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Is this reaction possible?

by soothsayer
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soothsayer
#1
Feb14-14, 06:55 PM
P: 398
The equation is balanced, but would it actually work?

NaOCH3 + CO2 + H2O → NaHCO3 + CH4O

Or would the reaction simply ignore the water in the left hand side and go,

2NaOCH3 + CO2 → Na2CO3 + H2O + C2H4

Or perhaps a little of both? I'm just starting to teach myself Chemistry, so any insight would be great!
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Simon Bridge
#2
Feb14-14, 11:36 PM
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You'd have to do the energetics - but at room temp, reactions are often done in solution ... i.e. the water does not take part.

In general - not every balanced chemical equation will actually happen.
Big-Daddy
#3
Feb17-14, 11:49 AM
P: 343
Well, at least a little bit of both, in the sense that every correctly balanced chemical reaction has to go to equilibrium, but if one of your products is really energetically unstable, you could have an incredibly low equilibrium constant for that reaction, in which case you may want to treat that reaction as negligible - you would certainly not be able to say it is the "main reaction". The other issue is kinetics - if one of the products is really kinetically unstable, the production of that product will be really slow and the consumption of any of that product which is produced would be really fast.

CH4O looks to me like it would probably be both very energetically unstable and very kinetically unstable. So, you might get some of it - but that amount would be ridiculously small (I would think) and take a really long time to form. I don't know if C2H4 is that likely to come out of this either but it is probably more likely - so you would get more of it, but still very little, and it would still take a while to form though it would be far faster than the formation of CH4O.

AFAIK in chemistry we do not hear of C making more than 4 bonds and H rarely makes more than 1 - but without doing one of these we cannot have CH4O.

Borek
#4
Feb17-14, 12:30 PM
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Is this reaction possible?

Quote Quote by Big-Daddy View Post
CH4O looks to me like it would probably be both very energetically unstable and very kinetically unstable
Does CH3OH look better?
Yanick
#5
Feb17-14, 12:36 PM
P: 383
Quote Quote by soothsayer View Post
The equation is balanced, but would it actually work?

NaOCH3 + CO2 + H2O → NaHCO3 + CH4O
What you have written out here is an overall reaction which requires several steps and equilibria. Ignoring the carbon dioxide and just looking at the sodium methoxide and protons, you are describing the protonation of sodium methoxide which is very favorable. The pKa of methanol (a very weak acid) is about 16, so any kind of acid in solution (even the water) will lead to an appreciable amount of methanol.

For example, NaOCH3 + H+ → Na+ + CH3OH

would be favorable in most circumstances unless the pH of the solution is extremely high. Consider that if the pH is ~16 (the pKa of methanol) you will have a 1:1 ratio of methoxide:methanol.

The reaction you have written above is slightly more complicated because the acid would be carbonic acid (and even the bicarbonate because methoxide is such a strong base) which depends on the equilibrium between carbon dioxide in the gas phase, carbon dioxide dissolved in the solution, carbonic acid, bicarbonate and, finally, carbonate. This all really depends on the conditions but in theory (and the short answer) is, yes, it should happen.

Quote Quote by soothsayer View Post
Or would the reaction simply ignore the water in the left hand side and go,

2NaOCH3 + CO2 → Na2CO3 + H2O + C2H4

Or perhaps a little of both? I'm just starting to teach myself Chemistry, so any insight would be great!
This reaction is more complicated because you are essentially dimerizing (kind of, I don't really know the technical name for such a reaction) methanol/methoxide. My organic chemistry is rusty (and my text is at my folks' place because I don't have room in the tiny hovel that is my apartment) so I don't want to comment too much about this process but my gut tells me that this would not be the main product(s) of the system.

Quote Quote by Big-Daddy View Post
CH4O looks to me like it would probably be both very energetically unstable and very kinetically unstable. So, you might get some of it - but that amount would be ridiculously small (I would think) and take a really long time to form. I don't know if C2H4 is that likely to come out of this either but it is probably more likely - so you would get more of it, but still very little, and it would still take a while to form though it would be far faster than the formation of CH4O.

AFAIK in chemistry we do not hear of C making more than 4 bonds and H rarely makes more than 1 - but without doing one of these we cannot have CH4O.
CH4O is methanol, a pretty stable molecule. We keep it in large jugs in the flammables closet without having to worry too much about it disappearing. The notation (CH4O) can be confusing if you don't have a good "chemistry vocabulary" which is why I prefer to write it as CH3OH. The latter helps people "see" the connectivity a little bit better.
Big-Daddy
#6
Feb17-14, 02:17 PM
P: 343
Quote Quote by Borek View Post
Does CH3OH look better?
... that was dumb.

Quote Quote by Yanick View Post
CH4O is methanol, a pretty stable molecule. We keep it in large jugs in the flammables closet without having to worry too much about it disappearing. The notation (CH4O) can be confusing if you don't have a good "chemistry vocabulary" which is why I prefer to write it as CH3OH. The latter helps people "see" the connectivity a little bit better.
Yup, sorry. For some reason I visualized the C atom having more than 4 bonds, even though I've seen CH4O used for methanol various times.


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