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Fusionfire
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Deuterated chemical explosives
Hello all,
I am trying to think of a chemical explosive that has all the <sup>1</sup>H replaced by <sup>2</sup>H.
The following factors are of interest:
1) High mole % of H in comparison to other elements.
2) Easy to synthesise on a laboratory and industrial scale.
3) Minimal wastage of valuable <sup>2</sup>H, supplied as heavy water, during the conversion process to the explosive.
4) Chemically stable.
5) Relatively insensitive.
The best I could think of was an oxy-hydrogen explosive mixture. It would have a 67% mole percentage of H, can be produced by the electrolysis of
heavy water (in theory with very little loss of <sup>2</sup>H) and is chemically stable + insensitive when the reactants are kept apart.
The problem is that this would be in gas or liquid phase and a solid state energetic is preferable. I am also not sure how the VoD and sensitivity
vary with pressure. If you keep raising the pressure of a oxy-hydrogen mixture isothermally, does a spontaneous reaction occur at some point?
What about frozen hydrogen peroxide crystals intimately mixed with lithium hydride (all deuterated) and kept frozen?
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hissingnoise
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Before initiating a self-sustaining fusion reaction on the global scale, I think you should consider, for a while, the numerous questions that will
forever remain unanswered . . .
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Fusionfire
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Oh, do calm down 
A bit of armchair hypothesising never led to a nuclear armageddon!
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Bot0nist
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Sorry to sound ignorant, but could you all elaborate on the theoretical apocalyptic outcome of such an undertaking.
Me no understand.
U.T.F.S.E. and learn the joys of autodidacticism!
Don't judge each day only by the harvest you reap, but also by the seeds you sow.
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hissingnoise
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We're doomed! Doomed, I tell you! All doooommmed . . .
Och Aye!
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Fusionfire
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Well to initiate nuclear fusion you would need to subject light nuclei to very high temperatures, so that their kinetic energies exceeds the
electrostatic repulsion of the protons in their nuclei. At close enough distances the strong nuclear force takes over and you get nuclear fusion + a
large release of energy.
Apart from rather expensive research using pulsed laser arrays and magnetic confinement, the only way this has ever been achieved is by gravitational
confinement in stars and a fission stage.
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Fusionfire
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Yes yes ha ha that is all very funny, but now back to the original post... 
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Nicodem
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By dissolving ammonium nitrate in D2O and rotavaping you will get deuterated ammonium nitrate. Doing this twice or trice will give you >90%
deuterium enrichment. You can then add those few % of whatever needs to be added to make the thing detonable. Totally nonsensitive secondary explosive
and as easy as it gets using only cheap D2O.
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hissingnoise
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OK! Go ahead - take out the entire solar system while you're at it . . .
Who really cares?
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Fusionfire
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Quote: Originally posted by Nicodem  | | By dissolving ammonium nitrate in D2O and rotavaping you will get deuterated ammonium nitrate. Doing this twice or trice will give you >90%
deuterium enrichment. You can then add those few % of whatever needs to be added to make the thing detonable. Totally nonsensitive secondary explosive
and as easy as it gets using only cheap D2O. |
Haha very funny, dissolving ammonium nitrate in heavy water and then evaporating it.
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hissingnoise
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But on the plus side, if you electrolyse 'heavy water' and spark the product-mixture, you'll end up with the original substrate.
Its collection, though, just might pose problems! 
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Nicodem
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What is so funny? That is the common method for deuteration of exchangeable protons. And besides D2O is relatively cheap if you buy it from chemical
suppliers. It is massively used and massively produced. Si*ma sells it for less than a $ per g (99% deuterium rich!). Of course, if you want to make
it yourself it will cost you way more.
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Fusionfire
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So are you saying it is possible to replace protium with deuterium in AN by just adding and evaporating heavy water? I.E.
N<sup>1</sup>H<sub>4</sub>NO<sub>3</sub> + <sup>2</sup>H<sub>2</sub>O <--->
N<sup>2</sup>H<sub>4</sub>NO<sub>3</sub> + <sup>1</sup>H<sub>2</sub>O
Where does the energy to break the N-<sup>1</sup>H & O-<sup>2</sup>H bonds come from, to replace them with
N-<sup>2</sup>H bonds?
What determines the equilibrium position?
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Nicodem
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| Quote: Originally posted by Fusionfire | So are you saying it is possible to replace protium with deuterium in AN by just adding and evaporating heavy water? I.E.
N<sup>1</sup>H<sub>4</sub>NO<sub>3</sub> + <sup>2</sup>H<sub>2</sub>O <--->
N<sup>2</sup>H<sub>4</sub>NO<sub>3</sub> + <sup>1</sup>H<sub>2</sub>O
|
Of course! Have you never heard of H/D exchange? That's how exchangeable protons are routinely detected in NMR spectroscopy. You add a drop of D2O or
deuterated methanol to the sample solution in DMSO-d6 or CDCl3 and check which 1H NMR peaks decrease or disappear.
| Quote: | | Where does the energy to break the N-<sup>1</sup>H & O-<sup>2</sup>H bonds come from, to replace them with
N-<sup>2</sup>H bonds? |
The protons in ammonium ions are exchangeable (their pKa in water is 9.3) so that their exchange is nearly instantaneous. You don't need to break any
bond, they continuously break themselves when the ammonium ion is solvated in protic solvents (proton exchange is a dynamic equilibrium).
| Quote: | | What determines the equilibrium position? |
It essentially boils down to statistics as this is by far the most contributing factor, so you can calculate a rough estimate of the deuterium
enrichment from the amounts of NH4NO3 vs. D2O used and the number of process repeatings (if you have some basic mathematical skills you will easily
find out how to do this). The boiling point difference between D2O and HOD is too low to contribute anything and even if it would it would contribute
positively. Thermodynamic factors such as the solvation difference between the ammonium ion and its deuterated counterparts are surely observable, but
can also be largely ignored in practice. So at the end it is all about statistics.
Of course you can recover most of the D2O by distillation instead of rotavaping it away, but it will obviously be less deuterated than the starting
one. Yet, you can use it for the first run for a fresh batch of NH4NO3 to increase efficiency in the second process with the non-recovered D2O. The
rationality of such recycling depends on the amount of ND4NO3 you want to prepare. For small amounts it is not worth the fuss.
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sternman318
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| Quote: Originally posted by Fusionfire | So are you saying it is possible to replace protium with deuterium in AN by just adding and evaporating heavy water? I.E.
N<sup>1</sup>H<sub>4</sub>NO<sub>3</sub> + <sup>2</sup>H<sub>2</sub>O <--->
N<sup>2</sup>H<sub>4</sub>NO<sub>3</sub> + <sup>1</sup>H<sub>2</sub>O
Where does the energy to break the N-<sup>1</sup>H & O-<sup>2</sup>H bonds come from, to replace them with
N-<sup>2</sup>H bonds?
What determines the equilibrium position? |
Well this is merely a guess, but both water and ammonium dissasociate in solution
NH4+ + H2O <--> NH3 + H3O+
and
H20 + H20 <--> OH- + H3O+
Of course the dissasocation is small for both, which is why you might need to run the procedure a few times to ensure a high amount of deuteration.
However statistics favors the hydrogen-1 being replaced by the hydrogen-2, since there is large excess deuterium/ hydrogen-2.
EDIT: Nevermind, Nicodem has this
[Edited on 10-8-2011 by sternman318]
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The WiZard is In
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Sorry this is done-did every day and we are all still here.
Accession Number : ADA468050
Title : Role of Thermochemical Decomposition in Energetic
Material Initiation Sensitivity and Explosive Performance
Descriptive Note : Conference paper (preprint)
Corporate Author : AIR FORCE RESEARCH LAB EDWARDS AFB CA
Personal Author(s) : Shackelford, Scott A.
Handle / proxy Url : http://handle.dtic.mil/100.2/ADA468050
Report Date : 05 FEB 2007
Pagination or Media Count : 31
Abstract : Catastrophic initiation of an energetic material
consists of a complex, interactive, sequential train of
mechanistic mechanical, physical, and chemical processes
which occur over a finite time period and proceed from
macroscopic into sub-microscopic composition levels (bulk >
crystalline > molecular > atomic). Initiation results when these
processes proceed at a rate which generates sufficient energy
(heat) to reach a threshold stage within this finite time period.
Thus, the rate at which these mechanistic processes occur
defines initiation sensitivity and affects performance.
Thermochemical decomposition processes regulate the rate at
which heat energy is released at the molecular level, and
therefore to some extent, control energetic material initiation
sensitivity and performance characteristics. Kinetic deuterium
isotope effect (KDIE) data, obtained during ambient pressure
thermochemical decomposition process, identifies the
mechanistic rate-controlling bond rupture which ultimately
regulates the energy release rate of a given energetic material.
This same rate-controlling bond rupture also appears as a
significant rate-limiting feature in higher order deflagration,
combustion, and explosion phenomena. The effect the KDIE
-determined rate-controlling bond rupture exerts on initiation
sensitivity, and its potential influence in combustion and
explosion performance is delineated.
Accession Number : ADA121652
Title : Explosive-Driven Hemispherical Implosions for Generating Fusion Plasmas
Descriptive Note : Interim rept.
Corporate Author : TORONTO UNIV DOWNSVIEW (ONTARIO) INST FOR AEROSPACE STUDIES
Personal Author(s) : Sagie, D. ; Glass, I. I.
Handle / proxy Url : http://handle.dtic.mil/100.2/ADA121652
Report Date : MAR 1982
Pagination or Media Count : 104
Abstract : The UTIAS explosive-driven-implosion facility was
used to produce stable, centered and focused hemispherical
implosions to generate neutrons from D-D reactions. A high
resolution scintillator-detection system measured the neutrons
and gamma-rays resulting from the fusion of deuterium.
Several approaches were used to initiate fusion in deuterium.
The simpliest and most direct proved to be in predetonated
stoichiometric mixture of deuterium-oxygen. The other
successful method was miniature Voitenko-type compressor
where a plane diaphragm was driven by the implosion wave
into a secondary small spherical cavity that contained pure
deuterium gas at one atmosphere. A great deal of work still
remains in order to measure accurately the neutron flux and its
velocity distribution as well as the precise interactions of the
neutrons with the steel chamber which produced the gamma
-rays. Nevertheless, this is the only known work where fusion
neutrons were produced by chemical energy only in a direct
and indirect manner.
And a SL of other refs.
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Fusionfire
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Crikey you just smacked my head with a realisation of a physical phenomenon I had under my nose but took for granted! Sorry for a moment I thought you
were taking the mickey out of me
| Quote: Originally posted by Nicodem | | Quote: Originally posted by Fusionfire | So are you saying it is possible to replace protium with deuterium in AN by just adding and evaporating heavy water? I.E.
N<sup>1</sup>H<sub>4</sub>NO<sub>3</sub> + <sup>2</sup>H<sub>2</sub>O <--->
N<sup>2</sup>H<sub>4</sub>NO<sub>3</sub> + <sup>1</sup>H<sub>2</sub>O
|
Of course! Have you never heard of H/D exchange? That's how exchangeable protons are routinely detected in NMR spectroscopy. You add a drop of D2O or
deuterated methanol to the sample solution in DMSO-d6 or CDCl3 and check which 1H NMR peaks decrease or disappear. |
No, unfortunately. They didn't cover that in engineering.
| Quote: |
| Quote: | | Where does the energy to break the N-<sup>1</sup>H & O-<sup>2</sup>H bonds come from, to replace them with
N-<sup>2</sup>H bonds? |
The protons in ammonium ions are exchangeable (their pKa in water is 9.3) so that their exchange is nearly instantaneous. You don't need to break any
bond, they continuously break themselves when the ammonium ion is solvated in protic solvents (proton exchange is a dynamic equilibrium).
|
What happens if you add heavy water to an alkane, e.g. paraffin wax? Are the C-H bonds here so tightly bonded that you don't get proton exchange?
What if you use a proton exchange polymer like Nafion?
| Quote: |
| Quote: | | What determines the equilibrium position? |
It essentially boils down to statistics as this is by far the most contributing factor, so you can calculate a rough estimate of the deuterium
enrichment from the amounts of NH4NO3 vs. D2O used and the number of process repeatings (if you have some basic mathematical skills you will easily
find out how to do this). |
Since it seems I am thankfully in the company of experts, I'll risk looking like an idiot and take a dive in hope of getting a bit of education
outside my usual realm...
1 mole of NH4NO3 has 4 moles of H.
2 moles of D2O has 4 moles of D.
So if you mix 1 mole of NH4NO3 with 2 moles of D2O, assuming proton exchange to be unbiased*, you'd end up with 1 mole of H2O and 1 mole of D2O, and
NH2D2NO3. Probably the reaction will maximise entropy so you wouldn't end up with 1 mole of H2O and 1 mole of D2O, but 2 moles of DHO.
| Quote: |
Of course you can recover most of the D2O by distillation instead of rotavaping it away, but it will obviously be less deuterated than the starting
one. Yet, you can use it for the first run for a fresh batch of NH4NO3 to increase efficiency in the second process with the non-recovered D2O. The
rationality of such recycling depends on the amount of ND4NO3 you want to prepare. For small amounts it is not worth the fuss. |
Understood. It is kind of like trying to increase the thermodynamic efficiency of a heat engine by using the waste heat to increase the temperature of
the gas prior to combustion.
BTW would anyone like to comment on the energetic properties of a crystalline hydrogen peroxide - lithium hydride mixture?
*Wikipedia says:
| Quote: |
The lone electron pair on the nitrogen atom (N) in ammonia, represented as a pair of dots, forms the bond with a proton (H+). Thereafter, all four N-H
bonds are equivalent, being polar covalent bonds |
So I understand that all four N-H bonds can be replaced by N-D bonds with unbiased probability
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Fusionfire
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It seems though, that work on a chemical energy -> nuclear fusion energy device stopped in the late 20th century.
http://en.wikipedia.org/wiki/Pure_fusion_weapon
| Quote: |
Accession Number : ADA121652
Title : Explosive-Driven Hemispherical Implosions for Generating Fusion Plasmas
Descriptive Note : Interim rept.
Corporate Author : TORONTO UNIV DOWNSVIEW (ONTARIO) INST FOR AEROSPACE STUDIES
Personal Author(s) : Sagie, D. ; Glass, I. I.
Handle / proxy Url : http://handle.dtic.mil/100.2/ADA121652
Report Date : MAR 1982
Pagination or Media Count : 104
|
and
| Quote: |
Despite the many millions of dollars spent by the U.S. between 1952 and 1992 to produce a pure fusion weapon, no measurable success was ever achieved.
In 1998, the U.S. Department of Energy (DOE) released a restricted data declassification decision stating that even if the DOE made a substantial
investment in the past to develop a pure fusion weapon, "the U.S. Is not known to have and is not developing a pure fusion weapon and no credible
design for a pure fusion weapon resulted from the DOE investment". |
I'm not just thinking about a pure fusion weapon but, for example, imagine the enormous potential of a pulsed or continuous (chemical) detonation
engine that releases clean nuclear energy by fusion as well.
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niertap
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This is stupid and not going to happen. You do realize you have to have a nuclear bomb(fission), "spark plug", and other things to start a fusion
reaction. An explosive cannot just apply that heat and pressure. The hottest chemical reaction possible is around 4500k and fusion requires
~100,000,000k.
The only way is to make an explosive flux power generator and focus multiple explosions into 1.
Ignorance is bliss
Outliers in life are modeled by chemical kinetics
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Fusionfire
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| Quote: Originally posted by niertap | This is stupid and not going to happen. You do realize you have to have a nuclear bomb(fission), "spark plug", and other things to start a fusion
reaction. An explosive cannot just apply that heat and pressure. The hottest chemical reaction possible is around 4500k and fusion requires
~100,000,000k.
The only way is to make an explosive flux power generator and focus multiple explosions into 1.
|
I didn't say I could make it happen. It was just armchair speculation, which was why I tried to focus on the chemistry and not the engineering (as is
the remit of posts in this sub-forum).
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The WiZard is In
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| Quote: Originally posted by niertap | This is stupid and not going to happen. You do realize you have to have a nuclear bomb(fission), "spark plug", and other things to start a fusion
reaction. An explosive cannot just apply that heat and pressure. The hottest chemical reaction possible is around 4500k and fusion requires
~100,000,000k.
The only way is to make an explosive flux power generator and focus multiple explosions into 1.
|
Or Star Trek .... Matter - antimatter.
djh
----
Many scientist, most notable Carl
Sagan, have believed the sheer
number of environments in the
universe makes it possible that life
had developed elsewhere. But as
Clancy points out, "It's one thing to
believe that life might exist on other
planets, and quite another to believe
that it is secretly examining your
private parts."
Stuart Vyse's review of :—
Susan A Clancy
Abducted : How People Come to Believe
They Were Kidnapped By Aliens
Science 310 [5752] 1280-81
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Sickman
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| Quote: Originally posted by Fusionfire | Hello all,
I am trying to think of a chemical explosive that has all the <sup>1</sup>H replaced by <sup>2</sup>H.
The following factors are of interest:
1) High mole % of H in comparison to other elements.
2) Easy to synthesise on a laboratory and industrial scale.
3) Minimal wastage of valuable <sup>2</sup>H, supplied as heavy water, during the conversion process to the explosive.
4) Chemically stable.
5) Relatively insensitive.
The best I could think of was an oxy-hydrogen explosive mixture. It would have a 67% mole percentage of H, can be produced by the electrolysis of
heavy water (in theory with very little loss of <sup>2</sup>H) and is chemically stable + insensitive when the reactants are kept apart.
The problem is that this would be in gas or liquid phase and a solid state energetic is preferable. I am also not sure how the VoD and sensitivity
vary with pressure. If you keep raising the pressure of a oxy-hydrogen mixture isothermally, does a spontaneous reaction occur at some point?
What about frozen hydrogen peroxide crystals intimately mixed with lithium hydride (all deuterated) and kept frozen? |
Deuterated Ammonium picrate is likely going to be one of the better answers, and it will still have one light hydrogen if the picric acid hasn't been
deuterated as well.
The ND3 is easily made from Mg3N2 (magnesium nitride) and heavy water.
As far as it igniting a fusion reaction: maybe, when hell freezes over!
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Dr.Bob
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| Quote: Originally posted by Fusionfire |
What happens if you add heavy water to an alkane, e.g. paraffin wax? Are the C-H bonds here so tightly bonded that you don't get proton exchange?
What if you use a proton exchange polymer like Nafion?
BTW would anyone like to comment on the energetic properties of a crystalline hydrogen peroxide - lithium hydride mixture?
|
The C-H bonds in an alkane are too strong to break easily without some strong reagent or strong base. They do not exchange with solvent protons in
normal conditions or unless they are activated by certain substituents. The proton exchange resins/polymers will not affect that.
I can't imagine any practical way to make any mixture of LiH and H2O2 that would not react to generate H2 and LiOOH or LiOH and O2. Even frozen,
once a few molecules got close enough to react, they would generate an exotherm which would lead to melting and reaction/explosion. Please don't try
this at home...
Bob
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Fusionfire
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| Quote: Originally posted by Dr.Bob | | Quote: Originally posted by Fusionfire |
What happens if you add heavy water to an alkane, e.g. paraffin wax? Are the C-H bonds here so tightly bonded that you don't get proton exchange?
What if you use a proton exchange polymer like Nafion?
BTW would anyone like to comment on the energetic properties of a crystalline hydrogen peroxide - lithium hydride mixture?
|
The C-H bonds in an alkane are too strong to break easily without some strong reagent or strong base. They do not exchange with solvent protons in
normal conditions or unless they are activated by certain substituents. The proton exchange resins/polymers will not affect that.
|
Presumably as the C-C bond energy (346 kJ/mol) is less than the C-H one (411 kJ/mol) if you heat up an alkane in the presence of deuterium it will
break up into shorter alkanes or alkenes or alkynes, and then the C-H bonds will get substituted for C-D. C=C bond energies are 602 kJ/mol and
C≡C are 835 kJ/mol. Please correct me if I am wrong but I guess this must be how deuterated methane is made.
http://www.caslab.com/Methane-D4_deuterated_methane_CAS_558-...
How would you go from here to get deuterated paraffin? (see below why you'd be interested in this...)
My reference to proton exchange polymers was separate. If you have a block of Nafion and soak it in heavy water, will you start to find that the H+
(technically sulphonic acid groups) start picking up deuterium so as to maximise the system's entropy?
| Quote: Originally posted by Dr.Bob |
I can't imagine any practical way to make any mixture of LiH and H2O2 that would not react to generate H2 and LiOOH or LiOH and O2. Even frozen,
once a few molecules got close enough to react, they would generate an exotherm which would lead to melting and reaction/explosion. Please don't try
this at home...
Bob |
You could mix the LiH with a long chain alkane (e.g. low MP wax, preferably also deuterated) and then cool/solidify it, so effectively you have a waxy
coating to passivate your LiH from runaway exothermic oxidation by the peroxide.
Blast your conc. H2O2 through an atomiser, freeze it to fine crystals and mix the H2O2 crystals with the passivated LiH.
Hence a high mole % deuterium chemical explosive, capable of acting as a fuel for nuclear fusion (both the deuterium and lithium). A slightest source
of external heat would set this off!
[Edited on 10-8-2011 by Fusionfire]
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Sickman
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| Quote: Originally posted by Fusionfire |
You could mix the LiH with a long chain alkane (e.g. low MP wax, preferably also deuterated) and then cool/solidify it, so effectively you have a waxy
coating to passivate your LiH from runaway exothermic oxidation by the peroxide.
Blast your conc. H2O2 through an atomiser, freeze it to fine crystals and mix the H2O2 crystals with the passivated LiH.
Hence a high mole % deuterium chemical explosive, capable of acting as a fuel for nuclear fusion (both the deuterium and lithium). A slightest source
of external heat would set this off!
[Edited on 10-8-2011 by Fusionfire] |
External heat would only achieve activation energy for a chemical reaction! Your hopes of fusion are kaput without a source of neutron bombardment.
Please don't clutter the forum with your "cold fusion mumbo jumbo"!
http://grooveshark.com/#/s/Tell+No+Tales/3mu0wU?src=5
[Edited on 10-8-2011 by Sickman]
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