For the system `S (s) + O_(2)(g) rarr SO_(2)(g)` ?

Options

A. `Delta H = Delta E`
B. `Delta H gt Delta E`
C. `Delta E gt Delta H`
D. `Delta H = 0`

Answer

To solve the question regarding the reaction \( S (s) + O_2 (g) \rightarrow SO_2 (g) \), we will analyze the relationship between the change in enthalpy (\( \Delta H \)) and the change in internal energy (\( \Delta E \)) for this reaction.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction we are considering is:
\[ S (s) + O_2 (g) \rightarrow SO_2 (g) \]

2. **Write the Expression for Enthalpy Change**:
The relationship between the change in enthalpy (\( \Delta H \)) and the change in internal energy (\( \Delta E \)) is given by the formula:
\[ \Delta H = \Delta E + \Delta N_g \cdot R \cdot T \]
where \( \Delta N_g \) is the change in the number of moles of gas, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.

3. **Calculate \( \Delta N_g \)**:
- **Moles of Gaseous Reactants**: In the reactants, we have 1 mole of \( O_2 \) (gaseous).
- **Moles of Gaseous Products**: In the products, we have 1 mole of \( SO_2 \) (gaseous).
- Therefore, the change in the number of gaseous moles is:
\[ \Delta N_g = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} = 1 - 1 = 0 \]

4. **Substitute \( \Delta N_g \) into the Enthalpy Equation**:
Since \( \Delta N_g = 0 \), we can substitute this into the enthalpy equation:
\[ \Delta H = \Delta E + 0 \cdot R \cdot T \]
This simplifies to:
\[ \Delta H = \Delta E \]

5. **Conclusion**:
From our calculations, we find that the change in enthalpy (\( \Delta H \)) is equal to the change in internal energy (\( \Delta E \)). Therefore, the correct answer to the question is that \( \Delta H = \Delta E \).

### Final Answer:
The correct option is **A: \( \Delta H = \Delta E \)**.