Which of the following reactions is `Delta H` less than `Delta E` ?

Options

A. `C_(12)H_(22)O_(11)(s)+6O_(2)(g)rarr 6CO_(2)(g)+6H_(2)O(l)`
B. `2SO_(2)(g)+O_(2)(g)rarr 2SO_(3)(g)`
C. `N_(2)O_(4)(g)rarr 2NO_(2)(g)`
D. `N_(2)(g)+O_(2)(g)rarr 2NO(g)`

Answer

To determine which reaction has \( \Delta H < \Delta E \), we need to analyze the reactions provided and calculate the change in the number of gaseous moles (\( \Delta N_G \)) for each reaction. This will help us use the relationship between enthalpy change (\( \Delta H \)) and internal energy change (\( \Delta E \)).

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The relationship between the change in enthalpy (\( \Delta H \)) and the change in internal energy (\( \Delta E \)) is given by:
\[ \Delta H = \Delta E + \Delta N_G \cdot R \cdot T \]
where \( R \) is the gas constant and \( T \) is the temperature.

2. **Calculating \( \Delta N_G \)**:
\( \Delta N_G \) is calculated as the difference between the number of moles of gaseous products and the number of moles of gaseous reactants.

3. **Analyzing Each Reaction**:

- **Reaction 1**:
\[ C_{12}H_{22}O_{11(s)} + 6 O_{2(g)} \rightarrow 6 CO_{2(g)} + 6 H_{2}O_{(l)} \]
- **Gaseous Products**: 6 (from \( CO_2 \))
- **Gaseous Reactants**: 6 (from \( O_2 \))
- \( \Delta N_G = 6 - 6 = 0 \)
- Thus, \( \Delta H = \Delta E \) (not valid for \( \Delta H < \Delta E \)).

- **Reaction 2**:
\[ 2 SO_{2(g)} + O_{2(g)} \rightarrow 2 SO_{3(g)} \]
- **Gaseous Products**: 2 (from \( SO_3 \))
- **Gaseous Reactants**: 3 (2 from \( SO_2 \) + 1 from \( O_2 \))
- \( \Delta N_G = 2 - 3 = -1 \)
- Since \( \Delta N_G \) is negative, \( \Delta H < \Delta E \) (valid).

- **Reaction 3**:
\[ N_{2(g)}O_{4(g)} \rightarrow 2 NO_{(g)} \]
- **Gaseous Products**: 2 (from \( NO \))
- **Gaseous Reactants**: 1 (from \( N_2O_4 \))
- \( \Delta N_G = 2 - 1 = 1 \)
- Thus, \( \Delta H > \Delta E \) (not valid).

- **Reaction 4**:
\[ N_{2(g)} + O_{2(g)} \rightarrow 2 NO_{(g)} \]
- **Gaseous Products**: 2 (from \( NO \))
- **Gaseous Reactants**: 2 (from \( N_2 \) and \( O_2 \))
- \( \Delta N_G = 2 - 2 = 0 \)
- Thus, \( \Delta H = \Delta E \) (not valid).

4. **Conclusion**:
The only reaction where \( \Delta H < \Delta E \) is the second reaction:
\[ 2 SO_{2(g)} + O_{2(g)} \rightarrow 2 SO_{3(g)} \]

### Final Answer:
The reaction for which \( \Delta H < \Delta E \) is:
**Option B: \( 2 SO_{2(g)} + O_{2(g)} \rightarrow 2 SO_{3(g)} \)**