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Author: Subject: Kolbe electrolysis -> ethane
overload
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[*] posted on 9-7-2011 at 10:44
Kolbe electrolysis -> ethane


Kolbe electrolysis or Kolbe reaction is an organic reaction named after Adolph Wilhelm Hermann Kolbe. The Kolbe reaction is formally a decarboxylative dimerisation and proceeds by a radical reaction mechanism. Generally, the reaction can be observed as:
R1COO− + R2COO− → R1-R2 + 2 CO2
If R1, R2 are different, then alkanes R1-R1 and R2-R2 are also formed.

As an example, electrolysis of acetic acid yields ethane and carbon dioxide:
CH3COOH → CH3COO− → CH3COO· → CH3· + CO2 2CH3· → CH3CH3

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A journal with pictures. (by woelen)

http://woelen.homescience.net/science/chem/exps/precision_el...

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(a post by not-important)

The Kolbe reaction is favored by high current densities and low temperatures, the non-Kolbe products are increased by addition of indifferent electrolytes such as KHCO3 (0.2 M) and Na2SO4 (1+ M).

Smooth electrodes of platinum, iridium, or rhodium give the best results, graphite electrodes give little to no Kolbe product. Most other common electrode materials, with the exception of vitreous carbon, also give poor yields.

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Bubbling the gas mixture (CO2 + C2H6 + ???) through sodium hydroxide solution should remove most of the CO2 without affecting the ethane.

[Edited on 9-7-2011 by overload]
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[*] posted on 9-7-2011 at 15:52


Do you have a point? You've taken material already available elsewhere on these forums and cut-n-pasted it without adding anything, as far as I can tell.
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[*] posted on 10-7-2011 at 13:18


The answer to your question can be found in the subject.
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[*] posted on 26-7-2011 at 17:21


There are many uses for ethane. Due to this fact I will continue to post information that I belive other people like myself may find useful.

-------------------------------------------
I gathered some information to share.
-------------------------------------------

m3 = 35.3146667 f3

ETHANE || Density 36.3305149 g/f3 || 1.282 kg/m3(15 C gas, 1 atm || 546.49 kg/m3(liquid, at b.p.,1 atm)

CO2 || Density 56.6 g/f3 1.562 g/mL (solid at 1 atm and −78.5 °C) || 0.770 g/mL (liquid at 56 atm and 20 °C) 1.977 g/L (gas at 1 atm and 0 °C)

ethane |C2H6 30.07 g/mol -1 H
Hydrogen |H 1.00794
Oxygen |O 15.9994
CO2 |CO2 43.9
Carbon |C 12.01

-------------------------------------------

I used the information I just shared to determine the contents of one cubic foot of gas produced by Kolbe electrolysis.

18.1 grams ethane
28.3 grams carbon dioxide

I now have an idea to share..

In my idea the gas is captured in a container filled with water. As the gas is bubbled into the container water is displaced. At the top of this container is a valve that is of such a size and shape that a balloon can be attached. Once the container is filled with gas a balloon is attached to the opening on the valve and the valve opened. The container is then pressed down into a bath of water causing pressure on the gas and the balloon. The balloon is filled and the valve closed.

Now the balloon is attached to a piece of thin glass tubing that has been submerged into a concentrated sodium hydroxide solution. (all of these steps will most likely contaminate the ethane with outside air but this is ok since it will be purer than what was originally produced). On a gas outlet tube is attached a smaller, easier to fill, baloon. This bubbler apparatus should be air tight before the ethane/CO2 in the balloon is passed through the sodium hydroxide solution. Once the gas has been bubbled through the Na solution a reducition in the amount of gas should be evident. The gas should be passed through the solution repeatedly until no more gas is consumed by the sodium hydroxide. You should now have a balloon full of very pure ethane gas.

One thing I haven't been able to figure out is the exact volume of the gasses. The reaction is supposed to produce ethane and CO2 in a 1:1 ratio but that doesn't mean the gases will take up the same amount of area. Input on this will be appreciated.
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[*] posted on 26-7-2011 at 18:44


Quote: Originally posted by overload  


One thing I haven't been able to figure out is the exact volume of the gasses. The reaction is supposed to produce ethane and CO2 in a 1:1 ratio but that doesn't mean the gases will take up the same amount of area. Input on this will be appreciated.


I thought the same molar amounts of gases took up the same amount of volume? That is, 1 mole of gas takes up 22.4 liters of space.




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[*] posted on 26-7-2011 at 21:22


Quote: Originally posted by redox  
Quote: Originally posted by overload  


One thing I haven't been able to figure out is the exact volume of the gasses. The reaction is supposed to produce ethane and CO2 in a 1:1 ratio but that doesn't mean the gases will take up the same amount of area. Input on this will be appreciated.


I thought the same molar amounts of gases took up the same amount of volume? That is, 1 mole of gas takes up 22.4 liters of space.


Molar Volume, Avogadro's Law

The molar volume is the volume occupied by one mole of a substance (chemical element or chemical compound) at a given temperature and pressure.

There are two standards, commonly used in schools:

STP (standard temperature and pressure) which is 0 ºC and 1 atmosphere.
RTP (room temperature and pressure) which is 20 ºC and 1 atmosphere.

Avogadro’s Law states that:

1 mole of every gas occupies the same volume, at the same temperature and pressure.

At STP (standard temperature and pressure), this volume is 22.4 liters

At RTP (room temperature and pressure), this volume is 24 dm3 (liters)

We can also say:

The molar volume of a gas is 22.4 liters at STP (standard temperature and pressure).

The molar volume of gas is 24 dm3 at RTP (room temperature and pressure).

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Cubic decimetre (dm3), a volume unit equivalent to a litre
1 cubic foot = 28.3168466 liters

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So at 20 deg C one mol of any gas will take up 24 liters of area. I feel dumb now.. lol

[Edited on 27-7-2011 by overload]
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[*] posted on 27-7-2011 at 00:35


There are many more valuable Kolbe electrolysis examples...like:
-mixed Kolbe electrolysis with NaNO2 to provide nitrocarbons (propanoic acid yields nitroethane aside with butane)
-electrolysis of O2N-CH2-CH2-CO2H (3-nitropropanoic acid) yields 1,4-dinitrobutane
-electrolysis of (CH3-O2C)2CH-NO2.NH3 (2-nitromalonic dimethyl ester, amonium salt (nitronate) yields (CH3-O2C)2C(NO2)-C(NO2)(CO2-CH3)2 (2,3-dinitro-2,3-dicarboxy-1,4-butandioic acid tetramethyl ester)
The later can be hydrolysed in dilute acid to get methanol and ((HO2C)2C(NO2)-)2 wich will decarboxylate into O2N-CH2-CH2-NO2 (dinitroethane)




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[*] posted on 27-7-2011 at 10:57


Quote: Originally posted by PHILOU Zrealone  
There are many more valuable Kolbe electrolysis examples...like:
-mixed Kolbe electrolysis with NaNO2 to provide nitrocarbons (propanoic acid yields nitroethane aside with butane)
-electrolysis of O2N-CH2-CH2-CO2H (3-nitropropanoic acid) yields 1,4-dinitrobutane
-electrolysis of (CH3-O2C)2CH-NO2.NH3 (2-nitromalonic dimethyl ester, amonium salt (nitronate) yields (CH3-O2C)2C(NO2)-C(NO2)(CO2-CH3)2 (2,3-dinitro-2,3-dicarboxy-1,4-butandioic acid tetramethyl ester)
The later can be hydrolysed in dilute acid to get methanol and ((HO2C)2C(NO2)-)2 wich will decarboxylate into O2N-CH2-CH2-NO2 (dinitroethane)


I was unaware of these variations. Would you happen to have a link where I can learn more?

[Edited on 27-7-2011 by overload]
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[*] posted on 28-7-2011 at 10:01


Another factor to consider is the partial pressure of water vapor when considering the volume of gas collected over water. This is well known to most, but the new chemists amongst us may not have realized it. A portion of the gas will actually be water vapor. The portion will correspond with the temperature, and will probably be at equilibrium with the water.

http://intro.chem.okstate.edu/1515sp01/database/vpwater.html

This chart shows the pressure at sea level, nominally 760mm of Hg. You would have to subtract this percentage Vp/760 x 100 to get the correct volume. I know I didn't explain it in detail, but those just starting will get an idea of the significant amount of space which is taken up by water vapor at equilibrium. At the boiling point of water (or any liquid) it's vapor pressure equals 760 mm Hg. Even at 20C the volume is over 2%, but due to the lower molecular weight of H2O (18) it is one of the lightest , by molecular weight, components of the air.
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[*] posted on 2-8-2011 at 10:43


Quote: Originally posted by overload  


I was unaware of these variations. Would you happen to have a link where I can learn more?


There is a good book into the library here on the forum about Electrolysis of Organic compounds...take a look at page 109 ;)

Also on organic synthesis website you will find the reference for dimerisation of nitroalkanoic carboxylate into dinitro hydrocarbon.

In book Organic Chemistry Barnes & Noble N°6 -College Outline Series (Ed. F. Degering) pages 343 and 344.
34. Crum-Brown and Walker Reaction:
Electrolysis of the monoester of succinic acid yields the diester of adipic acid (and CO2 and Sodium)
36.Miller-Hofer Electrolysis:
CH3-CH2-CO2K + KI --> CH3-CH2-I + CO2 + 2 K
CH3-CH2-CO2Na + NaNO2 --> CH3-CH2-NO2 + CO2 + 2 Na
37.Walker Reaction:
Electrolysis of potassium 3,3-diethoxypropanoate yields 1,1,4,4-tetraethoxybutane (thus diethyl acetal of potassium malonaldehydate turns into acetal of succinic aldehyde).

It is important that for the dimerisation the group to preserve and to dupplicate has to be distant of at least one CH2 from the -CO2H that reacts...
So for example Cl-CH2-CO2H will lead to nothing but hydrolysis and gazeous decomposition products while Cl-CH2-CH2-CO2H will lead to Cl-CH2-CH2-CH2-CH2-Cl...




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[*] posted on 13-8-2011 at 09:36


Quote: Originally posted by Mr. Wizard  
Another factor to consider is the partial pressure of water vapor when considering the volume of gas collected over water. This is well known to most, but the new chemists amongst us may not have realized it. A portion of the gas will actually be water vapor. The portion will correspond with the temperature, and will probably be at equilibrium with the water.

http://intro.chem.okstate.edu/1515sp01/database/vpwater.html

This chart shows the pressure at sea level, nominally 760mm of Hg. You would have to subtract this percentage Vp/760 x 100 to get the correct volume. I know I didn't explain it in detail, but those just starting will get an idea of the significant amount of space which is taken up by water vapor at equilibrium. At the boiling point of water (or any liquid) it's vapor pressure equals 760 mm Hg. Even at 20C the volume is over 2%, but due to the lower molecular weight of H2O (18) it is one of the lightest , by molecular weight, components of the air.


A good example of how water vapor can be removed from air is a desiccator box or bag. The box of bag is filled with a layer of a desiccant like calcium chloride (my fav) or a beaker with calcium chloride is inserted (a container to hold the calcium chloride while it absorbs the water). A compound that would normally take days to crystallize due to water content will now crystallize in one due to the constant leaching of water vapor from the air within the box or bag by the calcium chloride. These have other uses too like drying oily compounds that you do not want to heat up.

So after the step when the gas is passed through sodium hydroxide solution to purify the ethane further you need to pass it through a glass drying tube filled with a desiccant.
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