Question
Explain bond enthalpy of diatomic molecules.

Answer

In case of diatomic molecules, since there is only one bond, the bond enthalpy is equal to heat of atomisation. For example, heat of atomisation of $HCl _{( g )}$ is $431.9 kJ mol ^{-1}$.
$HCl _{( g )} \longrightarrow H _{( g )}+ Cl _{( g )} \quad \Delta_{ atm } H^0=431.9 kJ mol ^{-1}$
$\therefore$ Bond enthalpy of $HCl$,
$\begin{aligned}
\Delta_{\text {atm }} H^0 & =\frac{\Delta_{\text {atm }} H}{\text { Number of bonds broken }} \\
D_{ H - Cl } & =\frac{431.9}{1} \\
& =431.9 kJ mol ^{-1} \quad
\end{aligned}$
(Bond enthalpy is generally denoted by D).

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