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 find the enthalpy of atomization of \( NH_3(g) \) and the average bond enthalpy of the \( N-H \) bond, we can use the given data and apply Hess's law.

### Step 1: Understanding the Enthalpy of Atomization
The enthalpy of atomization is the energy required to break all the bonds in a compound to form individual atoms in the gas phase. For ammonia \( NH_3(g) \), the reaction can be represented as:
\[ NH_3(g) \rightarrow N(g) + 3H(g) \]
The enthalpy change for this reaction is the enthalpy of atomization of \( NH_3(g) \).
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