4
$\begingroup$

I read two reactions which I want to specify first:

  • $\ce{Na2O2 + O3 + H2O -> 2NaOH + 2O2}$
  • $\ce{2KOH + 5O3 -> 2KO3 + 5O2 + H2O}$

Does $\ce{NaOH}$ also undergo a further reaction to produce $\ce{NaO3}$ or not? Since both are alkali metals and with relatively similar sizes(?), I thought it could, but Google doesn't show me anything apart from a reaction balancer site which just shows me the reverse equilibrium of the above reaction.

$\endgroup$
1
8
$\begingroup$

My own Google search does show that sodium ozonide can form [1]. But it requires low temperature:

Sodium ozonide was produced by the ozonization of sodium hydroxide at -80 to -100°[C] followed by extraction with liquid ammonia and removal of the latter under vacuum at-50° [2]. Composition of the product obtained: $\ce{NaO3}=80.1, \ce{NaOH•H2O}=11.2, \ce{H2O}=8.6$ [%].

Upon warming it decomposes to give the superoxide, and then in the presence of water back to sodium hydroxide:

The process of thermal decomposition of sodium ozonide to $\ce{NaO2}$ occurs at −20 and −10°.

At temperatures above 0°, the process of thermal decomposition of sodium ozonide is accompanied by the process of interaction of the $\ce{NaO2}$ formed with $\ce{H2O}$, and decomposition proceeds to the formation of $\ce{NaOH}$.

References

1. Tokareva, S.A., Pilipenko, G.P. Thermal decomposition of sodium ozonide. Russ Chem Bull 13, 686–688 (1964). https://doi.org/10.1007/BF00845322.

2. S. A. Tokareva and M. S. Dobrolyubova, Izv. AN SSSR, Ser. Khim. 1964, No. 4.

$\endgroup$
7
$\begingroup$

Yes, indeed. Sodium ozonide is formed in high percentage by the reaction of sodium hydroxide and ozone.

The reaction of ozone and anhydrous alkali metal hydroxides were studied by I.A. Kazarnovskii which confirmed the formation of red colored products known as ozonides. He proposed the following reaction mechanism:

\begin{align} \ce{2MOH + O3 &-> 2MOH.O3} \tag{a} \\ \ce{2MOH.O3 +2O3 &-> 2MO3.2HO2 + 2O2} \tag{b} \\ \ce{2HO2 &-> H2O + 1.5O2} \tag{c} \\ \ce{MOH + H2O &-> MOH.H2O} \tag{d} \\ \hline \ce{3MOH + 4O3 &-> 2MO3 + MOH.H2O +3.5O2} \tag{e} \\ \end{align}

So, a mixture of both ozonide and alkali metal hydroxide hydrate is formed as products with the simultaneous transformation of a large amount of ozone to oxygen.

Some information is also there about the structure and chemical properties of sodium ozonide:

Kazarnovskii et al. investigating the reaction of anhydrous sodium hydroxide powder and ozone-oxygen mixtures at -50 to -60°C, observed that, indeed, the hydroxide acquired an intense yellow color which, when extracted with liquid ammonia, resulted in the formation of a dark red solution. The color fades at r.t. Upon evaporation of the ammonia, crystals were recovered. Analysis of the recovered product showed $\ce{90 wt.\% NaO3}$.

Kazarnovskii further reported that $\ce{NaO3}$ is unstable, and that at room temperature it decomposes.

$$\ce{2NaO3 -> 2NaO2 + O2}$$

They also noted, however, that the ozonization of $\ce{NaOH}$ can be done in room temperature but the $\ce{NaO3}$ obtained from the room temperature reaction is not soluble in liquid ammonia, and the ozonide-hydroxide mixture is quite stable. Subsequent studies have explained the difference in the solubility and stability characteristics of $\ce{NaO3}$ prepared at room temperature and low temperatures. The reason proposed that $\ce{NaO3}$ can exist in two crystalline forms, one being soluble in liquid ammonia but unstable at room temperature, and the other being insoluble and stable although not confirmed experimentally.

The $\ce{NaO3}$ lattice is tetragonal, with $a = \pu{11.61 Å}$ and $c = \pu{7.66 Å}$. The space group is 14/mmm Its density varies from $\pu{1.56-1.60 g cm-3}$ . The standard heat of formation of $\ce{NaO3}$ is estimated as $\pu{45 kcal/mole}$.

$\endgroup$
2
  • $\begingroup$ Lenntech suggests that when ozone is exposed to aqueous sodium hydroxide it forms dioxygen species (peroxide, superoxide). Might that be occurring in the room temperature reaction you quote here? $\endgroup$ – Oscar Lanzi May 10 at 10:01
  • $\begingroup$ @OscarLanzi Essentially yes. If the stoichiometry of the reaction is altered or the reaction is done in r.t., various other species is noted to form, especially the alkali hydroxide hydrate $\ce{MOH.H2O}$ which being unstable at r.t. form $\ce{HO2.}$ radical along with side products like superoxide and peroxides. So, for optimal results, the reaction is done in sub-zero temperatures where the majority of product is the ozonide and oxygen. $\endgroup$ – Nilay Ghosh May 10 at 10:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.