Balancing AN+NM => ???? formula

[Einstein]

I thought that someone could tell me how do I balance this reaction?

NH4NO3 + CH3NO2 => H2O + CO2 + NO!2

This is taken straight from the Ragnar Bensons C-4 video, and I don't know exactly, what that "!" thing is doing in the formula??? And PLEASE, IF someone has balanced this reaction, then he/she could write it here, ok? Thanks!

[mongo blongo]

I'm not sure if I believe you that you are Einstein. :D

[Mad Scientist]

You aren't going to get raw hydrogen gas as a product of detonating ANNM. You'll get carbon dioxide, carbon monoxide (if the oxygen balance for your composition is negative), water vapor, nitrogen, possibly slight traces of hydrocarbons (if your composition is very oxygen negative), and tiny amounts of nitrogen oxides (if the oxygen balance for your composition is positive). Even if you managed to detonate straight nitromethane, you won't get hydrogen gas as a product. Taking thermodynamics into consideration, I believe this is what would occur:

8CH3NO2 --> 10H2O + C2H4 + 6CO + 4N2

Edit:
And by the way, I'm hoping that people like Mick will see that it is pointless to try to flame my brains out for existing... :rolleyes:

<small>[ June 21, 2002, 02:16 PM: Message edited by: Mad Scientist ]</small>

[Mr Cool]

Mad Scientist: do you seriously believe that you're going to get significant amounts of ethene from detonating CH3NO2?
The approximate equation would be CH3NO2 --&gt; CO + H2O + 0.5 H2 + 0.5 N2.
You will get H2 from detonating a suitably oxygen defficient HE, C reduces H2O to get H2 and CO.
Another example: C6H2(NO2)3CH3 -~-&gt; 6 CO + 1.5 N2 + C + 2.5 H2 is the approximate detonation equation for TNT.

If you wanted a perfect OB for your ANNM, the approximate equation would be:
3 NH4NO3 + 2 CH3NO2 --&gt; 2 CO2 + 9 H2O + 4 N2
Meaning 66.3% AN, 33.7% NM.

[xoo1246]

NH4NO3 OB= +20
CH3NO2 OB = -39.3
(100-x) * 20 = x * 39.3 =&gt;

(CH3NO2) x= 33.7 %
(NH4NO3) 100 - x = 66.3 %

Yep, Mr. Cool you are right.

[DBSP]

A nother way to equal the OB would be to add some CH3OH. Thus you wouldn't need to use ad much CH3NO2.

[Mad Scientist]

8mol CH3NO2 --&gt; 10mol H2O + 1mol C2H4 + 6mol CO + 4mol N2 + 2376kJ

8mol CH3NO2 --&gt; 8mol CO + 8mol H2O + 4mol H2 + 4mol N2 + 2136kJ

The first decomposition reaction clearly is favorable.

Yes, some explosives do liberate hydrogen and pure carbon when they are detonated - aromatic compounds. Take trinitrotoluene, for example.

2C6H2(NO2)3CH3 --&gt; 9CO + 3H2O + 2C2H2 + C

Acetylene is not thermally stable; it decomposes into carbon and hydrogen.

9CO + 3H2O + 2C2H2 + C --&gt; 9CO + 3H2O + 5C + 2H2

[Mr Cool]

Hmmm... actually I think you're right, about the second reaction being more exothermic at least. I must confess that I did no calculations, I was just working with a general rule for dealing with detonations/combustions.
Although it is worth bearing in mind that it's probably not only thermodynamic favourability that decides... kinetic stability, reaction rates, Chatelier's (sp?) principles etc will play a part. I guess that's why there are so many different products in reality.

Could you please post the bond enthalpies that you used to work those figures out? I was going to work it out but it's pointless if you're using a different set of figures.

[Mad Scientist]

CC (single bond) 356Kj/mol
CC (double bond) 598Kj/mol
CC (triple bond) 813Kj/mol
CO (double bond - the one found in carbon monoxide) 1073Kj/mol
NN (triple bond) 946Kj/mol
OH (single bond) 467Kj/mol
CH (single bond) 416Kj/mol
HH (single bond) 436Kj/mol
CN (single bond) 285Kj/mol

The final one I had to calculate myself...

NO (double bond - the one found in nitric acid, nitrates, etceteras) 449Kj/mol

[kingspaz]

aren't all those bond enthalpies supposed to be negative?

[Pu239 Stuchtiger]

Too bad he can't contribute anymore.

I believe Mad Scientist has it right. :)

[Mr Cool]

Bond entahlpies are neither positive or negative, they're just magnitudes. This is because that is the amount of energy released when the bond forms, or it is the amount of energy needed to break the bond. So it will be positive or negative, depending on what it's being used for.

I wonder what figure he used for the N--&gt;O dative single bond...?

[kingspaz]

mr cool, that is correct. i've just been taught to work with the enthalpy of bond formation. don't dative bonds have exactly the same enthalpy as normal bonds since its the same thing as a normal bond just using electrons both from the same atom to fill the orbital?