Question

Why does (or "do"?) Barium Oxide + Sulfuric Acid yield Ozone?

From my textbook:

The texbook says that if you pre-cool the test tubes with the initial components in the snow, you'll get some ozone. But why ozone and not just `O_2`?

Answer 1

Here's what I got.

Explanation:

Ok, now this is a very interesting question.

First thing first, the reaction involves barium peroxide, `"BaO"_2`, not barium oxide, which is `"BaO"`.

Now, as far as I know, this reaction is used to produce hydrogen peroxide, `"H"_2"O"_2`.

If you plan on using dilute sulfuric acid, you should use barium peroxide octahydrate, `"BaO"_2 * 8"H"_2"O"`. The reason behind this is that barium sulfate, `"BaSO"_4`, which is an insoluble ionic compound, forms on the surface of the peroxide, essentially halting the reaction.

Also, the reaction is performed with an ice-cold acid solution because the low temperature slows down the decomposition of hydrogen peroxide to water and oxygen gas.

`2"H"_2"O"_text(2(aq]) -> 2"H"_2"O"_text((aq]) + "O"_text(2(g]) uarr`

The balanced chemical equation for when this reaction (I'll use anhydrous barium peroxide for the sake of simplicity) is performed at low temperature looks like this

`"BaO"_text(2(s]) + "H"_2"SO"_text(4(aq]) -> "BaSO"_text(4(s]) darr + "H"_2"O"_text(2(aq])`

At room temperature and catalyzed by potassium iodide, `"KI"`, this reaction should proceed like this

`2"BaO"_text(2(s]) + 2"H"_2"SO"_text(4(aq]) stackrel(color(red)("KI")color(white)(aa))(->) 2"BaSO"_text(4(s])` `darr + 2"H"_2"O"_text((aq]) + "O"_text(2(g]) uarr`

So my guess is that this reaction will produce ozone as a side product, maybe depending on a combination of catalyst, reaction temperature, and concentration of the acid.

I was able to find a YouTube video on this supposed reaction - the narration and the subtitles are in Russian, so that will not be very helpful to students who don't speak it.