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Vertox
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Calculating the energy of thermite
Hey guys, I wanted to make 50g of Thermite (magnesium+Copperoxide) but i dont know how to calculate the Energy released. Could anyone help me?
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j_sum1
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Where is blogfast when you need him?
I recall a lovely table of thermodynamic data for thermites. You might look in the exotic thermites thread.
In the mean-time, your Mg-CuO mix is going to be very energetic. More than enough to boil off the copper. Expect to see a whoosh and a brown copper
cloud and nothing remaining behind.
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Keras
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Look at formation enthalpies. ∆H0 and stuff.
If my computation is sound, for an Al/Fe2O3 thermite, you should get about 850 kJ for each 250 g of the mix (roughly).
[Edited on 21-8-2018 by Keras]
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MrHomeScientist
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The usual copper thermite, copper oxide and aluminum, is energetic enough to almost be explosive. Substituting in magnesium will make it even more
energetic, so this could be rather dangerous. How do you plan on igniting it?
Another useful tidbit: 50g is about the minimum size for a reliable thermite. Anything smaller is difficult to ignite and it tends to fizzle out
before burning all the way through. You need a large enough charge to generate the heat to keep it going.
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Keras
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∆H0 for copper oxide is only -150 kJ/mol so no wonder. You get about 1.3 MJ/mol released, which is a lot.
[Edit: this sounds like more than TNT detonation, at 1 MJ/mol. Looks like I’m wrong, but I wonder. The difference is mainly the kinetics and the
fact that thermite doesn’t release gases.]
[Edited on 21-8-2018 by Keras]
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Vertox
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jesus christ 1.3mJ/mol? Thats a lot. Its for a youtube video btw. And Im going to ignite it with a spark so I can ignite it from like 10 meters away
and only have to press a button (if that works)
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Keras
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Quote: Originally posted by Vertox  | | jesus christ 1.3mJ/mol? Thats a lot. Its for a youtube video btw. And Im going to ignite it with a spark so I can ignite it from like 10 meters away
and only have to press a button (if that works) |
Preferably use a fuse. That gives you ample time to seek shelter.
I'm not quite sure why aluminium oxide has such a low enthalpy of formation (-1.5 MJ/mol). The bonds between Al and O are really strong, perhaps
because of the 3s2 3p1 electronic configuration? Boron and gallium III oxides also have very low enthalpies of formation.
In contrast, copper II oxide has a minimal enthalpy of formation, meaning the bonds between Cu and O are weak and easily broken. That's why the
reaction produces so much energy.
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Vertox
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the 1.3mJ/mol is for copperoxide + Magnesium right?
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Keras
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Right. Iron oxide/hydroxide's enthalpy is around 800 kJ/mol, so you have to use much more heat to break the oxide into iron metal and oxygen, with
only about 800 kJ/mol dissipated.
Chromium III oxide's enthalpy is 1100 kJ/mol, that's why chromium thermites just glow gently and don't fuss and spark as the iron ones.
[Edited on 21-8-2018 by Keras]
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Vertox
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I still dont really understand how to calculate it... Can somebody please give me a step by step "tutorial"
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j_sum1
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Well you have been pointed in the right direction with the ∆H0 values – standard heats of formation.
You add up the ∆H0values of your products and subtract the ∆H0 values of your reactants – in their balanced stoichiometric
ration of course. If the result is negative you have an exothermic reaction. If positive you have an endothermic reaction.
This gives an approximate, ballpark figure. To get a more precise result you need to add on two endothermic processes that occur: the
∆H0 figures that you might look up are for standard conditions: 25°C. Some of the energy goes into heating your products and some to
phase changes.
The formula for the enthalpy change on heating is ∆H=m cp ∆T. The problem here is you don't actually know the change in temperature.
This is dependent on the insulation and geometry of your reaction vessel. Also cp is not constant but this is a lesser concern.
The second process concerns phase changes: you need to look up the latent heat of melting and the latent heat of vaporisation for each product that
goes to the liquid or gaseous phase. The enthalpies associated with these age generally greater than those associated with the temperature change.
The remaining energy from the exotherm is available energy that goes into heating the environment or shooting spectacular sparky stuff in different
directions.
TBH, i have never bothered doing these calculations -- although I might if I was attempting to refine a thermite process (such as giving a booster to
silicon or titanium thermites). A quick look at the ∆H0 values has been sufficient for me to know what to expect. Knowledge of what
phase the products will be in comes second. And then experience gives me a bit of the kinetics which is a major factor in the behaviour of the
reaction.
Like I said, with Mg/CuO, expect fast and dramatic and stand well back. I would be further than 10 metres if there was more than 100g or so.
[edit] formatting
[Edited on 22-8-2018 by j_sum1]
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Dan Vizine
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I needed to prepare a document that concerned the redox reaction of Fe2O3 with sodium. Here's an excerpt. Just change the reductant to Al and balance
the reaction. Then, apply this logic (the balanced equation allows you to determine the coefficients in front of each term):
"The balanced equation for the reduction of iron(III) oxide by sodium metal is:
6Na(s) + Fe2O3(s) ---> 3Na2O(s) + 2 Fe(s)
This represents an idealized description of the reaction under consideration. It neglects the degree of hydration, effects of impurities and other
unknowns. As a first step, the standard molar Gibbs free energy change is examined to determine its magnitude and to confirm the expected negative
sign which indicates that the above will be a spontaneous process.
The standard molar Gibbs free energy change is defined by the difference in the total Gibbs free energy of the products and the reactants
∆G° rxn = Σ ∆fG° (products) - Σ ∆fG° (reactants)
and is calculated using tabulated values of free energies of formation, ∆fG°
∆fG° Na = 0 (by definition) ∆fG° Fe = 0 (by definition)
∆fG° Fe2O3 = – 742.2 kJ/mol ∆fG° Na2O = – 377.1 kJ/mol
∆G° rxn = 3 ∆fG°(Na2O) + 2∆fG°(Fe) – 6 ∆fG°(Na) – ∆fG°(Fe2O3)
= 3(– 377.1 kJ/mol) – (– 742.2 kJ/mol)
∆G° rxn = – 389.1 kJ/mol
Thus, if this system receives sufficient activation energy, the reaction can be expected to proceed exothermically to completion.
The amount of thermal energy released (on a molar basis) is also given by comparing properties of the reactants and products.
∆H° rxn = Σ ∆fH° (products) - Σ ∆fH° (reactants)
= 3 ∆fH°(Na2O) + 2∆fH°(Fe) – 6 ∆fH°(Na) – ∆fH°(Fe2O3)
= 3(– 416 kJ/mol) – (– 822.2 kJ/mol)
∆H° rxn = – 425.8 kJ/mol
Again, values of the standard enthalpies of formation were obtained from tabulated values.
∆fH° Na = 0 (by definition) ∆fH° Fe = 0 (by definition)
∆fH° Fe2O3 = – 822.2 kJ/mol ∆fH° Na2O = – 416 kJ/mol
For the sake of comparison, the standard Goldschmidt reaction using Al as the fuel metal releases almost twice as much heat on a molar basis, –
847.6 kJ/mol vs. – 425.8 kJ/mol."
[Edited on 8/22/2018 by Dan Vizine]
"All Your Children Are Poor Unfortunate Victims of Lies You Believe, a Plague Upon Your Ignorance that Keeps the Youth from the Truth They
Deserve"...F. Zappa
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Vertox
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oh wow that looks complicated... But thank you so much!
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Dan Vizine
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It isn't, really. It just looks imposing because of the symbols.
The balanced equation for the reduction of iron(III) oxide by Al metal is: 2Al + Fe2O3 ---> Al2O3 + 2 Fe
The amount of thermal energy released (on a molar basis) is: ∆H° rxn = Σ ∆fH° (products) - Σ ∆fH° (reactants) = ∆fH°(Al2O3) +
2∆fH°(Fe) – 2 ∆fH°(Al) – ∆fH°(Fe2O3) = –1675.7 – (–822.2) = – 853.5 kJ/mol
standard enthalpies of formation were obtained from tabulated values:
∆fH° Al = 0 (by definition)
∆fH° Fe = 0 (by definition)
∆fH° Fe2O3 = – 822.2 kJ/mol
∆fH° Al2O3 = – 1675.7 kJ/mol
It's not uncommon for tables of thermodynamic data to vary from one to the next.
That's why we got – 853.5 kJ/mol this time and – 847.6 kJ/mol above.
[Edited on 8/24/2018 by Dan Vizine]
"All Your Children Are Poor Unfortunate Victims of Lies You Believe, a Plague Upon Your Ignorance that Keeps the Youth from the Truth They
Deserve"...F. Zappa
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MrHomeScientist
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Also you should take into account the state of the products: in the above thermite, everything will be molten at the end so you should use the heats
of formation for the liquid phase if you can find them. I vaguely remember doing this a while ago; I don't think it made too much difference, but it's
something to be aware of.
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Dan Vizine
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| Quote: Originally posted by MrHomeScientist | | Also you should take into account the state of the products: in the above thermite, everything will be molten at the end so you should use the heats
of formation for the liquid phase if you can find them. I vaguely remember doing this a while ago; I don't think it made too much difference, but it's
something to be aware of. |
Yes, I realize that and removed the (s)'s in my stoichiometric equation in a tacit admission of my thermodynamic shortcomings. I then (conveniently)
rationalized this shortcut by reminding myself that a simple answer was desired.
[Edited on 8/24/2018 by Dan Vizine]
"All Your Children Are Poor Unfortunate Victims of Lies You Believe, a Plague Upon Your Ignorance that Keeps the Youth from the Truth They
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Dan Vizine
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Vertox,
I know this reference has been posted earlier, but you probably haven't seen it. This explores all sorts of interesting thermites.
Attachment: A Survey of Combustible Metals, Thermites, and Intermetallics .pdf (5.9MB)
This file has been downloaded 373 times
"All Your Children Are Poor Unfortunate Victims of Lies You Believe, a Plague Upon Your Ignorance that Keeps the Youth from the Truth They
Deserve"...F. Zappa
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Vertox
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oh thanks! gonna check it out
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AJKOER
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I think thermobaric explosions are pretty cool too, however, before even considering one, I'm going to need some help answering some potential
questions from my neighbors like:
> What was that massive fireball in your backyard?
> And exactly why is my house on fire?
> Can a thermobaric explosion leave a trail of charred vegetation and over cooked birds?
> Could I argue that thermobaric explosions are a good means to starting backfires when the neighborhood is threatened by a forest fire?
> Are you nuts?
Thanks in advance for the help.
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TheMrbunGee
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Don't forget the efficiency of reaction, which depends on surface area of reactants. Which means - what is particle size of Mg and CuO?
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AJKOER
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| Quote: Originally posted by TheMrbunGee | | Don't forget the efficiency of reaction, which depends on surface area of reactants. Which means - what is particle size of Mg and CuO?
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Well, actually in the case of a metal oxide precipitate, perhaps not just surface area, but also a process of ageing of the cupric oxide, which can
impact reactivity (see, for example, interesting comments at http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.659... ).
While I have performed an electrosynthesis (see http://www.sciencemadness.org/talk/viewthread.php?tid=84047#... ) that works well in cheaply and efficiently creating CuO, I will leave the ageing
technique to others to explore for an application where it is may be deemed material (and obvious as in the case of a thermite, like perhaps Fe/CuO,
as I suspect the Mg/CuO thermite is already too dangerously exothermic).
[Edited on 13-9-2018 by AJKOER]
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MrHomeScientist
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A thermite reaction is nothing at all like a "thermobaric explosion". They aren't even remotely similar. Where did you get that idea? Have you ever
even seen a thermite reaction before?
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AJKOER
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| Quote: Originally posted by MrHomeScientist | | A thermite reaction is nothing at all like a "thermobaric explosion". They aren't even remotely similar. Where did you get that idea? Have you ever
even seen a thermite reaction before? |
No, of course not related, but relatedly cool! See these amazingly scary pictures of thermites at https://www.google.com/imgres?imgurl=https://upload.wikimedi... .
And my implied point still remains, the overly dangerously Mg/CuO thermite (likely explosive by the way) is perhaps not quite as nuts (or energetic)
as experimenting with a thermobaric explosive.
Obivously, my attempt at chemical hyperbole to reinstate sanity (note my last question, 'Are you nuts?') was not totally communicated to all. Perhaps
taking a closer look at the set of pictures (all based, I suspect, on less energetic thermites) may help.
[Edited on 13-9-2018 by AJKOER]
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MrHomeScientist
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Be careful just googling "thermite" and trusting that every picture returned is correct. The first two from that link certainly are thermites, the
third maybe is, the fourth looks like a missile strike or rocket explosion (from labanese-forces.com), the fifth is just fire, the sixth is "A large
explosion of confiscated mortar rounds, grenades, guns and other explosive devices", and the seventh is just a camp fire. My skepticism that you've
ever seen a thermite yourself increases...
I've done many, many thermite reactions in several different formulations. I do them all the time for chemistry demonstrations. The risk is very
minimal if you set it up properly (no moisture, delayed ignition method, audience sufficiently far away, etc.).
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AJKOER
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MrHomeScientist:
Yes, no upfront contact with a thermite, as I haven't seen much practical use, as do not perform public displays of energetic materials in general.
My last words of warning for those into energetic materials, one is not likely ever completely prepared for surprises!
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