Given the bond energies `N-N` , `N-H` and `H-H` bond are `945, 436` and `391KJmol^(-1)` respectively, the enthalpy change of the reaction
`N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g)` is

Explanation

To find the enthalpy change of the reaction \( N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \), we will use the bond energies provided for the bonds involved in the reaction.

### Step-by-Step Solution:

1. **Identify the Bonds Broken and Formed:**
- In the reactants, we have:
- 1 N≡N bond in \( N_2 \)
- 3 H-H bonds in \( 3H_2 \)
- In the products, we have:
- 6 N-H bonds in \( 2NH_3 \) (since each \( NH_3 \) has 3 N-H bonds)

2. **Calculate the Energy Required to Break the Bonds:**
- The bond energy for \( N≡N \) is 945 kJ/mol.
- The bond energy for \( H-H \) is 391 kJ/mol.
- Therefore, the total energy required to break the bonds in the reactants is:
\[ \text{Energy to break bonds} = \text{Bond energy of } N_2 + 3 \times \text{Bond energy of } H_2 \]
\[ = 945 \, \text{kJ/mol} + 3 \times 391 \, \text{kJ/mol} \]
\[ = 945 + 1173 = 2118 \, \text{kJ} \]

3. **Calculate the Energy Released from Forming Bonds:**
- The bond energy for \( N-H \) is 436 kJ/mol.
- The total energy released when forming the bonds in the products is:
\[ \text{Energy released} = 6 \times \text{Bond energy of } N-H \]
\[ = 6 \times 436 \, \text{kJ/mol} = 2616 \, \text{kJ} \]

4. **Calculate the Enthalpy Change (\( \Delta H \)):**
- The enthalpy change for the reaction can be calculated using the formula:
\[ \Delta H = \text{Energy of bonds broken} - \text{Energy of bonds formed} \]
\[ = 2118 \, \text{kJ} - 2616 \, \text{kJ} \]
\[ = -498 \, \text{kJ} \]

5. **Interpret the Result:**
- The negative sign indicates that the reaction is exothermic, meaning it releases energy.

### Final Answer:
The enthalpy change of the reaction \( N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \) is \( -498 \, \text{kJ} \).