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Topic: How does the concentration affect the equilibrium state?  (Read 841 times)

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Offline jackwhacky

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How does the concentration affect the equilibrium state?
« on: February 11, 2021, 08:58:23 AM »
Hello. I'm reading in my textbook that reversible reactions can be affected by the concentration of both products and reactants. Though I'm having a bit of difficulty understanding this in the context of effervescent tablets that dissolve in:
1. Normal water
2. Carbonated water

Because according to my textbook, the reaction of dissolving the tablet in carbonated water should take a longer time because the water already has carbon dioxide and the system is already close to equilibrium.

I understand that these tablets contain citric acid (which gives off oxonium ions) and sodium bicarbonate. The reactions go like this:


1-  NaHCO3(s)  ::equil:: Na+(aq)+HCO3-(aq)   
2-  HCO3-(aq)+ H3O+(aq) ::equil:: H2CO3(aq)+H2O(l)
3-  H2CO3(aq)  ::equil:: CO2(g)+H2O(l)

The problem is that I'm trying to come up with an explanation to why it dissolves slower in the carbonated water but the closest thing that I came up with and that I'm not even sure is right is that: In carbonated water, when we open the bottle, the reaction #3 above goes to the right to build carbon dioxide and water, but when the effervescent tablet is added, the reaction #3 above goes slower because there is already carbon dioxide in the carbonated water and the system is close to equilibrium, therefore the effervescent tablet doesn't break up as easily because the reactions in #2 and #3 go to the left instead.

Am I thinking right or am I completely off track?

Thank you for your time!

Offline Meter

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Re: How does the concentration affect the equilibrium state?
« Reply #1 on: February 11, 2021, 11:22:07 AM »
I think you got it.

In a heavily carbonated solution, reaction #3 will favor the reactant side, and since more H2CO3 is being formed, this will "feed back" into the other two reactions if that makes any sense. This is just Le Chatelier's principle applied to multiple reactions at once.  ;)

You can quantify these things using equilibrium constants and reaction kinetics.

Offline jackwhacky

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Re: How does the concentration affect the equilibrium state?
« Reply #2 on: February 12, 2021, 10:57:48 AM »
I think you got it.

In a heavily carbonated solution, reaction #3 will favor the reactant side, and since more H2CO3 is being formed, this will "feed back" into the other two reactions if that makes any sense. This is just Le Chatelier's principle applied to multiple reactions at once.  ;)

You can quantify these things using equilibrium constants and reaction kinetics.

I see! I think I got it now.

Interesting how when I repeated this experiment at home and dissolved another tablet in a cup of water that already had a dissolved tablet in it the reaction didn't go slower. When I dissolved a third one though it took 10 more seconds or so. I came up with a possible explanation that the second tablet's concentration was not enough to give a noticeable effect until the 3rd tablet was added. Never occurred to me even though I have been studying this for a while now, but I guess you learn along the way  :)

Thanks a lot for your help, much appreciated!

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