At `25^(@)C`, the bond dissociation energy of `N_(2)(g)` is 946 `kJ*mol^(-1)`. What does it mean ? What would be the standard atomisation enthalpy of `N_(2)(g)` at `25^(@)C` ?

Explanation

At `25^(@)C`, the standard bond dissociation energy of `N_(2)(g)` is `946kJ*mol^(-1)`. This means that the energy required to break 1 mol of `N-=N` bonds completely in gaseous state to form gaseous nitrogen atoms in 948 kJ. At `25^(@)C`, the standard state of nitrogen is `N_(2)(g)`. now, the formation of 1 mol of N(g) takes place by the following process: `(1)/(2)N_(2)(g)toN(g)`.
Since 1 mol of N(g) is produced from `N_(2)(g)` in process [1], the enthalpy change in this process at `25^(@)C` will be equal tot he standard atomisation enthalpy of nitrogen.
In process [1], change in enthalpy `=(1)/(2)xx`bond dissociation energy of `N-=N` bond`=(1)/(2)xx946=473kJ*mol^(-1)`
Therefore, at `25^(@)C` the standard atomisation enthalpy of nitrogen is `473kJ*mol^(-1)`.