# Chemistry - Why doesn't hydrogen peroxide decompose into hydrogen and oxygen?

We have two theoretical decomposition pathways:

$$\ce{H2O2(l) -> H2O(l) + 1/2 O2(g)} \tag{1}$$

$$\ce{H2O2(l) -> H2(g) + O2(g)} \tag{2}$$

The required data (all at $298~\mathrm{K}$) is taken from the appendix of Atkins' Physical Chemistry, 9th ed:

$$\begin{array}{cccc} \hline \text{Species} & \Delta_\mathrm{f} H^\circ\mathrm{~/~kJ~mol^{-1}} & S^\circ_\mathrm{m} \mathrm{~/~J~K^{-1}~mol^{-1}} & \Delta_\mathrm{f} G^\circ\mathrm{~/~kJ~mol^{-1}} \\ \hline \ce{H2O2(l)} & -187.78 & 109.6 & -120.35 \\ \ce{H2O(l)} & -285.83 & 69.91 & -237.13 \\ \ce{H2(g)} & 0 & 130.684 & 0 \\ \ce{O2(g)} & 0 & 205.138 & 0 \\ \hline \end{array}$$

## The short way

The standard Gibbs free energy change of a reaction ($\Delta_\mathrm{r} G^\circ$) can be calculated from the individual standard Gibbs free energies of formation exactly analogously to the enthalpy change:

$$\Delta_\mathrm{r} G^\circ = \sum_J \nu_J[\Delta_\mathrm{f} G^\circ(\ce{J})]$$

So, for reaction 1,

\begin{align} \Delta G^\circ_1 &= [1(-237.13) + \frac{1}{2}(0) - 1(-120.35)]~\mathrm{kJ~mol^{-1}} \\ &= -116.78~\mathrm{kJ~mol^{-1}} \end{align}

For reaction 2,

\begin{align} \Delta G^\circ_2 &= [1(0) + 1(0) - 1(-120.35)]~\mathrm{kJ~mol^{-1}} \\ &= +120.35~\mathrm{kJ~mol^{-1}} \end{align}

Clearly reaction 1 is favourable and reaction 2 isn't.

## The long way

There's absolutely no need to do this, but you could use the "extra" data given above to calculate $\Delta H^\circ$ and $\Delta S^\circ$ for both reactions. You'd get:

\begin{align} \Delta H^\circ_1 &= [1(-285.83) + \frac{1}{2}(0) - 1(-187.78)]~\mathrm{kJ~mol^{-1}} \\ &= -98.05 ~\mathrm{kJ~mol^{-1}} \\ \Delta H^\circ_2 &= [1(0) + 1(0) - 1(-187.78)]~\mathrm{kJ~mol^{-1}} \\ &= +187.78 ~\mathrm{kJ~mol^{-1}} \\ \Delta S^\circ_1 &= [1(69.91) + \frac{1}{2}(205.138) - 1(109.6)]~\mathrm{J~K^{-1}~mol^{-1}} \\ &= +62.879 ~\mathrm{J~K^{-1}~mol^{-1}} \\ \Delta S^\circ_2 &= [1(130.684) + 1(205.138) - 1(109.6)]~\mathrm{J~K^{-1}~mol^{-1}} \\ &= +226.222 ~\mathrm{J~K^{-1}~mol^{-1}} \end{align}

If you're only interested in the final result, you don't need to do all this at all. But it does give you some insight into the difference between the two reactions: reaction 1 is enthalpically favoured whereas reaction 2 is entropically favoured.

Generally speaking, enthalpy usually plays a larger role in determining $\Delta G^\circ$. However, it is not a very good idea to generalise this statement and compare reactions solely on the basis of their $\Delta H^\circ$.

Since $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$, and taking $T = 298~\mathrm{K}$,

\begin{align} \Delta G^\circ_1 &= -98.05 ~\mathrm{kJ~mol^{-1}} - (298~\mathrm{K})(+62.879 ~\mathrm{J~K^{-1}~mol^{-1}}) \\ &= -116.79~\mathrm{kJ~mol^{-1}} \\ \Delta G^\circ_2 &= +187.78 ~\mathrm{kJ~mol^{-1}} - (298~\mathrm{K})(+226.222 ~\mathrm{J~K^{-1}~mol^{-1}}) \\ &= +120.37~\mathrm{kJ~mol^{-1}} \\ \end{align}

consistent with our earlier findings.