Bond energies of `N-=N, H-H` and `N-H` bonds are 945,463 & 391 kJ `mol^(-1)` respectively, the enthalpy of the following reactions is :
`N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g)`

Explanation

To calculate the enthalpy change for 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. The bond energies are as follows:

- Bond energy of \( N \equiv N \) (N triple bond N) = 945 kJ/mol
- Bond energy of \( H-H \) = 463 kJ/mol
- Bond energy of \( N-H \) = 391 kJ/mol

### Step-by-Step Solution:

1. **Identify Bonds Broken and Formed:**
- In the reactants, we have:
- 1 bond of \( N \equiv N \) (1 N2 molecule)
- 3 bonds of \( H-H \) (3 H2 molecules)
- In the products, we have:
- 6 bonds of \( N-H \) (2 NH3 molecules, each containing 3 N-H bonds)

2. **Calculate the Total Energy of Bonds Broken:**
- Energy required to break the bonds in the reactants:
\[ \text{Total Energy of Bonds Broken} = \text{Energy of } N \equiv N + 3 \times \text{Energy of } H-H \]
\[ = 945 \text{ kJ/mol} + 3 \times 463 \text{ kJ/mol} \]
\[ = 945 + 1389 = 2334 \text{ kJ/mol} \]

3. **Calculate the Total Energy of Bonds Formed:**
- Energy released when forming the bonds in the products:
\[ \text{Total Energy of Bonds Formed} = 6 \times \text{Energy of } N-H \]
\[ = 6 \times 391 \text{ kJ/mol} \]
\[ = 2346 \text{ kJ/mol} \]

4. **Calculate the Enthalpy Change (\( \Delta H \)):**
- The enthalpy change for the reaction is given by:
\[ \Delta H = \text{Total Energy of Bonds Broken} - \text{Total Energy of Bonds Formed} \]
\[ = 2334 \text{ kJ/mol} - 2346 \text{ kJ/mol} \]
\[ = -12 \text{ kJ/mol} \]

### Final Answer:
The enthalpy change (\( \Delta H \)) for the reaction \( N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \) is \(-12 \text{ kJ/mol}\).