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.

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.

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Answer 2

Barium oxide does not react with sulfuric acid to yield ozone. The reaction between barium oxide and sulfuric acid typically produces barium sulfate and water. Ozone is not involved in this reaction.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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