For the reaction,
`N_(2)(g)+3H_(2)(g)to2NH_(3)(g)`
Heat of reaction at constant volume exceeds the heat reaction at constant pressure by the value of xRT. The va lue x is:

Answer

To solve the problem, we need to determine the value of \( x \) in the equation that relates the heat of reaction at constant volume (\( \Delta U \)) and the heat of reaction at constant pressure (\( \Delta H \)). The reaction given is:

\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction involves nitrogen gas and hydrogen gas reacting to form ammonia gas.

2. **Understand the Relationship**:
We know from thermodynamics that:
\[ \Delta H = \Delta U + \Delta n \cdot R \cdot T \]
where \( \Delta n \) is the change in the number of moles of gas, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.

3. **Calculate \( \Delta n \)**:
- **Reactants**: 1 mole of \( N_2 \) + 3 moles of \( H_2 \) = 4 moles of reactants.
- **Products**: 2 moles of \( NH_3 \).
- Therefore, the change in moles \( \Delta n \) is:
\[ \Delta n = \text{moles of products} - \text{moles of reactants} = 2 - 4 = -2 \]

4. **Substitute \( \Delta n \) into the Equation**:
Substituting \( \Delta n \) into the relationship gives:
\[ \Delta H = \Delta U + (-2)RT \]
This can be rearranged to:
\[ \Delta U = \Delta H + 2RT \]

5. **Express the Difference**:
From the problem statement, we know that:
\[ \Delta U = \Delta H - xRT \]
Setting the two expressions for \( \Delta U \) equal to each other gives:
\[ \Delta H - xRT = \Delta H + 2RT \]

6. **Solve for \( x \)**:
Rearranging the equation:
\[ -xRT = 2RT \]
Dividing both sides by \( RT \) (assuming \( RT \neq 0 \)):
\[ -x = 2 \]
Therefore, we find:
\[ x = -2 \]

### Conclusion:
The value of \( x \) is \( 2 \).