Identify the molecule for which the enthalpy of atomization $\left(\Delta_{\mathrm{a}}…
Identify the molecule for which the enthalpy of atomization $\left(\Delta_{\mathrm{a}} \mathrm{H}^{-}\right)$and bond dissociation enthalpy $\left(\Delta_{\mathrm{BOND}} \mathrm{H}^{-}\right)$are not equal)
$\mathrm{H}_2$
$\mathrm{Cl}_2$
$\mathrm{F}_2$
$\mathrm{CH}_4$
Solution
1. For $\mathrm{H}_2$
- Atomization $\mathrm{H}_2 \rightarrow 2 \mathrm{H}$
Bond dissociation $\mathrm{H}_2 \rightarrow 2 \mathrm{H}$
so $\Delta \mathrm{H}$ atomization $=\Delta \mathrm{H}$ dissociation
2. For $\mathrm{Cl}_2$
- Atomization $\mathrm{Cl}_2 \rightarrow 2 \mathrm{Cl}$
- Bond dissociation $\mathrm{Cl}_2 \rightarrow 2 \mathrm{Cl}$
so $\Delta \mathrm{H}_{\text {atom }}=\Delta \mathrm{H}_{\text {dissociation }}$
3. For $\mathrm{F}_2$
Atomization $\mathrm{F}_2 \rightarrow 2 \mathrm{~F}$
Bond dissociation $\mathrm{F}_2 \rightarrow 2 \mathrm{~F}$
so $\Delta \mathrm{H}_{\text {atomization }}=\Delta \mathrm{H}_{\text {dissociation }}$
4. $\mathrm{CH}_4$
Atomization $\mathrm{CH}_4 \rightarrow \mathrm{C}+4 \mathrm{H}$
Bond dissociation - Each $\mathrm{C}-\mathrm{H}$ bond in $\mathrm{CH}_4$ has its own dissociation enthalpy and these are not necessarily the same as enthalpy of atomization. $\Delta \mathrm{H}$ atomization $\neq \Delta \mathrm{H}$ dissociation.