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 question 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 Components**:
- The reaction involves solid silver oxide (\( \text{Ag}_2\text{O}(s) \)), solid silver (\( \text{Ag}(s) \)), and gaseous oxygen (\( \text{O}_2(g) \)).

2. **Determine the Change in Moles of Gaseous Products**:
- The only gaseous product in the reaction is \( \frac{1}{2}\text{O}_2(g) \).
- The change in the number of moles of gas (\( \Delta N_G \)) is calculated as:
\[ \Delta N_G = \text{moles of gaseous products} - \text{moles of gaseous reactants} \]
- Here, we have:
\[ \Delta N_G = \frac{1}{2} - 0 = \frac{1}{2} \]

3. **Use the Relationship Between \( \Delta H \) and \( \Delta E \)**:
- The relationship is given by:
\[ \Delta H = \Delta E + \Delta N_G RT \]
- Substituting \( \Delta N_G \):
\[ \Delta H = \Delta E + \left(\frac{1}{2}\right) RT \]

4. **Analyze the Equation**:
- From the equation \( \Delta H = \Delta E + \frac{1}{2} RT \), we can see that \( \Delta H \) is greater than \( \Delta E \) because \( \frac{1}{2} RT \) is a positive term (assuming \( R \) and \( T \) are positive constants).

5. **Conclusion**:
- Therefore, we conclude that \( \Delta H > \Delta E \).

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