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| | From:  MikeKL5 (Original Message) | Sent: 6/17/2005 9:35 PM |
Hey Steve, I have a question for you that I should probably know the answer to, but I can't come up with a good explanation for this. Carbon dioxide and water are very low-energy, stable compounds. Why, even at high pressure, when carbon dioxide is dissolved in water, does carbonic acid (an unstable compound) form? What is the driving force behind this rxn? MikeKL5 |
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| | | From: ·Steve· | Sent: 6/18/2005 10:51 PM |
Hi Mike, how have you been? I am teaching a Chem I class this summer, and in our last lab, metathesis reactions, we just talked about the instability of H2CO3 when it forms. So, good timing!
The behavior of CO2 and water is rather complicated. First, we have an equilibrium of gaseous CO2 with loosely hydrated CO2 in solution:
CO2(g)  CO2(aq)
and then a small fraction of this dissolved CO2 can form H2CO3:
CO2(aq) + H2O(l)  H2CO3(aq) Further, a small fraction of this small amount of H2CO3 can dissociate, which perhaps we can say, "ties up", a small amount of the dissolved CO2: H2CO3(aq)  H+(aq) + HCO3-(aq)
The equilibrium constant for this dissociation, Ka1, is normally given as about 4.2 X 10-7, but this is not really correct because it assumes that all of the dissolved CO2 is in the form of H2CO3. It is a "fudged" value, in other words, but as such it still works if you are calculating, say, H+ concentration of a solution of some amount of CO2 dissolved in water, so it is convenient to use. The true value is about 2 X 10-4, more in keeping with theoretical expectations.
And to an almost insignificant extent, a tiny fraction of HCO3- ion dissociates as well:
HCO3-(aq)  H+(aq) + CO32-(aq)
So the explanation really comes down to the fact that CO2 and water can react with each other reversibly, and thus it is better to look at the process in terms of equilibrium. "Thermodynamically", formation of H2CO3 is unfavorable because, as you noted, CO2 and H2O are much more stable. From the standpoint of entropy change, the process is also unfavorable (DS will be negative). Nevertheless, since the reaction is reversible, then statistically some amount of CO2, H2O, and H2CO3 can be present at equilibrium. The dissociation of H2CO3 in solution may also favor its formation just a bit.
Steve
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| | | From: MikeKL5 | Sent: 6/20/2005 8:31 PM |
Hey Steve, Sorry that it took me so long to get back to you. We had just got back from vacation when I e-mailed you that question, and I had to spend the weekend doing yard/house work. :-) Your explanation makes sense. I would not have thought that product removal by loss of an acidic proton would be able to drive the formation of carbonic acid. I just figured that since the ionization reaction is usually so easily reversible in an acid that it wouldn't have been significant. Thanks for answering my question. I appreciate it. Have fun with you class this summer. I'm sure I'll be asking you more questions soon. Now that I'm back from vacation, and refreshed, I'm ready to get back to my research. Take care bud, MikeKL5 |
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| | | From: ·Steve· | Sent: 6/21/2005 4:44 AM |
Ahh, the chemistry of yard work!  >> I would not have thought that product removal by loss of an acidic proton would be able to drive the formation of carbonic acid. << Well, I don't think this is THE driving force, but it is perhaps a contributing factor! This site, http://www.wwnorton.com/chemistry/overview/ch15.htm, has a little Java applet that is helpful. Not anything new, but it shows how the equlibrium distribution of products and reactants depend on the value of DG of the reaction. Click on "Equilibrium and Thermodynamics Tutorial" near the bottom of the page, wait for it to load, and click on play a couple of times to get to the adjustable free energy profile. It lets you vary the G of the product side, showing the difference in the profile (note how the equlibrium arrow changes - cute!). In the formation of H2CO3 from CO2(aq) and H2O(l), we have a highly endergonic reaction with a large, positive DG. In the energy profile, CO2(aq) and H2O(l) are low on the reactant side and H2CO3(aq) is high on the product side. The dissociation of H+ ion from a weak acid has a similar profile; thus the products, H+(aq) + A-(aq), are, as we know, not favored at equilibrium either. The explanation of how any H2CO3 forms lies in these profiles. There is enough thermal energy present for a small percentage of the molecules of CO2 and H2O to react and get over the activation hump to the H2CO3 side, but the majority of them don't make it over. Hope this helps, without raising too many more questions than it answers! Steve |
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