For the reaction `N_(2)O_(4)(g)to2NO_(2)(g)`

Options

A. `DeltaHgtDeltaE`
B. `DeltaHltDeltaE`
C. `DeltaH=DeltaE`
D. `DeltaH=0`

Explanation

To solve the question regarding the reaction \( N_2O_4(g) \rightarrow 2NO_2(g) \), 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 given is \( N_2O_4(g) \rightarrow 2NO_2(g) \).

2. **Determine the Change in Moles of Gas (\( \Delta N_G \))**:
- The number of moles of gaseous products = 2 (from \( 2NO_2 \)).
- The number of moles of gaseous reactants = 1 (from \( N_2O_4 \)).
- Therefore, \( \Delta N_G = \text{moles of products} - \text{moles of reactants} = 2 - 1 = 1 \).

3. **Use the Relationship Between \( \Delta H \) and \( \Delta E \)**:
The relationship is given by the equation:
\[ \Delta H = \Delta E + \Delta N_G RT \]
where \( R \) is the universal gas constant and \( T \) is the temperature in Kelvin.

4. **Substitute \( \Delta N_G \) into the Equation**:
We have \( \Delta N_G = 1 \), so we can rewrite the equation as:
\[ \Delta H = \Delta E + 1 \cdot RT \]
or simply:
\[ \Delta H = \Delta E + RT \]

5. **Analyze the Relationship**:
From the equation \( \Delta H = \Delta E + RT \), it is clear that:
\[ \Delta H > \Delta E \]
because \( RT \) is a positive term (as long as \( T \) is greater than 0).

6. **Conclusion**:
Therefore, the correct treatment for the reaction is that \( \Delta H \) is greater than \( \Delta E \).

### Final Answer:
The correct treatment for the reaction \( N_2O_4(g) \rightarrow 2NO_2(g) \) is that \( \Delta H > \Delta E \).