Why doesn't hydrogen peroxide decompose into hydrogen and oxygen?
We have two theoretical decomposition pathways:

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

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

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

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

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,

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

For reaction 2,

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

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:

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

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 $T = 298~\mathrm{K}$,

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

consistent with our earlier findings.