RDX using K process? - Archive File

[megalomania]

-A-
Frequent Poster
Posts: 100
From:
Registered: JAN 2001
posted 04-11-2001 03:29 AM
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I have made rdx using the usual method involving hexamine and nitric acid. However there are three things I don't like about this method:
1- The formation of formaldehide
2- Too much nitric acid used
3- Low yield (sometimes)

Another way is using acetic anhydride, that I don't have. So, I begin searching and heard something about the k process, involving ammonium nitrate, hexamine and nitric acid. The hexamine is reacted with ammonium nitrate and nitric acid to form hexamthylenedinitrate, and then rdx. However, I am not clear as how the proportions should be. According to a past post, the equation is:

C6H12N4 + 3HNO3 + 3NH4NO3-->2C3H6N6O6 + 6H2O + NH3

If I am right, that sould be 140 grams of hexamine with 189 grams of nitric acid and 240 grams of ammonium nitrate. But then, the ratios of the K process as posted by Uberchlor should be 1 part hexamethylenetetramine, 10 parts nitrate of ammonia and 18 parts of nitric acid. What are the correct weight ratios?????.

[inFinie]

You were correct
If the equation is:
C6H12N4 + 3HNO3 + 3NH4NO3-->2C3H6N6O6 + 6H2O + NH3

140gr + 189gr + 240gr -->222gr + 108gr + 17gr

All compound considered pure.

[Mr Cool]

That is in theory, assuming that the reaction doesn't stop when the concentration of nitric acid gets below about 85%, which it does.
So, from those amounts, 108g of water is produced. If you want the acid to be 85% at this point, then that water is the remaining 15%. So, we need an extra (85*108/15) = 612g of nitric acid, to ensure that the reaction goes as far to completion as possible. Meaning that for 140g (1 mole) of hexamine, you should use no less than 800g of pure nitric acid (about 13 moles). It'd probably be a good idea to use even more if you want the highest yields based on hexamine, since the reaction is inefficient at 85% acid concentration.
The ammonium nitrate should also be in excess to help the reaction along at a decent rate.
Therefore it corresponds quite nicely, though not exactly, with the ratios in the original post, of 1:10:18, HMTA:AN:NA (I'm assuming that's molar ratio, otherwise that really is a huge excess of acid!).