I'm trying to understand why hydrogen peroxide doesn't decompose into $\ce{H2}$ and $\ce{O2}$, but instead into $\ce{H2O}$ and $\ce{O2}$.


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.


The actual equation for the decomposition is:

$\ce{2H2O2}$ $\ce{->}$ $\ce{2H2O}$ + $\ce{O2}$

Yet another form of the equation that you think is supposed to happen is:

$\ce{2H2O2}$ $\ce{->}$ $\ce{2H2}$ + $\ce{2O2}$

Using bond enthalpies, it can be calculated that the energy needed to break all the bonds in gaseous hydrogen peroxide ($\ce{H2O2}$)is the summation of the energies need to break the two ($\ce{O-H}$) bonds and the ($\ce{O-O}$)bond. The energies are $\ce{2 * 465 kJ/mol}$ and $\ce{213 kJ/mol}$ respectively.

The extra energy added to the system required to do so for 2 moles of $\ce{H2O2(g)}$ = 2286 kJ

The total energy required to form 2 mol each of $\ce{H2}$ and $\ce{O2}$ = 1848 kJ

For $\ce{2H2O2}$ $\ce{->}$ $\ce{2H2}$ + $\ce{2O2}$ to occur, 438 kJ must be absorbed in to the system or reaction vessel, i.e. ∆H = +438kJ — an endothermic process.

The energy required to form $\ce{2H2O}$ and $\ce{O2}$ = 2358 kJ

If $\ce{2H2O2}$ $\ce{->}$ $\ce{2H2O}$ + $\ce{O2}$ occurs, 72 kJ will be absorbed by the reaction vessel, i.e. the reaction has ∆H = -36 kJ/mol of $\ce{2H2O2}$an exothermic process.

By employing the Gibbs free energy equation (and general rules of thumb regard entropy and temperature), exothermic reactions tend to be spontaneous while endothermic reactions are not. As the reaction producing water and oxygen gas is exothermic, it is also the more "favoured reaction". Even though the change of entropy of $\ce{2H2O2}$ $\ce{->}$ $\ce{2H2}$ + $\ce{2O2}$ is likely to be higher than that of the other reaction, ∆S has less of an impact on ∆G than ∆H does.

I used bond enthalpy data from a university chemistry class data sheet online.

  • $\begingroup$ I agree that in general the $-T\Delta S$ term contributes less than $\Delta H$, but the data (i.e. $S^\circ_m$ values) that you need to calculate the entropy change is so easily available, there's really no excuse to not calculate $\Delta S$ and $\Delta G$. $\endgroup$ – orthocresol Jan 4 '16 at 10:26
  • $\begingroup$ Understood :) I'll keep that in mind. $\endgroup$ – aspiringSD Jan 5 '16 at 15:08

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