT \]
Substituting the value of \( \Delta N_G \):
\[ \Delta H = \Delta E + 0.5 RT \]

4. **Analyze the Expression**:
From the expression \( \Delta H = \Delta E + 0.5 RT \), we can conclude:
- \( \Delta H \) is greater than \( \Delta E \) because \( 0.5 RT \) is a positive term added to \( \Delta E \).

5. **Conclusion**:
Therefore, the correct statement regarding the relationship between \( \Delta H \) and \( \Delta E \) in this reaction is:
- \( \Delta H > \Delta E \)

### Final Answer:
The correct option is that \( \Delta H \) is greater than \( \Delta E \).