Identify the reaction for which `Delta H != Delta E`

Options

A. `s("rhombic") + O_(2)(g) rarr SO_(2)(g)`
B. `N_(2)(g) + O_(2)(g) rarr 2NO(g)`
C. `H_(2)(g) + Cl_(2)(g) rarr 2HCl(g)`
D. `CO(g) + 1/2 O_(2)(g) rarr CO_(2)(g)`

Explanation

To solve the problem of identifying the reaction for which ΔH ≠ ΔE, we need to understand the relationship between enthalpy (ΔH) and internal energy (ΔE) in the context of chemical reactions. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Definitions
- **ΔH (Enthalpy Change)**: This is the heat change at constant pressure.
- **ΔE (Internal Energy Change)**: This is the heat change at constant volume.

### Step 2: Identify the Relationship
The relationship between ΔH and ΔE can be expressed with the formula:
\[ \Delta E = \Delta H - \Delta N_g RT \]
where:
- \( \Delta N_g \) = change in the number of moles of gas (moles of gaseous products - moles of gaseous reactants)
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

### Step 3: Determine Conditions for ΔH ≠ ΔE
For ΔH to not equal ΔE, the term \( \Delta N_g RT \) must be non-zero. This occurs when there is a change in the number of moles of gas during the reaction.

### Step 4: Analyze the Given Reactions
We need to analyze the four options provided to find which one has a non-zero \( \Delta N_g \).

1. **First Reaction**:
- Reactants: 1 mole of solid + 1 mole of gas
- Products: 1 mole of gas
- \( \Delta N_g = 1 - 1 = 0 \)

2. **Second Reaction**:
- Reactants: 1 mole of gas + 1 mole of gas
- Products: 2 moles of gas
- \( \Delta N_g = 2 - 2 = 0 \)

3. **Third Reaction**:
- Reactants: 1 mole of gas + 1 mole of gas
- Products: 2 moles of gas
- \( \Delta N_g = 2 - 2 = 0 \)

4. **Fourth Reaction**:
- Reactants: 1 mole of CO (gas) + 1 mole of O2 (gas)
- Products: 1 mole of CO2 (gas)
- \( \Delta N_g = 1 - (1 + 1/2) = 1 - 1.5 = -0.5 \) (which is not zero)

### Step 5: Conclusion
From the analysis, the fourth reaction is the only one where \( \Delta N_g \) is not equal to zero. Therefore, for this reaction, ΔH ≠ ΔE.

### Final Answer
The reaction for which ΔH ≠ ΔE is the fourth option.

---