For the reaction : `N_(2(g))+3H_(2(g))hArr2NH_(3(g))`, `DeltaH=-ve`, the correct statement is :

Options

A. `DeltaH=DeltaU=0`
B. `|DeltaH| lt |DeltaU|`
C. `|DeltaH| gt |DeltaU|`
D. `DeltaH = DeltaU ne 0`

Answer

To solve the question regarding the reaction \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \) with \( \Delta H < 0 \), we will analyze the thermodynamic properties of the reaction step by step.

### Step-by-Step Solution:

1. **Identify the Reaction and Its Enthalpy Change**:
- The reaction given is \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \).
- It is stated that the enthalpy change \( \Delta H \) for this reaction is negative, indicating that the reaction is exothermic.

2. **Calculate the Change in Moles of Gas (\( \Delta N \))**:
- We need to calculate \( \Delta N \), which is the change in the number of moles of gas during the reaction.
- Moles of products = 2 (from \( 2NH_3 \))
- Moles of reactants = 1 (from \( N_2 \)) + 3 (from \( 3H_2 \)) = 4
- Therefore, \( \Delta N = \text{Moles of products} - \text{Moles of reactants} = 2 - 4 = -2 \).

3. **Relate Enthalpy Change to Internal Energy Change**:
- The relationship between enthalpy change (\( \Delta H \)) and internal energy change (\( \Delta U \)) can be expressed as:
\[ \Delta H = \Delta U + P\Delta V \]
- For ideal gases, \( P\Delta V \) can be expressed in terms of \( \Delta N \):
\[ P\Delta V = \Delta NRT \]
- Hence, we can rewrite the equation as:
\[ \Delta H = \Delta U + \Delta NRT \]

4. **Substitute the Value of \( \Delta N \)**:
- Since \( \Delta N = -2 \):
\[ \Delta H = \Delta U - 2RT \]

5. **Analyze the Sign of \( \Delta H \) and \( \Delta U \)**:
- Given \( \Delta H < 0 \) (since the reaction is exothermic), we can infer that:
\[ \Delta U - 2RT < 0 \]
- This implies that:
\[ \Delta U < 2RT \]
- Since \( R \) (universal gas constant) and \( T \) (temperature) are positive, \( 2RT \) is positive, meaning \( \Delta U \) must also be a negative value but less than \( 2RT \).

6. **Conclusion**:
- The correct statement regarding the relationship between \( \Delta H \) and \( \Delta U \) is that \( \Delta H < \Delta U \) because \( \Delta H \) is negative and \( \Delta U \) is less negative (or more positive) than \( 2RT \).

### Final Answer:
The correct statement is that \( \Delta H < \Delta U \).