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Al Koholic
April 6th, 2003, 11:42 PM
Ok...this may sound a little weird at first and I may be off track totally here but bear with me.

By increasing the pressure of the environment in which a detonation/explosion takes place, shouldn't this decrease the reaction rate (gas production) of the compound (which is a first order reaction) and shift the equilibrium position towards the reactant slightly? I know this effect if it exists would be unbelieveably slight and most of you may not care at all...but from a theoretical standpoint shouldn't this be true? Also, increasing the temperature of the environment in which the explosion takes place ought to have a slight effect as well if one treats heat as a product of the reaction. Simple LeChatliers here.

Now that I've posed this question (which also has other variations that I need not discuss because I think the point has been made), I wonder what would be any of these potential effects? Would VoD be lower for instance? If you got thoughts or I'm just babbling again :) lemme know!

<small>[ April 06, 2003, 10:43 PM: Message edited by: Al Koholic ]</small>

Microtek
April 7th, 2003, 02:59 AM
I think you are right but as you said the effect of pressure on HE must be exceedingly slight. Temperature on the other hand does make an appreciable amount of difference which is why artillery crews must take the temp of the propellant into account when computing trajectories ( unless they can measure muzzle velocity directly, many modern units do this instead ).

Tuatara
April 7th, 2003, 04:47 AM
I see what you are driving at, and it is certainly true for deflagration processes (eg rocket fuel), but aren't detonations a shockwave process? This means the reaction rate would depend primarily on the velocity of sound in the explosive - density, temperature, rigidity.

Einsteinium
April 7th, 2003, 03:14 PM
LeChâtelier describes reversible processes, detonation ins't reversible: carbon dioxide, water, oxygen, nitrogen, nitrogen oxides etc... (common detonation products) dosen't reacts together producing back the explosives (that would be nice though <img border="0" title="" alt="[Wink]" src="wink.gif" /> ). Also, LeChâtelier dosen't care about time, the principe is time-independant thus speaking of reaction "rates" isn't correct if it's based on this principe.

However, we can predict that the detonation wouldn't be as favorized as it would be at a lower pressure but when the bomb actually explodes, the pressure dramasticly drops to atmospheric and the process then goes on normally. Another logic applies to temperature as it's known that temperature has an influence on explosions. This behavior is predicted by thermodynamics : at high temperature the entropic factor is predominant (since the amount of gas produced) and this favors the reaction. That is, the higher the temperature, the greater the explosion (this is easily understood as mostly all explosives detonates when heated under vacuum (without oxygen...)).

vulture
April 7th, 2003, 04:04 PM
Do not confuse kinetics with thermodynamics!
Le chatelier is thermodynamics and only affects the position of the equilibrium.
Explosive substances equilibrium is usually FAR to the right, hence they are explosive, so the effect will be limited.
Le chatelier has no effect on reaction speed.
Reaction speed only says something about the speed at which a reactant forms, but nothing about it's concentration compared to others (atleast not directly).

</font><blockquote><font size="1" face="Verdana, Arial, Helvetica">quote:</font><hr /><font size="2" face="Verdana, Arial, Helvetica">This means the reaction rate would depend primarily on the velocity of sound in the explosive - density, temperature, rigidity </font><hr /></blockquote><font size="2" face="Verdana, Arial, Helvetica">True. This is essentially the main difference between detonation and deflagration;
Deflagration = reaction propagation through heat conductance
Detonation = reaction propagation through shockwave transmission and breaking of bonds

Al Koholic
April 7th, 2003, 04:29 PM
I'm aware that the detonation isn't typically thought of as reversible in the typical sense of something like a salt dissociation in solution but there will still be a reaction quotient and equillibrium even though like vulture said...will be FAAAARRRRRRRR to the right hehe :) . What's been said is pretty much what I was thinking would be the case just making sure. It's verrrrrry interesting that temperature has such a large effect on the position!!! I had no idea that one would be required to re-calculate trajectories due to temp changes. I would have thought that both temp and pressures effects would have been very slight...

Interesting discussion going on at least!

<small>[ April 07, 2003, 03:30 PM: Message edited by: Al Koholic ]</small>

vulture
April 7th, 2003, 05:39 PM
Temperature has a HUGE effect on the position of the equilibrium!
I won't go into detail here, but there's an integration formula for calculating enthalpy at a different temperature than standard conditions.
To make it even more difficult, the heat capacity, which is a component of the integral, is usually a temperature dependent function too, which means things can get really nasty mathematically.

Also, the equilibrium constant is also function of the temperature, as is the Gibbs free energy and the entropy....

If you can actually understand my babbling, congrats! <img border="0" title="" alt="[Wink]" src="wink.gif" />

Einsteinium
April 7th, 2003, 08:07 PM
You shouldn't see equilibriums in all chemicals transformations. This is a reaction with equilibrium in most conditions is farrrrr to the right: NaOH + HCl &lt;-&gt; NaCl + H20

When a substance detonates, the change is irreversible as other products would be formed (instead of the original explosive) if the products of the detonation ever comes to reacts together. There are many simpler organic reactions which aren't reversible, for example acid-catalyzed hydrolysis of an ester is reversible but base-catalyzed hydrolysis isn't because the resulting acid further reacts with the base. That is, the final product cannot reverts back to the ester. You can't write a reaction quotient for this reaction, same logic applies in detonations, N<sub>2</sub> won't react with CO<sub>2</sub> and O<sub>2</sub> etc.. to produce back the original TNT.

Vulture: There are links between kinetics and thermodynamics! Look for Arrhenius equation and the activated complex theory. Arrhenius equations is thermodynamic since it contains elemental properties as the Gibb's energy (enthalpy and entropy..) but contains k, the kinetic constant. It implies that the reaction rate grows up proportionally with the temperature with any reactions.

<small>[ April 07, 2003, 07:11 PM: Message edited by: Einsteinium ]</small>

Al Koholic
April 7th, 2003, 10:14 PM
I suppose this is why we're having this conversation! :)

I am still failing to see why one cannot write a reaction quotient for a detonation. Even in a detonation you will have some concentration of reactant that varies with time and a proportional change in the concentration of the products over time. This should be all you need to write A --&gt; B + C [B][C]/[A] = Q

I think this should be applicable to detonations...lead azide for example. Pb(N3)2 --&gt; Pb + 3N2 [N2]^3[Pb]/[Pb(N3)2] = Q

The way I imagine lead azide detonating based on my current knowledge is an extremely rapid chemical reaction essentially. I have an idea that DURING the detonation there could be any number of things going on such as the presence of one or more transition states, lead azide molecules that reach the activation energy and don't proceed on to product but simply fall back to the reactant state, and the possibility that the mechanism of detonation allows slight decomposition of the explosive but a reversal of the process before it proceeds too far. Things like this are observed in every enzyme kinetic known. A certain number of activated enzyme/substrate complexs, no matter how high the molecular efficiency of the enzyme, will form and then even though reaction activation energy has been reached, fall back to the state of enzyme and substrate without forming product. I know this is a far cry from the complexity of a detonation but I can't warrent the fact that something similiar to this is not occuring in the detonating aether. Based on this model, I don't think it is statistically possible to ever achieve a completely, absolute 100% detonation where NO original explosive exists....not even a molecule. Perhaps this clarify's my admittedly self generated and most likely inaccurate personal interpretation of a detonating explosive but thats why I posted the topic...to see whatchyall think! :)

Marvin
April 8th, 2003, 08:40 AM
Technically all reactions have an equilibrium. In the case of base catalysed hydrolysis you dont prevent the reversability, you simply provide a second much favoured equilibrium that forms the salt, and pushes the primary reaction furthur to completion by the resulting low concentration of free acid.

There are links between thermodynamics and rate of reaction, as an example, the general rule of thumb at room temperature is an increase in temp by 10C will double the rate of any reaction, this follows from the thermal population of the exited transition states. If the explosive isnt confined, its doing no work, and all the energy remains, if its allowed to expand against the atmosphere, pressure is dropping, and thus concentration of reactants and products is dropping, its doing work against the air, temperature is dropping and as a result of both the rate of reaction should fall.

Detonation certainly is an equilibrium, but the reformation of TNT isnt an issue simply becuase its such a complicated molecule, and the energy difference is large. What is very much an issue however, is relative amounts of CO2, CO, O2, H2, NH3, CN, N2 etc produced in the reaction. The higher the temp of combustion, the less complete the combustion. LeChatlier applies here very well simply as a result of forcing energy into a reaction thats producing it.

Al Koholic
April 9th, 2003, 05:27 PM
Well said Marvin...you sure have a way with explaining things very clearly on this board.

What I would like to hear you're input on would be something I alluded to above. Do you think that the issue of "the reformation of TNT" could actually be a present issue during the detonation itself? I realize what you are saying about the relative abundancies of the products of the overall detonation but I am just trying to deduce what could be happening in the actual detonation aether. I consider it at least a possibility that there could in fact be TNT being partially decomposed only to immediately reform again during the detonation itself. I think it might even be possible for a number of molecules to exhibit this behavior due to what would be in all liklihood an extremely tansient event. I don't want to speculate with unnecessary detail here but what I am thinking is along the lines of a TNT molecule reaching an energy state at which it begins to lose (for examples sake) a hydrogen or something and then immediately recaptures that hydrogen or another one and goes back to being TNT for a very short time. I know that this really wouldn't matter one bit for the effect of the explosion because it is a transient state but I wonder what people here think of that?

Einsteinium
April 9th, 2003, 06:09 PM
This is true from a theoretical point of view; all particles/chemicals are in equilibriums with other particles/chemicals therefore can be described with an appropriate reaction quotien. However, thermodynamic isn't exact nor appropriate in many cases and detonation is one of them. Detonation involves simultaneous bond cleavage, there is a probability that all bonds thus broken immediatly recovers but if that happens, the results is the same as if there have been no detonation at all. If the bonds does recovers partially, new products are formed which decomposes ultimatly to elemental carbon, simple oxygen/nitrogen/carbon gases and water. When we take accounts of the amount of products formed (CO2, CO, O2, H2, NH3, CN, N2 etc) we are taking of other equilibriums as N2 + O2 &lt;-&gt; 2NO which cannot be coupled with the first detonation for obvious reason. Some of theses equilibriums would favors the detonation, some wouldn't. If you want to go further in theoretical arguments, I do know that there is an equilibrium between the air I breathe and the very noxious coumpound palytoxin or with TNT (all are farrrrrrrr in the favors of air) but I wouldn't start applying LeChâtelier directly here as it is nonsense IMO. In no way pressure/heat changes would start a gas-phase chemical synthesis of TNT by itself, as it would rather form something else. The probabilities are there though, it *could* happen, but I'd rather say that there is no equilibrium at all between TNT and air.

Al Koholic: I haven't thought about very simple detonating explosives such as Lead Azide, for which I do agree a reaction quotient is definitly appropriate, even in practice, because the fragments has a way to recombine into Lead Azide without necessarly doing it *immediatly* after the bonds has been broken or having one chance over the Avogadro's number, if you see what I mean.

Al Koholic
April 10th, 2003, 12:24 AM
Oh I completely see what you mean. That's the primary reason I chose something so simple as an example there. :)
At the end of the reaction it doesn't really matter either way because all I am thinking about here is the intermediate mechanism. I wonder if there is a way to study the actual mechanism of this reaction? Unfortunately there are probably 20 labs asking the same question and not coming up with answers either so we obviously can't come up with definitive answers here but this is fun stuff to think about.