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Endimion17
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Quote: Originally posted by Poppy  | Yea considering the density you would float as if you were not on earth... Useing a wooden boat could be dangerous, as if it breaks apart is no
safekeeping. Why not just sail aroud with a reinforced boat like those made of aluminium used for fishing? Dude, no need to scuba dive around duh
...you crazy?
I wud put that the dissolved salts may or may not contribute to the overral destructive power of the lake's water. Once I serendiptously mixed 98%
sulfuric, 53% nitric and 35% hydrochloric acids to some special proportion stainsless stel dissolved in it vigorously, without heating. So vigorous
the few mL of what had been prepared bubbled non stop until make its way out the reaction vessel. Seems some equilibrium allowed for the highest
corrosive state(s) of the component acid to be over-actived, something like a constinuously fed excited state.
So, taking 5% percent steps, how many solutions of both 3 acids would be required to probe the once prepared superacid formed, rediscovering it? about
20 tries?
Indeed, on the dissolution of the aluminium can inside the lake a lot of porrly described reactants could raise the very same strange smell of rotten
eggs from the reaction and the site doesn't support enough to tell how.
As for contribution of dissolved salts:
Ca(2+) + 2H2O --> Ca(OH)2 + 2H+ (pH raises)
2H+ + Fe + --> Fe(2+) + H2
Fe(2+) + H2O --> Fe(OH)2 + 2H+
Ca(OH)2 + 2H+ --> Ca(2+) + 2H2O (pH decreases)
In fact metallics cations add's a lot to the corrosive power of the solution, although it doesn't change its acidity.
Seems Al is sometimes contaminated too, its smells horribly whn etched
[Edited on 9-9-2012 by Poppy] |
I don't know the lake's density. It would be easy to calculate it, but I highly doubt it's extremely dense.
Who said anything about diving? I'm talking about swimming, while keeping my head above surface. If you smear yourself with grease and wear waterproof
swimming pants, the experience would be pretty much like fun in the Dead Sea, except for the different density.
A wooden boat, freshly greased with tar, would be a lot more resistant than any metallic one.
Metallic cations which hydrolize raise pH and some have catalytic properties in some reactions, but I sincerely doubt they add much reactivity to the
lake. Aluminium sulphate, mentioned in that "official" quote, hydrolizes readily.
If you want, you can do some math and calculate the molarity of salts and HCl in the lake. I'm sure you'll come up with very dilute solutions as I did
with sulphuric acid. Heck, if I was in my lab now, I'd make an artificial sample of the lake and try to taste a drop. Then I'd heat it up and submerge
aluminium, flower, iron. Just for fun.
Some aluminium alloys give off nasty smell when dissolved, yes.
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Poppy
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| Quote: Originally posted by blogfast25 | Ca2+ doesn't noticably hydrolyse. You can boil a CaCl2 solution till dryness and all you get is the CaCl2 hydrate.
[Edited on 9-9-2012 by blogfast25] |
True.
I did some calculations for the value of K for the reaction:
CaO + H2O --> Ca(OH)2
It's inf act really low.
But it seems uncertain, here is the review:
Enthalpy of formation, standard conditions:
CaO: -635.5kJ/mol
H2O: -285.8kJ/mol
Ca(OH)2: -63.7kJ/mol
Standard conditions Entropy:
CaO: 40 J/mol.K
H2O: 70 J/mol.K
Ca(OH)2: 76.1 J/mol.K
Reaction enthalpy:
-63.7 - (-635.5 -285,8) = 857.6 kJ/ mol (I)
Reaction entropy:
76,1 - (40 + 70) = -33.9 j/ mol.K (II)
Gibbs free energy (useing I and II)
G = H -TS (d stands for delta)
G = (I) -T.(II)
G = 857600 J/ mol - 298K( -33.9 J/molK)
G = (857600 + 10100)J/ mol
G = 867.7kJ/mol
Finally, calculating K:
K = [products]/[reagents] = e^(-G/RT)
K = e^(-342.2306)
K = 6.73x10^(-153)
Some problem occurred indeed. If its a very likely reaction to happen, K should be great, no that order of 10^(-153) bullshit!
[Edited on 9-9-2012 by Poppy]
By useing Wiki's table for stadard Gibbs free energy:
Ca(OH)2: -897.5 kJ/mol
CaO: -603.3 kJ/mol
H2O: -237,14 kJ/ mol
applaying to the final K formula gives
K = 9,36x10^9
Which totally disagress all of the above calculations. I give up this is total bullshit for me.
Could some expert ellucidate the mistake ??? 
Its similar to the problem faced here:
https://sites.google.com/a/lowellcardinals.org/enthalpy-entropy-and-spontaneity/Home/enthalpy/entropy/free-energy-and-spontaneity/connecting-it-all-ce
ll-potential-and-equilibrium
[Edited on 9-10-2012 by Poppy]
[Edited on 9-10-2012 by Poppy]
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blogfast25
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| Quote: Originally posted by Poppy |
True.
I did some calculations for the value of K for the reaction:
CaO + H2O --> Ca(OH)2
It's inf act really low.
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… Except of course that:
CaO(s) + H2O(l) → Ca(OH)2(s)
… has zilch, nada, nothing to do with:
Ca(2+)(aq) + 2 H2O(l) → Ca(OH)2(s) + 2 H+(aq) [sic]
Nor is this ‘total bullshit’: you just used a total bullshit value for HoF [Standard Enthalpy of Formation] of Ca(OH)2. It’s well known that
quicklime(CaO) reacts vigorously and exothermally with water, to form slaked lime (Ca(OH)2). Your second value for HoF of Ca(OH)2 is likely to be
correct. It helps to cite the source of your values.
General hint when trying this kind of calculations: the HoF of an solid ionic substance is only applicable when that product is present as a solid
ionic substance, NOT AS A SOLUTION IN WATER. For instance, when you neutralise an alkali with an acid you don’t obtain the HoF of the resulting
salt, only the neutralisation enthalpy of H3O+(aq) + OH-(aq) → 2 H2O(l). It’s those little suffixes (aq), (l), (s) that make all the
difference here!
[Edited on 10-9-2012 by blogfast25]
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Poppy
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| Quote: Originally posted by blogfast25 |
… Except of course that:
CaO(s) + H2O(l) → Ca(OH)2(s)
… has zilch, nada, nothing to do with:
Ca(2+)(aq) + 2 H2O(l) → Ca(OH)2(s) + 2 H+(aq) [sic]
[Edited on 10-9-2012 by blogfast25] |
I know it has nothing to do: If the Ca(OH)2 reaction is reversible, to some degree, back to CaO, then the acidification of Ca(H2O)6, would have, at
least, a similar equilibrium product (not mistaking K with Ksp here), probabily lower.
Both products are quite insoluble. The presence of water can come into things because it doesn't really counts as a solvent here.
| Quote: Originally posted by blogfast25 |
you just used a total bullshit value for HoF [Standard Enthalpy of Formation] of Ca(OH)2. It’s well known that quicklime(CaO) reacts vigorously and
exothermally with water, to form slaked lime (Ca(OH)2). Your second value for HoF of Ca(OH)2 is likely to be correct. It helps to cite the source of
your values.
[Edited on 10-9-2012 by blogfast25] |
Wiki says: Ca(OH)2 HoF source is Wiki
The only source I found for the value of it, which was apparently taken from a patent on solar heating system. So yes, its probably messed up.
Let us suppose the specif heat capacity of CaO to be 1, same as water. Putting that, 1 mol CaO plus 1 mol H2O sums to a mass of 74g. Applying the
formula Q = m.c.dT, those 63kJ would be enought to heat the ingredients for 200° more.
Conclusion: It was the heat of reaction all the way. The Gibbs free energy calculation involves only HoF.
My mistake.
Thanks for the insight Blogfast25,
With corrections the value for K goes to: K = 3,98x10^10
---------------------------------------
Then back to the logics, the reverse reaction of Ca(OH)2 to CaO is still more likely even than the self ionization of water!! 
Theorically, but I don't think all the surround water would allow for this, unless its a intrinsic property of the recluse HO-Ca-OH bonds. For that,
first, the acidification of the cation aquocomplexes must take place. Assuming its acidity is very low, the reaction may or not become unfeasible.
But that may be different in the neighborhood of a free metallic surface, where pH may suffer sudden raises.
[Edited on 9-10-2012 by Poppy]
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blogfast25
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Best source of Standard Enthalpies (and Entropies) of Formation is NIST Webbook, IMHO. It also lists Shomate Equations for Heat Capacity over wide
ranges of temperatures.
It lists a value of - 986.1 for HoF of Ca(OH)2.
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Poppy
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You have access to those? They're not free!
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arsphenamine
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NIST WebBook link (click here)
While spotty beyond the second row elements, it has served me freely and well
over the past two years in a rather single-minded search of Group 15 thermo properties.
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unionised
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It seems to have been some time since anyone mentioned anything volatile (apart from water) so I wonder what it may have to do with things that smell.
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Poppy
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Could you post a screen shot of that page you gather the information?
Really, some dumb head did the major favor of translating the page I guess they done a tremendous mistake by vanishing with the information.
First on clicking that link are just some information regarding the books publication, further clicking leads to official site. Where you grabbing the
data?
Please post that with printscreen I feel just like some1 should die for being so "helpful" (like the guy who translated the page)
| Quote: Originally posted by unionised | | It seems to have been some time since anyone mentioned anything volatile (apart from water) so I wonder what it may have to do with things that smell.
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Seems its going into the unlike probabilities of unconventional reactions..
[Edited on 9-10-2012 by Poppy]
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blogfast25
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Poppy:
On that page (http://webbook.nist.gov/chemistry/), choose ‘formula’, that takes you to: http://webbook.nist.gov/chemistry/form-ser.html. Type in formula, tick ‘condensed phase’ for HoF and other thermochemical data, then click
search.
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Poppy
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Yes!
Thank you !
IT WORKS HAHHHAAHH IT WORKSSS!!!!!!!!!
*shinny world destruction*
-------------------------------
Power, unlimited. 
Its so compelte, hope they don't start taxing for making that data public. Better download it all.
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SM2
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Well, I'm pretty sure when dissolving Fe tailings in crude muriatic acid, you will be getting all sorts of foul smelling carbonyls and sulphides, as
well as selenides, all in small amounts. They impart a somewhat nasty smell to your HCL/Fe. Hope I'm not too off here, but this is what I was
taught,.
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blogfast25
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For the hydrolysis of solvated cations there’s no real need to invoke HoFs etc. A generic solvated cation will act as a weak Bronsted acid according
to:
[M(H<sub>2</sub>O)<sub>n</sub>]<sup>m+</sup>(aq) + H<sub>2</sub>O(l) ↔
[M(H<sub>2</sub>O)<sub>n-1</sub>OH]<sup>(m-1)+</sup>(aq) + H<sub>3</sub>O<sup>+</sup>(aq)
with equilibrium constant K<sub>a1</sub> and possible second deprotonation with K<sub>a2</sub> etc.
In the case of dilute watery solutions of calcium salts (of strong acids), the solutions run close to neutral, so
[H<sub>3</sub>O<sup>+</sup>] ≈ 10<sup>-7</sup> and K<sub>a1</sub> <<< 1.
The smaller magnesium (II) cation has a smaller ionic radius, stronger central electrical field and thus more capable of expelling a proton. Mg salts
do show some mild hydrolysis.
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gutter_ca
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Slightly OT, but of interest: http://www.scienceblog.com/community/older/archives/E/usgs02...
An abandoned mine in northern CA (I used to do some analytical work on the drainage), that has the lowest recorded pH in the world: -3.6 in drippings.
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