Carbon Dioxide solubility in water and Le Chatliers principal (1 Viewer)

[Shoom]

Hi
Why is it that when we open a soft drink bottle CO2 gas is produced in terms of Le Chatliers principal.

Also can someone write the overall equation involving CO2 gas CO2 aqeuoes and then Carbonic Acid.


Which part of this reaction is endothermic, THANKS

 

[cwag]

CO_2(g) --> CO_2(aq)
CO_2(aq) + H₂O --> H₂CO_3(aq)
H₂CO_3(aq) --> H⁺ + HCO^-
HCO_3(aq)^- --> H⁺ + CO_3(aq)^2-

mainly..
CO_2(g) + H₂O_(aq) --> H2CO3(aq)
when pressure is reduced (opening bottle) equilibrium will shift left to minimise this change (Le chateliers principle) thus creating CO_2(g)
the above reaction is exothermic in the forward direction therefore increasing temp will aslo shift eq to left

 

[Shoom]

So its ok if I use this equation

CO2+ H2O ( equilibrium symbol) H2CO3

When the CO2 is dissolved in water to produce H2CO3 heat is given of?


I still dont understand how shifting the equation to the left favours the ( increased pressure)

 

[danz90]

Well by opening up the bottle, your essentially increasing the volume, which ultimately reduces the concentration of CO₂ (since pressure is proportional to 1/V , V=volume). As a result, the equilibrium will shift to the production of CO₂(g) so it partially offsets the reduced [CO₂].
The following reaction will be favoured:

CO₂(aq) ----> CO₂(g)

 

[Shoom]

So, when I open the bottle I decrease pressure hence the eq goes to the left?


Please explain it in terms of Le Chats principal.

 

[danz90]

So, when I open the bottle I decrease pressure hence the eq goes to the left?


Please explain it in terms of Le Chats principal.

I pretty much did, above.

The pressure decreasing leads to a DECREASE in concentration of any gaseous CO₂ in the system. As a result, the equilibrium will shift such that more CO₂ (g) is produced to partially offset the decreased [CO₂(g)]. Hence, the equation i gave above is favoured.

 

[Shoom]

Ok

IF I open a bottle, pressure is decreased. So the solubility of CO2 degreases , so to balance that out shouldnt the eq go to the right?

 

[danz90]

Ok

IF I open a bottle, pressure is decreased. So the solubility of CO2 degreases , so to balance that out shouldnt the eq go to the right?

Well, think about it physically. In equilibrium, there are ALWAYS some amounts of both products and reactants present. Obviously, the amount of each changes.

Inside a soft drink bottle there is still some gaseous CO₂ in the air just above the liquid. So, when you open, the pressure is GREATLY decreased, thus decreasing the concentration of CO₂ (because there is a greater volume that gaseous CO₂ can occupy). Also, some CO₂ will escape, thus removing some CO₂(g) from the system.

As a result, the equilibrium will shift in such a manner, that more CO₂(g) is present at equilibrium, to partially offset both the loss of CO₂(g) (because some of it has escaped from the bottle once you open it), and also the decreased concentration of CO₂(g), as you greatly decrease pressure by opening the bottle (you are essentially increasing the volume that the given amount of CO₂(g) occupies).

Hence, all the equations in equilibrium are favoured to result in more gaseous CO₂(g) present:

H⁺ + HCO₃^-(aq) ---> H₂CO₃(aq)

H₂CO₃(aq) ---> H₂O(l) + CO₂(aq)

CO₂(aq) ---> CO₂(g)

Hope that helps

 

[Shoom]

Thankyou So Much!!!!!!!!!!!!!!!!!!!!

 

[danz90]

Thankyou So Much!!!!!!!!!!!!!!!!!!!!

lol no prob. but did u understand it.. did it answer ur question?

 

[Shoom]

Ok, by opening the bottle the pressure is decreased.


So the equations are shifted to the left.


So whats the point of shkaing it, why does it speed up the reaction etc?

 

[danz90]

Ok, by opening the bottle the pressure is decreased.


So the equations are shifted to the left.


So whats the point of shkaing it, why does it speed up the reaction etc?

Well, I'd imagine shaking the bottle would cause greater rate of collisions between CO₂(aq) particles, as well as other particles involved in the equilibrium. By you shaking it, you're providing kinetic energy which also provides activation energy for the reaction to occur. As a result of increased rate of collisions, that would increase the rate of the following reaction:

CO₂(aq) ---> CO₂(g)

Not sure if that's correct though lol

 

[JasonNg1025]

I'd say that was correct. That's why the "Healed and Sealed" trick works Look it up on youtube if you don't know it.

 

[Zippora]

the reaction moves to the right when you open it! to produce more gas!

 

[danz90]

the reaction moves to the right when you open it! to produce more gas!

lol depends at which way u look at it.. the main and correct thing is, which we all agree on.. is that when the bottle is opened, equilibrium shifts so that more gas is produced/released etc.

 

[Zippora]

lol depends at which way u look at it.. the main and correct thing is, which we all agree on.. is that when the bottle is opened, equilibrium shifts so that more gas is produced/released etc.

sure, and if we look at it they way that you wrote it above, and the way that every other person on this thread has written it, then the equilibrium is shifting right when the pressure is released.
I dont understand why everyone is writing

CO2(aq) --> CO2(g) and then saying it shifts to the left????

 

[JasonNg1025]

I think when they're referring to "left" they're referring to the equation which is usually

CO_{2 (g)} + H₂O _(l) <=> H₂CO_{3 (aq)}

Which is another of writing

CO_{2 (g)} --> CO_{2 (aq)}

And so equilibrium goes left

 

[danz90]

I think when they're referring to "left" they're referring to the equation which is usually

CO_{2 (g)} + H₂O _(l) <=> H₂CO_{3 (aq)}

Which is another of writing

CO_{2 (g)} --> CO_{2 (aq)}

And so equilibrium goes left

Ohhh I seeeee lol