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[*] posted on 27-1-2015 at 11:59
Does this reaction happen at all?


Hello,

I'd like to know if the following reaction happens (or if there are other preferred side reactions) and at which temperature.

2Na2CO3 + 8C + 3N2 -> 4NaCN + 6CO

Thank you
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hissingnoise
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[*] posted on 27-1-2015 at 12:43


No, is the short answer ─ as N2 bonds are extremely strong.

NaCN is prepared by reacting sodamide with carbon at high temp. or NaOH with HCN . . .

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[*] posted on 27-1-2015 at 14:06


But there's the cyanamide process:

From Fixation of Atmosperic Nitrogen Page 14.

Quote:

Calcium carbide is produced through the
reaction between lime and coke in an electric furnace. By
the interaction of calcium carbide and pure nitrogen at a
red heat, the nitrogen is fixed in the form of calcium cyanamide.

Sodium cyanide can be made from sodium carbonate and calcium cyanamide.
From Wikipedia:
Quote:

Calcium cyanamide was used to produce sodium cyanide by fusing with sodium carbonate, which was used in cyanide process in gold mining:

CaCN2 + Na2CO3 + 2C → 2 NaCN + CaO + 2CO


[Edited on 27-1-2015 by Molecular Manipulations]




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[*] posted on 27-1-2015 at 14:14


Quote: Originally posted by hissingnoise  
No, is the short answer ─ as N2 bonds are extremely strong.

Energy of bonding in CO is higher than in N2.
This:
2Na2CO3 + 8C + 3N2 -> 4NaCN + 6CO
cannot work becuse it violates mass conservation law.
Reaction of alkali carbonates + carbon + nitrogen at ~1000 C is old industrial process of cyanides preparation.

So - go (back ?) to school, instead of posting crap.




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[*] posted on 27-1-2015 at 14:35


Quote: Originally posted by kmno4  

2Na2CO3 + 8C + 3N2 -> 4NaCN + 6CO
cannot work becuse it violates mass conservation law.

Removing the acidic overtone, that means the equation is not balanced.

6N = 4N. error.

Out of interest, why did you choose this particular 'reaction' ?

Edit:

Attached is something i typed up for someone regarding Equation Balancing.


Attachment: Balancing Equations.pdf (44kB)
This file has been downloaded 517 times





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[*] posted on 27-1-2015 at 15:02


Quote: Originally posted by kmno4  
Quote: Originally posted by hissingnoise  
No, is the short answer ─ as N2 bonds are extremely strong.

Energy of bonding in CO is higher than in N2.


http://www.sciencemadness.org/talk/viewthread.php?tid=61264#...

Acc. the first link in the page above:

C=O: 749 kJ/mol
N triple N: 946 kJ/mol




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[*] posted on 27-1-2015 at 15:09


Quote: Originally posted by aga  

Attached is something i typed up for someone regarding Equation Balancing.


The 'algebraic route' is the way most of us do it, but short cutting the algebra in our heads. And it only really works for those who can handle simultaneous linear equations.

And a big boy like you having trouble with 'nasty fractions'? Go away! :D

[Edited on 27-1-2015 by blogfast25]




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[*] posted on 28-1-2015 at 06:09


Quote:
Reaction of alkali carbonates + carbon + nitrogen at ~1000 C is old industrial process of cyanides preparation.

Indeed, you're right ─ I didn't give enough thought to the question, but the temp. required makes it so difficult as to be of little but academic interest to the home chemist . . .

Quote:
So - go (back ?) to school, instead of posting crap.

So, but aren't you quite the snarky little smart-alec?


Quote:
Energy of bonding in CO is higher than in N2.

And stating that double bonds are stronger than triple bonds just might be the most "intelligent" thing ever said on this board, Sherlock?

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[*] posted on 28-1-2015 at 07:10


Quote: Originally posted by hissingnoise  
Indeed, you're right ─ I didn't give enough thought to the question, but the temp. required makes it so difficult as to be of little but academic interest to the home chemist . . .




How long is a piece of string? I'm constantly pushing the temperature envelope of my hobby laboratory. A 1000 C will soon be available to me. Even w/o advanced heating technology, charcoal fired furnaces reach that temperatures quite easily.

Leading nitrogen gas through a molten mixture of soda and an excess (?) of carbon heated on a brisk BBQ fire would not be outside the capability of quite a few here.


[Edited on 28-1-2015 by blogfast25]




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[*] posted on 28-1-2015 at 10:36


Quote: Originally posted by blogfast25  
And a big boy like you having trouble with 'nasty fractions'? Go away! :D

They ARE scary !

After a day with fractions i get all fractious, see halves everywhere and need to drink to the point of double vision simply to restore normality.




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[*] posted on 28-1-2015 at 12:45


Are you one of these people that has problems cutting a round cake into 7 equal portions?



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[*] posted on 28-1-2015 at 13:34


Quote: Originally posted by blogfast25  
Are you one of these people that has problems cutting a round cake into 7 equal portions?

One never has to worry about those problems if they divide the cake in eight pieces, and enjoying before showing up to the party :)
Why's everyone so angry? Gas prices are low.
Regardless, there are easier methods to the purchase of cyanides than their production.
But why the anger, everyone? I didn't think this forum was like that till recently. Maybe I'll go over to the *other* online chemistry forum... (The THREATS are flying!)




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[*] posted on 28-1-2015 at 14:24


Quote: Originally posted by The Volatile Chemist  
Why's everyone so angry? Gas prices are low.
.But why the anger, everyone? I didn't think this forum was like that till recently.


All I can see is one member stepping up his efforts at mind reading. Try reading tea leaves, much more effective!

[Edited on 28-1-2015 by blogfast25]




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[*] posted on 28-1-2015 at 14:45


Quote: Originally posted by blogfast25  
Quote: Originally posted by kmno4  
Quote: Originally posted by hissingnoise  
No, is the short answer ─ as N2 bonds are extremely strong.

Energy of bonding in CO is higher than in N2.

C=O: 749 kJ/mol
N triple N: 946 kJ/mol


Quote: Originally posted by hissingnoise  

Quote:
Energy of bonding in CO is higher than in N2.

And stating that double bonds are stronger than triple bonds just might be the most "intelligent" thing ever said on this board, Sherlock?


Molecule of carbon monoxide contains triply bonded atoms of carbon and oxygen. Its length is slightly longer ( by few pm) than bond in N2 molecule, but bonding energy is larger (by more than 100 kJ/mol).

Look in the mirror and seek a dunce in there, I am sure you find.....




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[*] posted on 28-1-2015 at 15:33


Quote: Originally posted by kmno4  

Molecule of carbon monoxide contains triply bonded atoms of carbon and oxygen. Its length is slightly longer ( by few pm) than bond in N2 molecule, but bonding energy is larger (by more than 100 kJ/mol).



You are indeed correct.




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[*] posted on 29-1-2015 at 08:43


Quote:
You are indeed correct.

True, blogfast ─ he beat us to the dang Wiki page . . .

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[*] posted on 29-1-2015 at 09:04


Quote: Originally posted by hissingnoise  
Quote:
You are indeed correct.

True, blogfast ─ he beat us to the dang Wiki page . . .



Funny how you just forget things though: I could have sworn to C=O... till I looked it up.




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[*] posted on 29-1-2015 at 09:25


Can we blame that troublesome O=C=O, do you think?

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[*] posted on 29-1-2015 at 09:36


I think I'll blame ketones/aldehydes.



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[*] posted on 29-1-2015 at 09:54


Fewer late-night sherries for me then . . .

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[*] posted on 29-1-2015 at 11:10


Quote: Originally posted by blogfast25  
Are you one of these people that has problems cutting a round cake into 7 equal portions?

Not at all.

Just cut it into 8 pieces and promptly eat one.

Job done.




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[*] posted on 29-1-2015 at 11:14


That's a fattening cheat. Behave.

How are you going to serve 7 people the same amount from 4 cakes, huh?




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[*] posted on 29-1-2015 at 14:08


Quote: Originally posted by aga  
Quote: Originally posted by blogfast25  
Are you one of these people that has problems cutting a round cake into 7 equal portions?

Not at all.

Just cut it into 8 pieces and promptly eat one.

Job done.

To quote "One never has to worry about those problems if they divide the cake in eight pieces, and enjoying before showing up to the party" from a past post on this thread.....
Sorry for trying to 'read minds', everyone, to quote blogfast.




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[*] posted on 1-2-2015 at 12:16


Thanks for your replies. I didn't expect to stir up such an animated discussion with my question :)
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[*] posted on 1-2-2015 at 16:37


Quote: Originally posted by math  
Thanks for your replies. I didn't expect to stir up such an animated discussion with my question :)

Everything's animated at the great Sciencemadness!




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