[AWK]

Concentrated H2SO4 (over 40 %) reacts with KI according to reactions:
H2SO4 + 2KI = K2SO4 + 2HI
then H2SO4 oxidise HI to iodine
H2SO4 + 2HI = I2 + SO2 + 2H2O
When reaction cannot equilibrate one can obtain mixture of I2 and HI in different molar ratio depending on the reaction conditions, mixing, and so on.

Slighty more concentrated H2SO4 (~70 %) reacts analogously with KBr.

[woelen]

Concentrated H2SO4 (over 40 %) reacts with KI according to reactions:
H2SO4 + 2KI = K2SO4 + 2HI
then H2SO4 oxidise HI to iodine
H2SO4 + 2HI = I2 + SO2 + 2H2O
When reaction cannot equilibrate one can obtain mixture of I2 and HI in different molar ratio depending on the reaction conditions, mixing, and so on.

Slighty more concentrated H2SO4 (~70 %) reacts analogously with KBr.

With concentrated H2SO4 (>= 95%) and KI you get a brown and very dirty smoke/fume of HI/I2, SO2 and H2S. I did the experiment and it really stinks. The smell of H2S is overwhelming. The reason that the fume is not purple probably is due to fuming of HI in moist air, with the I2 dissolving in the little droplets and forming H(+) and brown I3(-).

Solid very impure I2 and yellow S remains in the slurry. On dilution, the reaction only partially is reversible. The HI/I2 and SO2 part indeed are reversible, the H2S and S part are not reversible. You'll end up with a milky liquid, containing solid S when a small amount of Na2SO3  is added. Without Na2SO3 you end up with a brown turbid liquid, containing I3(-) and solid S.

The difference with KBr is quite strong. With KBr the reaction is less dirty. With KBr you get red/brown Br2 vapor and fuming HBr, but there is no H2S and solid sulphur. On dilution a yellow liquid is obtained (more SO2 than Br2 escapes, so the yellow liquid can be explained), which on addition of a pinch of Na2SO3 becomes totally clear and colorless.

Wilco


PS: Please do not do this inside and if you do this, use very small amounts. As the OP already noticed, the fumes are not good at all for you!

[jdurg]

Well put woelen.  The iodide ion is VERY easily oxidized to elemental iodine, and in order for it to be oxidized, something else has to be reduced.  Hence you get elemental iodine and hydrogen sulfide.  (Comparitively speaking, I2 is a fairly weak oxidizer in comparison to the rest of the halogens and H2SO4, so the I- ion becomes a moderately good reducing agent compared to the sulfate/bisulfate ion and the other halides).  In reality, the vast majority of the product obtained from anhydrous KI and H2SO4 is elemental iodine or solutions of iodine.  Very little HI is produced from that reaction.  That's why concentrated phosphoric acid is used to produce HI on an industrial scale.  The bromide ion is not oxidized nearly as easily as the iodide ion, so the anlagous reaction with KBr and H2SO4 produces nearly pure HBr and very, very little bromine vapor.  With chloride salts, you just get pure HCl as the Cl- ion is fairly difficult to oxidize.

[woelen]

Well put woelen.  The iodide ion is VERY easily oxidized to elemental iodine, and in order for it to be oxidized, something else has to be reduced.  Hence you get elemental iodine and hydrogen sulfide.  (Comparitively speaking, I2 is a fairly weak oxidizer in comparison to the rest of the halogens and H2SO4, so the I- ion becomes a moderately good reducing agent compared to the sulfate/bisulfate ion and the other halides).  In reality, the vast majority of the product obtained from anhydrous KI and H2SO4 is elemental iodine or solutions of iodine.  Very little HI is produced from that reaction.  That's why concentrated phosphoric acid is used to produce HI on an industrial scale.

Indeed, very little HI and a lot of I2/H2S/S/SO2, all mixed. With other words, a lot of crap.

The bromide ion is not oxidized nearly as easily as the iodide ion, so the anlagous reaction with KBr and H2SO4 produces nearly pure HBr and very, very little bromine vapor.

Well, I do not totally agree. I did the experiment with KBr and you can easily see the color of bromine. I took a small spatula of KBr and added approximately 0.5 ml of concentrated H2SO4 and loosely stoppered the test tube. Within a minute or so, the whole test tube was filled with a clearly visible red/brown vapor of bromine. The major product indeed is HBr, but saying that only "very, very little bromine vapor" is produced is not exactly what is happening. Of course, it depends on your definition of "very, very little"  

With chloride salts, you just get pure HCl as the Cl- ion is fairly difficult to oxidize.

Yes, you get pure and colorless HCl.

Wilco