Explain the bond enthalpy.

(N/A) Bond enthalpy is the enthalpy change when one mole of a covalent bond in a gaseous molecule is broken to form products in the gaseous phase.
Diatomic Molecules:
For diatomic molecules like $H_{2_{(g)}} \rightarrow 2 H_{(g)}$,the enthalpy change $\Delta_{H-H} H^{\ominus} = 435.0 \ kJ \ mol^{-1}$ is the bond dissociation enthalpy.
Similarly,for $Cl_{2_{(g)}} \rightarrow 2 Cl_{(g)}$,$\Delta_{Cl-Cl} H^{\ominus} = 242 \ kJ \ mol^{-1}$ and for $O_{2_{(g)}} \rightarrow 2 O_{(g)}$,$\Delta_{O=O} H^{\ominus} = 428 \ kJ \ mol^{-1}$.
Polyatomic Molecules:
In polyatomic molecules,bond dissociation enthalpy varies for the same type of bond in different steps. For example,in methane $(CH_{4})$:
$CH_{4_{(g)}} \rightarrow CH_{3_{(g)}} + H_{(g)}$ ; $\Delta_{bond} H^{\ominus} = +427 \ kJ \ mol^{-1}$
$CH_{3_{(g)}} \rightarrow CH_{2_{(g)}} + H_{(g)}$ ; $\Delta_{bond} H^{\ominus} = +439 \ kJ \ mol^{-1}$
$CH_{2_{(g)}} \rightarrow CH_{(g)} + H_{(g)}$ ; $\Delta_{bond} H^{\ominus} = +452 \ kJ \ mol^{-1}$
$CH_{(g)} \rightarrow C_{(g)} + H_{(g)}$ ; $\Delta_{bond} H^{\ominus} = +347 \ kJ \ mol^{-1}$
In such cases,we use the mean bond enthalpy. For $CH_{4}$,the total enthalpy of atomization is $\Delta_{a} H^{\ominus} = 1665 \ kJ \ mol^{-1}$.
The mean $C-H$ bond enthalpy is $\Delta_{C-H} H^{\ominus} = \frac{1}{4}(1665) = 416 \ kJ \ mol^{-1}$.
General formula for reaction enthalpy: $\Delta_{r} H^{\ominus} = \Sigma \text{bond enthalpies}_{\text{reactants}} - \Sigma \text{bond enthalpies}_{\text{products}}$.