`DeltaHltDeltaU` for the reaction(s) :

Options

A. `N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g)`
B. `Ag_(2)O(s)rarr2Ag(s)+(1)/(2)O_(2)(g)`
C. `CO(g)+(1)/(2)O_(2)(g)rarrCO_(2)(g)`
D. `C(s)+O_(2)(g)rarrCO_(2)(g)`

Explanation

To determine for which reactions \( \Delta H < \Delta U \), 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:
- \( \Delta N_g \) = change in the number of gaseous moles (moles of products - moles of reactants)
- \( R \) = gas constant
- \( T \) = temperature

From this equation, we can infer that \( \Delta H < \Delta U \) when \( \Delta N_g \) is negative. This is because if \( \Delta N_g \) is negative, it will subtract from \( \Delta U \), making \( \Delta H \) less than \( \Delta U \).

### Step-by-Step Solution:

1. **Analyze the Reactions**: We will analyze each reaction to determine \( \Delta N_g \).

2. **First Reaction: \( N_2 + 3H_2 \rightarrow 2NH_3 \)**
- Moles of gaseous products = 2 (from \( 2NH_3 \))
- Moles of gaseous reactants = 4 (from \( N_2 + 3H_2 \))
- \( \Delta N_g = 2 - 4 = -2 \)
- Since \( \Delta N_g \) is negative, \( \Delta H < \Delta U \).

3. **Second Reaction: \( Ag_2O (s) \rightarrow 2Ag (s) + \frac{1}{2}O_2 (g) \)**
- Moles of gaseous products = \( \frac{1}{2} \) (from \( \frac{1}{2}O_2 \))
- Moles of gaseous reactants = 0 (no gaseous reactants)
- \( \Delta N_g = \frac{1}{2} - 0 = \frac{1}{2} \)
- Since \( \Delta N_g \) is positive, \( \Delta H > \Delta U \).

4. **Third Reaction: \( CO (g) + \frac{1}{2}O_2 (g) \rightarrow CO_2 (g) \)**
- Moles of gaseous products = 1 (from \( CO_2 \))
- Moles of gaseous reactants = \( 1 + \frac{1}{2} = \frac{3}{2} \)
- \( \Delta N_g = 1 - \frac{3}{2} = -\frac{1}{2} \)
- Since \( \Delta N_g \) is negative, \( \Delta H < \Delta U \).

5. **Fourth Reaction: \( C (s) + O_2 (g) \rightarrow CO_2 (g) \)**
- Moles of gaseous products = 1 (from \( CO_2 \))
- Moles of gaseous reactants = 1 (from \( O_2 \))
- \( \Delta N_g = 1 - 1 = 0 \)
- Since \( \Delta N_g \) is zero, \( \Delta H = \Delta U \).

### Conclusion:
From the analysis, we find that \( \Delta H < \Delta U \) for the first and third reactions. Therefore, the correct options are:

- **Option A**: \( N_2 + 3H_2 \rightarrow 2NH_3 \)
- **Option C**: \( CO (g) + \frac{1}{2}O_2 (g) \rightarrow CO_2 (g) \)