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Corvettaholic

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Stability of compounds in general
« on: April 26, 2004, 02:04:39 PM »
So its known that some compounds are really stable, and some stuff explodes if look at it wrong. What determines how stable a compound is? There was talk of the ammonium triiodide, which explodes if you talk to it, as opposed to atmospheric nitrogen which typically never explodes. Is there an easy answer to the stability question, or is this something I'll figure out when I learn a lot more about chemistry?

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Re:Stability of compounds in general
« Reply #1 on: April 26, 2004, 02:14:01 PM »
solids that can make very stable gases are usually very energetic. NI3 probably makes N2 and I2 as byproducts of the reaction and that's why it goes off at the slightest touch. I don't know what the byproducts of the reaction actually are, but thats what my gut chem instinct is telling me.
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Corvettaholic

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Re:Stability of compounds in general
« Reply #2 on: April 26, 2004, 02:48:21 PM »
Makes sense to me, do you have any examples of a solid that doesn't make a stable gas? Just for my own info, if you can't think of any don't worry about it  :)

The only thing that pops into my head is stuff based on silicon, cause I think of rocks.

[edited to new thread: Iodine]
« Last Edit: April 26, 2004, 06:27:13 PM by hmx9123 »

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Re:Stability of compounds in general
« Reply #3 on: April 26, 2004, 05:39:00 PM »
Quote
jdurg wrote:
It's pretty easy to grasp why nitrogen triiodide is so unstable if you just look at the size of the atoms.  The atomic radius of nitrogen is only about 56 pm while that of iodine is greater than twice that size at 115 pm.  In the NI3 molecule, you have three "gigantic" iodine atoms trying to all bond to one fairly "small" nitrogen atom.  The electronic repulsion is quite large as the numerous electrons on each of the iodine atoms wants to be as far apart from each other as possible.  The polarity of water molecules and the adduct of an ammonia molecule quite possibly is able to stabilize the charge on the iodine atoms thus keeping them in place.  (Which would explain why NI3 is stable when wet, though it can still detonate if struck properly).  When it dries, those polar molecules have moved away from the nitrogen triiodide thus destablizing the iodine atoms.  The natural repulsion kicks in and the slightest movement causes the molecule to rip apart.  This is what I believe happens as it does seem to make sense.  (Though nobody has really done any in-depth studies on NI3 since there really hasn't been a need to).

I remember reading that not much is know of NI3 at all, since its so explosive. I don't even think there is a crystal structure of it.
« Last Edit: April 26, 2004, 06:29:13 PM by hmx9123 »
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Re:Stability of compounds in general
« Reply #4 on: April 26, 2004, 06:40:57 PM »
Probably the biggest contributing factor to the stability of a molecule is the strength of the weakest bond.  Peroxides are notoriously unstable, but that's because the O-O bond is so weak.  Gasses as a product can sometimes be used as a guide to stability, but it doesn't always correllate.  The problem with this is two-fold:

1. Many very stable explosives (like TNT, TATB, etc.) form gasses upon decompositon (explosive or thermal).  If you just looked at these and said "They're unstable because they produce lots of gasses" you wouldn't be correct.

2. A few strange explosives (like cuprous acetylide) are very unstable but do not produce gasses upon explosion.  This debunks the idea that things that do not form gasses upon decomposition are stable.

Generally, all it takes is the weakest bond to break in an unstable molecule, and then the entire thing goes.  This is, of course, talking about explosively unstable molecules.  This generally means that the molecule has a high amount of energy to begin with.

If you want to talk about other highly unstable but non-explosive molecules, you can look at a lot of Pt and Pd catalysts, and other weird stuff like certain boranes and borohydrides.  There's a ton of highly unstable organometallic molecules that simply 'die' as we say in chemistry.  They become unreactive and turn into junk.  This again usually comes from them being reactive because of weak bonds or the small amount of energy needed to promote some electrons to an excited level where they can rearrange the molecule or some such.

And, for nitrogen triiodide, no one really knows the structure.  It's debated whether it's nitrogen triiodide or ammonium triiodide.
« Last Edit: April 26, 2004, 06:41:56 PM by hmx9123 »

Corvettaholic

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Re:Stability of compounds in general
« Reply #5 on: April 26, 2004, 06:50:06 PM »
So if its unstable, and non-explosive, then when it gets "set off" it just turns into something else right? Any good experiments I can try with lead, cause palladium is waaaay too expensive.

You mentioned in peroxides that the O-O bond is very weak. Maybe I'm not understanding this right, but I thought oxygen bonds with itself fairly well as in atmospheric oxygen. Or is it because since there's hydrogen involved, the oxygen doesn't form the double bond cause there's a hydrogen atom in the way, and therefore its really easy to bust the whole thing apart.

So to sum up my understanding of this subject, the weaker the weakest bond in a molecule, the more likely it is that the whole molecule will fall to pieces (explosively or not) and become something more stable (like maybe its components in elemental form, a couple compounds, whatever). Sound good to you guys?

Anyone have a good link where I can read about to how to figure out bond strength and stuff like that? If not, thats cool, I'll see if I can fish out my chemistry book from my shed (I wonder if its still in there?)  ::)

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Re:Stability of compounds in general
« Reply #6 on: April 27, 2004, 08:21:35 AM »
the strength of a chain is only as strong as its weakest link
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

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« Last Edit: April 27, 2004, 08:28:17 AM by AWK »
AWK

Offline hmx9123

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Re:Stability of compounds in general
« Reply #8 on: April 27, 2004, 07:48:33 PM »
AWK, that JChemEd article is fantastic!  Thanks!  I never knew about that study done on nitrogen triiodide.  I just read about it not being known in some books (including the Greenwood and Earnshaw) and figured it hadn't been discovered yet.

As for bond dissociation energies, I've compiled a table from several sources that I use for myself from time to time for simple calclulations.  Here it is:

Bond Energies:

Bond:   kJ/mol:   kcal/mol:
C-H       415        99
C-O       356        85
C-N       292        70
C-C       348        83
C-Br      276        66
C-I       238        57
C-Cl      338        81
C-F       484       116
C-S       259        62
C=C       607       145
C=N       619       148
C=O       724       173
C=C      833       199 (triple)
C=N       879       210

O-H       463       111
O-O       146        35
O-F       190        45
O-Cl      203        48
O-I       234        56

N-H       391        93
N-N       160        38
N-F       272        65
N-Cl      200        48
N-Br      243        58
N-O       201        48
N=O       607       145
N=N       418       100
N=N       941       225

F-H       563       134
F-F       154        37
F-Br      237        57

Cl-H      432       103
Cl-Cl     239        57
Cl-Br     218        52
Cl-S      253        60

Br-H      366        87
Br-Br     193        46
Br-S      218        52

I-H       299        71
I-I       149        36
I-Cl      208        50
I-Br      175        42

H-H       436       104

S-H       338        81
S-F       327        78
S-S       266        64

Si-H      376         90
Si-Si      340          81
Si-C      360          86
Si-O      452        108
Si-F       543        130

Corvettaholic

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Re:Stability of compounds in general
« Reply #9 on: April 28, 2004, 11:42:41 AM »
Hmx: major kudos for that table!

To make sure I'm reading it right, lets take C-H for instance. Bond energy of 415 kJ/mol, which is a measure of how strong it is. That would be a lot stronger than O-O of 146 kJ/mol, meaning the O-O bond is a lot easier to break and therefore less stable (providing all other variables are the same). So lets pretend it would take a sharp whack with a hand to deliver enough energy (over 146kJ/mol) to break the O-O bond (we're pretending here), therefore you would have to huck a cinderblock at the C-H bond, because its almost 4 times as strong. Given my simple terms I use, does this sound right to you?

Corvettaholic

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Re:Stability of compounds in general
« Reply #10 on: April 28, 2004, 12:41:08 PM »
Got the perfect way of cementing this in my head! Here we go:

We have one molecule of H2O. There are two H-O bonds present, and each H-O bond's strength (is this called enthalpy?) is 463 kJ/mol. So if we have 1 mol of H2O, the total strength of the bonds would be 926 kJ/mol. Doesn't enthalpy have something to do with either of those numbers?

Now, lets say I wanted to blow this molecule into little pieces. I would have to apply energy exceeding ONE of the bonds, to seperate the hydrogen atom, then I would be left with a H-O molecule. That molecule would have an unpaired electron (courtesy oxygen) and therefore try to grab something else. Lets say I applied enough enough energy to exceed both bond strength's put together, then that would destroy the whole thing?

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Re:Stability of compounds in general
« Reply #11 on: April 28, 2004, 02:08:08 PM »
Glad you liked the table.  It's served me well over the years.

These terms are all enthalpy, and are used to calculate rough heats of formation.  Yes, your rough calculation for water is correct.

Your second question, about blowing the whole molecule apart, theoretically works, too.  Apply enough energy, and the whole thing breaks apart.

As for smacking the molecule with the O-O bond to break it, that really does happen and is why peroxides are so dangerous.  It takes a small blow to break the O-O bond, and this releases energy to the rest of the molecule, which in turn breaks other bonds, and pretty quickly, the whole thing is disintigrating and rearranging to give the explosive decompositon products (this is a pretty simplistic picture, but you get the idea).

Corvettaholic

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Re:Stability of compounds in general
« Reply #12 on: April 28, 2004, 03:06:30 PM »
A few more things about the table (I'm learning a lot about enthalpy now  ;D )

I'm looking at the C-C bond that is a triple bond. Table says the enthalpy value is 833 kJ/mol, is there per bond or for the entire triple bond? If its per bond, that is going to be TOUGH to break, which is why I guess diamonds are so strong. So if lets say you take the enthalphy value of a H-H bond, which is 436 kJ/mol. What that means is, it took 436 kJ worth of energy to get 1 mole of H and to bond with another mole of H, or is it 436 kJ of energy to get .5 mole of H to bond with .5 of more H?

Last thing, if you exceed the enthalpy of formation for any particular bond, why does it break?

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Re:Stability of compounds in general
« Reply #13 on: April 28, 2004, 03:41:18 PM »
I think I can answer your last question.  The energy of a bond is how much energy is released when the bond forms.  So when a C-C triple bond is formed, it releases 833 kJ/mol.  The release of energy is thermodynamically favorable, so atoms try to bond in ways that will release energy.  If a bond strength is 500 kJ/mol, that means that the energy of the two atoms bonded is 500 kJ less than when they are apart.  When the bond forms, 500 kJ/mol is given off.  If you supply 600 kJ of energy to a mol of those bonds, suddenly the two atoms being bound to each other isn't energetically favorable.  With that much energy being put into the system, the atoms are put into a more stable situation by being apart from each other.  So basically what you're doing when you apply energy to a bond is you're just changing what is the most thermondynamically favorable situation for the atoms involved in the bond.  This is where entropy, enthalpy and Gibbs Free Energy really starts to work its wonders.  ;)


(And please feel free to correct me if I'm incorrect.  It's been a few years since I last had a thermodynamics course).
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Corvettaholic

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Re:Stability of compounds in general
« Reply #14 on: April 28, 2004, 04:48:54 PM »
So to expand on that a little bit, we have water which has a total enthalpy of formation of 926 kJ/mol. We have 1 mol of water. If I heat up water, it boils into steam, but the bonds are still intact because its still water. This means I didn't exceed 926 kJ of energy, but gave it enough to juice to do a state change. How does melting/boiling points factor into enthalpy? So if I apply 2000 kJ of energy, the bonds WILL break because its more thermodynamically favorable for the two hydrogen atoms and single oxygen atom to be by themselves. I maintain this 2000 kJ of energy, so it doesn't cool back down and go back to being water. Then I would have hydrogen and oxygen in seperate gaseous states.

So, applying energy to the bond replaces the energy the bond dissipated during formation, and it goes back to being normal until the added energy leaves the system (2nd law of thermo I think, I can't imagine a closed system). Once the bonds break, won't that constant energy input blow up the hydrogen gas? And why wouldn't the oxygen gas combine back with the hydrogen to make water again? Or would it?

And how much IS 2000 kJ anyway? Is that like dropping a bowling ball off my house, oxy-acetyline torch, nuclear weapon?
« Last Edit: April 28, 2004, 04:49:27 PM by Corvettaholic »

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