# Sulfur-free black powder equation and burning temperature?

1. 6 KNO3 + C8H4O --> 3 K2CO3 + 8 CO + 4 H2O + 3 N2

Oxygen is unbalanced.

2. 6 KNO3 + C7H8O --> 3 K2CO3 + 4 CO2 + 2 H2O + 3 N2

Hydrogen is unbalanced.

3. 6 KNO3 + C7H4O --> 3 K2CO3 + CO2 + 6 CO + 2 H2O + 3 N2

Carbon is unbalanced.

None of these equations have 70:30 by weight saltpeter to charcoal, and none is balanced. To fix the weight: it is possible that the charcoal is not to be completely dry, and that there are a few water molecules hiding out, but there would have to be 8:1 water to charcoal molecules for equations 1 and 2, and 9:1 for equation 3. To maybe fix the balance and masses: if the charcoals had a coefficient of 2, equation 1 would need 1:2 water to charcoal, equation 2 would need 2:2, and 3 would need 3:2. This is assuming that the saltpeter is carrying no water.

Potential equations with 2 moles charcoal and carrying water to achieve the right weight ratios and balances are following:

(I offer two potential equations each because I lack a reactivity series table and know not how extra atoms would come together. Please alert me if the following equations are just B.S.)

4. a. 6 KNO3 + 2 C8H4O . H2O --> 3 K2CO3 + 3 N2 + 11 CO + 4 H2 + H2O + 2 C ?

4. b. 6 KNO3 + 2 C8H4O . H2O --> 3 K2CO3 + 3 N2 + 7 CO + 5 H2O + 6 C ?

Excess carbon would take oxygen from water, or stay carbon, or would the charcoal not be all used up? I assume that the reactant water molecules will remain unchanged.

5. a. 6 KNO3 + 2 C7H8O . 2H2O --> 3 K2CO3 + 3 N2 + 11 CO + 8 H2 + 2 H2O ?

5. b. 6 KNO3 + 2 C7H8O . 2H2O --> 3 K2CO3 + 3 N2 + 10 H2O + 3 CO + 8 C ?

Same problem as #4

6. a. 6 KNO3 + 2 C7H4O . 3H2O --> 3 K2CO3 + 3 N2 + 11 CO + 4 H2 + 3 H2O ?

6. b. 6 KNO3 + 2 C7H4O . 3H2O --> 3 K2CO3 + 3 N2 + 3 CO + 7 H2O + 8 C ?

Same problem as #4

If anyone can confirm any of these equations or provide a more accurate one, please do. Also, anyone who can tell me the burning temperature of the correct equation, please do.

I think you probably really want to test this, because fast heterogeneous reactions with radicals, temperature-curves, pressure-curves, thermal-transfers to cylinder-walls etc are rather difficult to model.

-What are you doing...?

L

These reactions are about as far from the ideal gas law as gases get.

You would do best to look for data that already exists, e.g. http://www.frfrogspad.com/intballi.htm

L

Professionals sometimes use PROPEP or GUIPEP or another modelling programme to predict results of propellant combustion.

http://en.wikipedia.org/wiki/Gunpowder#Sulfur-free_gunpowder

I think this equation

6 KNO3 + C7H4O --> 3 K2CO3 + 4 CO2 + 2 H2O + 3 N2

[7C,4H,6K,6N,19O]

is balanced.

And this one too:

4 KNO3 + C7H4O --> 2 K2CO3 + 5CO + 2 H2O + 2 N2

[7C,4H,4K,4N,13O]

And the mass ratios of grams KNO3 to KNO3 and charcoal together are:

606/(606+104) = 0.85 for the first reaction,

and 404/(404+104) = 0.80 for the second one,

with the main difference between those reactions being the first one has enough oxygen to make CO2, and the second one just enough to make CO.

I admit neither of those numbers is exactly equal to 0.70, but they're

kind ofclose. I think for this kind of reaction, there is going to be a lot of guesswork here anyway.I mean you might get more meaningful results from just mixing stuff together and lighting it on fire.

Which makes me think: you

areplaying with fire here. So you should take those precautions necessary to keep from blowing yourself up.Wearing safety glasses is probably a good idea.