Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers, and students in the field of chemistry. It only takes a minute to sign up.
Sign up to join this community
Will this reaction proceed under standard conditions? I don't really care if it is quite slow, as long as it takes less than a month to finish:
$$\ce{S8(s) + 24 H2O2(aq) -> 8 H2SO4(aq) + 16 H2O(l)}$$
My rationale for this happening is that the elemental sulfur can be oxidized by hydrogen peroxide to form sulfur dioxide, which is further oxidized by hydrogen peroxide to form sulfur trioxide. This then reacts with water to form sulfuric acid.
You will likedly find it faster to perform the first oxidation by burning the elemental sulphur in a little tinfoil boat to produce gaseous sulphur dioxide. then set up some aperatus sucking the gasses produced to bubble up through the hydrogen peroxide to produce sulphuric acid. The peroxide does not oxidize sulphur dioxide to sulphur trioxide because it forms sulphuric acid directly.
Does require some aperatus but will be much faster than waiting for the peroxide. Calculate the gibs free energy if you are not sure if a reaction will be spontaneous.
Actually, here is a modification of your procedure involving some advanced radical chemistry with sulfur (see Scheme 1 graphics and pages 6 to 7) and the use of a photocatalyst with sunlight and oxygen from air.
My scheme is to burn a mix of Aluminum foil and sulfur. To the reaction product add dilute H2O2 (usually acidic for stability) in excess to capture all generated H2S and create a suspension of sulfur in dilute H2O2. Decant to remove Al(OH)3, but keep the sulfur suspension. Add a photocatalyst (like TiO2 or perhaps even Selenium sulfide sourced from SelSun shampoo, see this source). Apply sunlight and pump in air. Unclear if periodic dosing of dilute H2O2 is even required.
Logic: Per the source paper, the one-electron reduction potential of H2S to create $\ce{.HS}$ requires a relatively strong one-electron oxidants such as hydroxyl radical (which is sourced from the action of the photocatalyst acting on H2O2). Some expected reactions assuming the photocatalyst has produced a solvated electron:
$\ce{H2O2 + e-(aq) -> •OH + OH-}$
$\ce{•OH + H2S -> H2O + •HS}$ (Source Page 7, Eq (9) )
$\ce{•HS/•S- + O2 -> •SO2- (+ H+)}$ (Source Page 7, Eq (7) )
$\ce{•SO2- + O2 -> SO2 + •O2-}$ (Source Page 7, Eq (8) )
$\ce{H+ + •O2- -> •HO2}$ (pH > 5)
$\ce{•HO2 + •HO2 -> H2O2 + O2}$
Or: $\ce{•HO2 + e-(aq) -> HO2-}$
and even water can apparently supply a needed H+ to form H2O2.
So in essence, an in situ formation of •HO2/H2O2 employing a starting solution of H2O2, colloidal sulfur, oxygen and a photocatalyst to eventually convert the colloidal sulfur to H2SO4.
Just came across a 2006 reference ('Spectrum of ·HS and its reactions with oxygen in aqueous solution) that appears to mirror the science I suggested above involving sulfur. To quote:
The 266 nm laser photolysis formation and chemical behavior of ·HS in aqueous solution have been studied by laser photolysis-transient absorption technique. The 266 nm laser photolysis of H2O2 leads to the formation of ·OH, which reacts with H2S to produce ·HS, while solutions of HS- can also be directly photolyzed with 266 nm laser pulses to produce ·HS radical. The transient UV-Vis absorption spectrum of ·HS has been attributed to 220-300 nm with a maximum absorption at 220 nm and a shoulder at 250-270 nm. The chemical behavior of ·HS is very active in oxygen saturated aqueous solution: ·HS will first react with oxygen to form ·SO2-, which further reacts with another oxygen molecule with the production of SO 2 and ·O2-. In acid solution, ·O2- immediately protonates to form ·HO 2
