Talk:Sulfur-iodine cycle

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[edit] Cycle

Where do you get the 2nd H2SO4 for the second reaction? —Preceding unsigned comment added by 71.111.17.150 (talk) 23:19, 24 December 2007 (UTC)

From the first reaction: the amount of molecules present in these reactions are on the order of 10²⁴.

--Sealicus

[edit] Separation

How is O₂ separated from SO₂, and HI separated from H₂SO₄? --JWB 23:14, 8 May 2007 (UTC)

• O2 and SO2 can be separated by their different boiling points, H2SO4 and HI to (at 120 degrees HI is gas, while H2SO4 is liquid).

Both separations have to happen in the presence of water, though. Isn't HI water soluble as well? --JWB 16:07, 24 May 2007 (UTC)

• Yes, HI is soluble in water, but it is very volatile, while H2SO4 isn't. Only the problem is that concentrated H2SO4 may react with HI, giving I2, SO2 and H2O (backward reaction). But many of industrial chemical processes are reversible (such as an NH3 production from N2 and H2), so it isn't major problem.Student BSMU 16:09, 25 May 2007 (UTC)

Hydrogen iodide#Preparation notes that HI can also be distilled from a solution of NaI or other alkali iodide in concentrated phosphoric acid (note that sulfuric acid will not work for acidifying iodides as it will oxidize the iodide to elemental iodine). Nevertheless, I've added your explanation to the article, and it can be further refined. Also, won't some SO₂ accompany the volatilized HI, and if so does it have bad effects (e.g. reacting with the HI or H) and how is it separated?

Also, Sulfur trioxide#Preparation says that SO₂ can be oxidized to SO₃ between 400 and 600 °C. I guess this is another reversible reaction? At least, it sounds like step 2 will only disassociate part of the H₂SO₄/SO₃ on each heat and condense cycle. --JWB 18:47, 25 May 2007 (UTC)