Write notes on diatomic molecules and polyatomic molecules with respect to bond dissociation enthalpy.

(N/A) Diatomic Molecules:
For diatomic molecules,the bond dissociation enthalpy is equal to the enthalpy of atomization.
$H_{2(g)} \rightarrow 2H_{(g)} ; \Delta_{H-H} H^{\ominus} = 435.0 \ kJ \ mol^{-1}$
$Cl_{2(g)} \rightarrow 2Cl_{(g)} ; \Delta_{Cl-Cl} H^{\ominus} = 242 \ kJ \ mol^{-1}$
$O_{2(g)} \rightarrow 2O_{(g)} ; \Delta_{O=O} H^{\ominus} = 428 \ kJ \ mol^{-1}$
Bond dissociation enthalpy is the change in enthalpy when one mole of a covalent bond in a gaseous molecule is broken to form products in the gaseous phase.
Polyatomic Molecules:
In polyatomic molecules,the bond dissociation enthalpy varies for the same type of bonds within the molecule due to the changing chemical environment.
Example: Methane $(CH_4)$
$CH_{4(g)} \rightarrow C_{(g)} + 4H_{(g)} ; \Delta_{a} H^{\ominus} = 1665 \ kJ \ mol^{-1}$
The individual steps for breaking $C-H$ bonds are:
$CH_{4(g)} \rightarrow CH_{3(g)} + H_{(g)} ; \Delta_{bond} H^{\ominus} = +427 \ kJ \ mol^{-1}$
$CH_{3(g)} \rightarrow CH_{2(g)} + H_{(g)} ; \Delta_{bond} H^{\ominus} = +439 \ kJ \ mol^{-1}$
$CH_{2(g)} \rightarrow CH_{(g)} + H_{(g)} ; \Delta_{bond} H^{\ominus} = +452 \ kJ \ mol^{-1}$
$CH_{(g)} \rightarrow C_{(g)} + H_{(g)} ; \Delta_{bond} H^{\ominus} = +347 \ kJ \ mol^{-1}$
Since the energies differ,we use the mean bond enthalpy:
$\Delta_{C-H} H^{\ominus} = \frac{1}{4} (1665) = 416 \ kJ \ mol^{-1}$
General formula for reaction enthalpy: $\Delta_{r} H^{\ominus} = \Sigma \text{bond enthalpies of reactants} - \Sigma \text{bond enthalpies of products}$.