For which reaction, `(DeltaH > DeltaU)`? ( `Delta H`= change in enthalpy,`Delta U` = change in internal energy)

Options

A. `[Fe(s) + 5 CO(g) rarr Fe(CO)_5(l)]`
B. `[C("graphite") + O_2(g) rarr CO_2(g)]`
C. `[H_2O(l) rarr H^+(aq) + OH^(-)(aq)]`
D. `[CaCO_3(s)rarr CaO(s) + CO_2(g)]`

Answer

To determine for which reaction \( \Delta H > \Delta U \), we need to analyze the relationship between the change in enthalpy (\( \Delta H \)) and the change in internal energy (\( \Delta U \)). The key relationship is given by:

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

where \( \Delta N_g \) is the change in the number of moles of gaseous products minus the number of moles of gaseous reactants.

### Step-by-Step Solution:

1. **Understand the relationship**: We know that:
- If \( \Delta N_g > 0 \), then \( \Delta H > \Delta U \).
- If \( \Delta N_g = 0 \), then \( \Delta H = \Delta U \).
- If \( \Delta N_g < 0 \), then \( \Delta H < \Delta U \).

2. **Calculate \( \Delta N_g \) for each reaction**:
- **Reaction 1**:
- Products: 0 moles of gas
- Reactants: 5 moles of \( \text{CO} \) (gas)
- \( \Delta N_g = 0 - 5 = -5 \) (negative)
- **Reaction 2**:
- Products: 1 mole of \( \text{CO}_2 \) (gas)
- Reactants: 1 mole of \( \text{O}_2 \) (gas) + 1 mole of \( \text{C} \) (solid)
- \( \Delta N_g = 1 - 1 = 0 \) (zero)
- **Reaction 3**:
- Products: 0 moles of gas
- Reactants: 0 moles of gas
- \( \Delta N_g = 0 - 0 = 0 \) (zero)
- **Reaction 4**:
- Products: 1 mole of \( \text{CO}_2 \) (gas)
- Reactants: 0 moles of gas
- \( \Delta N_g = 1 - 0 = 1 \) (positive)

3. **Determine which reaction has \( \Delta N_g > 0 \)**:
- From our calculations, only **Reaction 4** has \( \Delta N_g = 1 \), which is greater than 0.

4. **Conclusion**:
- Therefore, for **Reaction 4**, we have \( \Delta H > \Delta U \).

### Final Answer:
The reaction for which \( \Delta H > \Delta U \) is **Reaction 4**.