# Which energy change corresponds to the enthalpy change of atomization of H_2 at 298 K ?

Apr 15, 2018

Atomization is simply the breaking of bonds to yield $\text{1 mol}$ of the atoms in the gas phase. So, if you have

$\text{H"_2(g) -> 2"H} \left(g\right)$,

then you have twice $\Delta {H}_{\text{atom}}^{\circ}$, i.e. you've broken one $\text{mol}$ of $\text{H"-"H}$ bonds for every one $\text{mol}$ of ${\text{H}}_{2} \left(g\right)$ molecules to form two $\text{mols}$ of $\text{H}$ atoms.

$\Delta {H}_{\text{atom}}^{\circ}$ is based on $\text{1 mol}$ of atoms, so really represents:

$\frac{1}{2} \text{H"_2(g) -> "H} \left(g\right)$

So, $\Delta {H}_{\text{atom}}^{\circ}$ is half the bond energy of a $\text{H"-"H}$ bond.