Chemical Forums
Chemistry Forums for Students => Inorganic Chemistry Forum => Topic started by: ZOB on November 19, 2007, 10:30:47 AM
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Hello, this is my first post with regards to a problem I have. I am currently attempting to synthesise (legally) hydroiodic acid from elemental iodine and hydrazine. The Chemistry in its basic form says:
2I2 + N2H4 = 4HI + N2
I synthesise the supposed HI and get a clear yellow solution which i then distil off at 126 deg C. However I then get upon cooling large crystals of NH4I. Does any one know why? All the books and research papers I have looked at say that HI is formed between the pH's 0.5 - 8. I am now at a quandry.
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Why exactly do you need HI? It's commonly used in illicit drug manufacture.
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I am making high purity iodides LiI, FeI2 and BaI and I want to make HI from elemental iodine to keep the costs down instead of buying in HI which as you probably know costs a small fortune. I have a licence to buy HI on the open market but from a business perspective I will make substantial savings if I make my own HI.
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I'm a little skeptical about your interest in preparing your own HI, but I don't think my answer is going to help you if you plan to use it for illicit purposes. I answer only to prevent you from blowing your or someone else's head off.
If you really do have a license to purchase HI and the capability and training to make it safely and properly, you should have ample access to resources sufficient to sort this out. I recommend that you DO sort it out before you make another attempt to prepare it.
Your reaction is producing ammonia. Ammonia + HI ---> NH4I. The ammonia comes from the disproportionation of hydrazine - 3 moles of hydrazine produce 1 mole of N2 and 4 moles of NH3.
This reaction is often explosive. The prospect of being showered in scalding hot HI should deter you from trying again.
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Thanks for the replies. Firstly you are right to be skeptical about mme producing HI and I respect your vigilance. However, if you have access to my e-mail you will see that it is a company e-mail address. I am investigating methods of making HI with the full blessing of my managers for the reasons I have stated. Furthermore it did I did manage to work the problem out for myself as you said.
HI + N2H4 = N2H5+I-
2N2H5+I- = NH3 + NH4I + HI
the greater the HI concentration the greater the decomposition.
By the way I am using iodine and Red Phosphorous now if this proves to be unworkable or too expensive I will be wheeling out the electrochemical cell.
thanks for the help
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You looking for anhydrous HI or aqueous?
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By the way I am using iodine and Red Phosphorous now if this proves to be unworkable or too expensive I will be wheeling out the electrochemical cell.
thanks for the help
When you say "iodine and red phosphorus" are you referring to these reactions?
2P + 3I2 --> 2PI3
PI3 + 3H2O --> 3HBr + H3PO3
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I'm assuming you mean HI and not HBr?
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Just use P + I2 + H2O to produce HI
N2H4 is carcinogenic
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I know but we did use about 100MT this year on other processes so we do know how to handle the material. By the way I2 + Red P looks to have worked. Just need to work out the economics.
thanks again
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There are other, cheaper ways to make HI, which is why I asked if you needed anhydrous or aqueous.
BTW, hydrazine is much more of an acute toxin than a carcinogen.