[ZOB]

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.

[hmx9123]

Why exactly do you need HI?  It's commonly used in illicit drug manufacture.

[ZOB]

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.

[Shaun]

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.

[ZOB]

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

[hmx9123]

You looking for anhydrous HI or aqueous?

[lost and found]

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

[hmx9123]

I'm assuming you mean HI and not HBr?

[puppy8800]

Just use P + I₂ + H₂O to produce HI
N₂H₄ is carcinogenic

[ZOB]

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

[hmx9123]

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.