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

From my textbook:

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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#?

1 Answer
Feb 14, 2016

Here's what I got.


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.