Bond dissociation enthalpies of `H_(2)(g)` and `N_(2)(g)` are `436.0 kJ mol^(-1)` and `941.8 kJ mol^(-1)`, respectively, and ethalpy of formation of `NH_(3)(g)` is `-46kJ mol^(-1)`. What is the enthalpy fi atomisation of `NH_(3)(g)`?. What is the avergae bond ethalpy of `N-H` bond?

Explanation

To solve the problem, we need to find the enthalpy of atomization of \( NH_3(g) \) and the average bond enthalpy of the \( N-H \) bond. We will use the given bond dissociation enthalpies and the enthalpy of formation of \( NH_3 \).

### Step 1: Write the Reaction for Formation of \( NH_3 \)
The formation of ammonia can be represented as:
\[ N_2(g) + \frac{3}{2} H_2(g) \rightarrow 2 NH_3(g) \]
The enthalpy change for this reaction (enthalpy of formation) is given as \( \Delta H_f = -46 \, kJ/mol \).
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