# What is the difference between "atomization energy" and "bond dissociation energy"?

Mar 6, 2017

Well, $\text{the atomization energy is the energy associated...........}$

#### Explanation:

$\text{the atomization energy is the energy associated with the }$
$\text{formation of 1 mole of gaseous atoms from 1 mole of gaseous}$
$\text{element in its standard state under standard conditions......}$

$\text{and the bond dissociation energy is the energy associated with}$ $\text{the CLEAVAGE of 1 mole of gaseous bonds from 1 mole of}$
$\text{gaseous molecules to form 2 gaseous radicals.........}$

So there are our definitions; let's see if we can represent each process by an equation, using a bimolecular species, ${X}_{2}$, as our exemplar:

$\text{ATOMIZATION:}$

$\frac{1}{2} {X}_{2} \left(g\right) + {\text{E}}_{1} \rightarrow \dot{X} \left(g\right)$

$\Delta {E}_{1} = \text{something}$

$\text{DISSOCIATION:}$

$X - X \left(g\right) + {\text{E}}_{2} \rightarrow 2 \dot{X} \left(g\right)$

$\Delta {E}_{2} = \text{some other thing}$

So given the definition, if we have a homonuclear diatomic, i.e. dihydrogen, dihalogen, etc, then $\Delta {E}_{2} = 2 \times \Delta {E}_{1}$. That is the $\text{energy of atomization}$ is HALF the $\text{bond dissociation energy}$ for gaseous homonuclear diatomic molecules.