For which one of the following systems `DE lt DH`

Options

A. `2SO_(2(g)) + O_(2(g)) rarr 2SO_(3(g))`
B. `N_(2(g)) + O_(2(g)) rarr 2NO_((g))`
C. `2NH_(3(g)) rarr N_(2(g)) + 3H_(2(g))`
D. `H_(2(g)) + I_(2(g)) rarr 2HI_((g))`

Answer

To determine for which system \( \Delta E < \Delta H \), we need to analyze the reactions given and calculate the change in the number of moles of gas (\( \Delta N_g \)) for each reaction. The relationship between the change in internal energy (\( \Delta E \)) and the change in enthalpy (\( \Delta H \)) can be expressed as:

\[ \Delta H = \Delta E + \Delta N_g RT \]

Where:
- \( \Delta N_g \) is the change in the number of moles of gas (moles of products - moles of reactants).
- \( R \) is the universal gas constant.
- \( T \) is the temperature in Kelvin.

If \( \Delta N_g > 0 \), then \( \Delta H > \Delta E \), which implies \( \Delta E < \Delta H \).

### Step-by-Step Solution:

1. **Identify the Reactions**:
- Reaction 1: \( 2 \text{SO}_2 \) (g)
- Reaction 2: \( \text{N}_2 + \text{O}_2 \rightarrow 2 \text{NO} \) (g)
- Reaction 3: \( 2 \text{NH}_3 \rightarrow \text{N}_2 + 3 \text{H}_2 \) (g)
- Reaction 4: \( \text{H}_2 + \text{I}_2 \rightarrow 2 \text{HI} \) (g)

2. **Calculate \( \Delta N_g \) for Each Reaction**:
- **Reaction 1**:
- Products: 2 moles of SO2
- Reactants: 2 moles of SO2
- \( \Delta N_g = 2 - 2 = 0 \)

- **Reaction 2**:
- Products: 2 moles of NO
- Reactants: 1 mole of N2 + 1 mole of O2 = 2 moles
- \( \Delta N_g = 2 - 2 = 0 \)

- **Reaction 3**:
- Products: 1 mole of N2 + 3 moles of H2 = 4 moles
- Reactants: 2 moles of NH3
- \( \Delta N_g = 4 - 2 = 2 \)

- **Reaction 4**:
- Products: 2 moles of HI
- Reactants: 1 mole of H2 + 1 mole of I2 = 2 moles
- \( \Delta N_g = 2 - 2 = 0 \)

3. **Analyze the Results**:
- For Reaction 1: \( \Delta N_g = 0 \) → \( \Delta H = \Delta E \)
- For Reaction 2: \( \Delta N_g = 0 \) → \( \Delta H = \Delta E \)
- For Reaction 3: \( \Delta N_g = 2 \) → \( \Delta H > \Delta E \) (since \( \Delta N_g > 0 \))
- For Reaction 4: \( \Delta N_g = 0 \) → \( \Delta H = \Delta E \)

4. **Conclusion**:
- The only reaction where \( \Delta E < \Delta H \) is Reaction 3: \( 2 \text{NH}_3 \rightarrow \text{N}_2 + 3 \text{H}_2 \).

### Final Answer:
The system for which \( \Delta E < \Delta H \) is \( 2 \text{NH}_3 \rightarrow \text{N}_2 + 3 \text{H}_2 \).