For which of the following change `Delta H!=Delta U`?

Options

A. `H_(2)(g)+I_(2)+(g)rarr2HI(g)`
B. `HCl(aq)+NaOH(aq)rarrNaCl(aq)+H_(2)O(l)`
C. `C(s)+O_(2)(g)rarrCO_(2)(g)`
D. `N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g)`

Answer

To determine for which of the following changes \( \Delta H \neq \Delta U \), we will use the relationship between enthalpy change (\( \Delta H \)) and internal energy change (\( \Delta U \)) given by the equation:

\[ \Delta H = \Delta U + \Delta n_g RT \]

Where:
- \( \Delta n_g \) = moles of gaseous products - moles of gaseous reactants
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

We will analyze each option to find the value of \( \Delta n_g \) and see if \( \Delta H \) is equal to \( \Delta U \).

### Step-by-step Solution:

1. **Identify the Reaction Changes**:
- For each option, identify the number of moles of gaseous products and reactants.

2. **Calculate \( \Delta n_g \)**:
- Use the formula \( \Delta n_g = n_{\text{products}} - n_{\text{reactants}} \).

3. **Evaluate Each Case**:
- **Option 1**: If \( n_{\text{products}} = 2 \) and \( n_{\text{reactants}} = 2 \):
\[ \Delta n_g = 2 - 2 = 0 \]
Thus, \( \Delta H = \Delta U \).

- **Option 2**: If there are no gaseous products:
\[ \Delta n_g = 0 \]
Thus, \( \Delta H = \Delta U \).

- **Option 3**: If \( n_{\text{products}} = 1 \) and \( n_{\text{reactants}} = 1 \):
\[ \Delta n_g = 1 - 1 = 0 \]
Thus, \( \Delta H = \Delta U \).

- **Option 4**: If \( n_{\text{products}} = 2 \) and \( n_{\text{reactants}} = 4 \):
\[ \Delta n_g = 2 - 4 = -2 \]
Thus, \( \Delta H = \Delta U - 2RT \), which means \( \Delta H \neq \Delta U \).

4. **Conclusion**:
- The only case where \( \Delta H \neq \Delta U \) is in **Option 4**.

### Final Answer:
The correct answer is **Option 4**.