# Balancing the reaction equation for the oxidation of lead(II) sulfide with ozone

While I was reading about the chemical reactions of ozone, I came across this reaction. $$\ce{PbS + 4O3 -> PbSO4 + 4O2}$$

Though the reaction seems well balanced, but when I attempted to balance it myself, I did this as $$\ce{PbS + 2O3 -> PbSO4 + O2}$$

I understand that there must be some reason that the 1st reaction is given in the book. I'm just not able to figure out why.

Can somebody give an explanation as to why is it so?

## 2 Answers

This is a semi-standard example of why 'naive' balancing does not work.

Ozone is a source of atomic oxygen, producing free oxygen molecules. So, in 'mild' conditions only one oxygen per ozone molecule would react as a strong oxidizer, and the remaining molecular oxygen would require elevated temperatures to react. Thus the equation would be

$\ce{PbS + 4O3 -> PbSO4 + 4 O2}$

Similarly, in excess of reducer at elevated temperature (though why would you want it with ozone is another matter) all atoms in ozone would react. Thus, the reaction should be

$\ce{3PbS + 4O3 -> 3PbSO4 }$

Because remember ozone always gives nascent oxygen, that O single atom, which is a very strong oxidizing agent. Your mistake here is that you are involving all 3 oxygen atoms in the reaction which isn't the case.