Question

How could sulfur be oxidized to `SO_3` by the action of `NO_3^(-)`?

Answer 1

Sulfur is oxidized to sulfur trioxide. Nitric acid is reduced to nitric oxide, `NO_2`. Here, we do NOT use oxygen gas as a reactant, and use standard redox processes.

Explanation:

Oxidation:

`S +3H_2O rarr SO_3 + 6H^+ + 6e^-` `(i)`

Zerovalent sulfur is oxidized to `S(VI^+)`.

Reduction:

`NO_3^(-) + 2H^(+) + e^(-) rarr NO_2 + H_2O` `(ii)`

So, `6xx(ii) + (i)=`

`S + 6NO_3^(-) + 6H^(+)rarr SO_3 + 6NO_2 + 3H_2O`; alternatively:

`S + 6HNO_3 rarr SO_3 + 6NO_2 + 3H_2O`

This is balanced with respect to mass and charge. It would not be a feasible reaction in practical terms. Industrially, sulfur trioxide is produced from sulfur dioxide and oxygen directly with some form of supported catalysis. It must be an incredible dirty and smelly process.