How Does Le Chatelier's Principle Explain a Soft Drink Going Flat?

[iRaid]

Homework Statement

1) Use Le Chatelier's prinicile to explain how a shift in the equilibrium H₂CO₃(aq) ⇌ H₂O(l) + CO₂(g) causes a soft drink to go flat when its container is left open to the atmosphere.

2) How might the addition of a noble gas to the reaction vessel affect this equilibrium?
2N₂H₄(g) + 2NO₂(g) ⇌ 3N₂(g) + 4H₂O(g)
Assume that the volume of the reaction vessel does not change.

Homework Equations

The Attempt at a Solution

1) My answer is: The more air that the soft drink comes into contact with the atmosphere the equilibrium will shift to the side where H₂O and CO₂ are because of the CO₂ turning into a gas.

2) My answer is: It may cause the reactions equilibrium to shift towards the products.


This maybe easy for some people, but it's hard for me to understand this concept and I really need a good grade on my tomorrows test.

Thank you.

 

[gabriels-horn]

Homework Statement

1) Use Le Chatelier's prinicile to explain how a shift in the equilibrium H₂CO₃(aq) ⇌ H₂O(l) + CO₂(g) causes a soft drink to go flat when its container is left open to the atmosphere.

2) How might the addition of a noble gas to the reaction vessel affect this equilibrium?
2N₂H₄(g) + 2NO₂(g) ⇌ 3N₂(g) + 4H₂O(g)
Assume that the volume of the reaction vessel does not change.

Homework Equations

The Attempt at a Solution

1) My answer is: The more air that the soft drink comes into contact with the atmosphere the equilibrium will shift to the side where H₂O and CO₂ are because of the CO₂ turning into a gas.

2) My answer is: It may cause the reactions equilibrium to shift towards the products.


This maybe easy for some people, but it's hard for me to understand this concept and I really need a good grade on my tomorrows test.

Thank you.

I'm assuming that your answers need to be purely of the qualitative nature.

1) Le Chatelier's priniciple basically says that a reversilble reaction will shift in the forward or reverse direction to maintain the equilibrium of the reaction. Think about the composition of air, it contains a small amount of water and carbon dioxide. Adding a component to one side of a reaction generally shifts the reaction to the opposite side, while taking away a component generally shifts the reaction to the side where the component was removed. These things happen to maintain the eqiulibrium of the reaction. It is of course more complicated than this in your question because you have pressure change in the reaction due to the presence of a gas which will also affect the equilibrium shift. The volume of solids and liquids have minor changes which are generally so small to the overall contribution that they are ignored.

2) In this reaction you have 2 reactive gases, hydrazine and nitrogen dioxide, forming two products that are more stable at STP. Adding a noble gas, which are usually unreactive except to the fluoride ion, will increase the pressure of the vessel. The important thing is that the volume does not chage (ΔV = 0). Think about what this will do to your reactans and products, will this affect their overall pressures by adding something non-reactive. Are any of your reagents or products noble gases?

 

[iRaid]

I understand 1, but for 2 it's weird because we never went over that in class. I understand it, just not how I will know they're reactive :s Well it's too late now, thanks anyway :)

 

[Borek]

1 - air contains so low amounts of carbon dioxide and water it can be treated as not containing these, hence equilbrium shifts far to the right.

2 - I have posted it long ago at CF, but I think it is worth of repeating. Introduction of the inert gas doesn't shift equilibrium of the gaseous reaction.

What counts here are partial pressures, not the total system pressure. Let's assume we have a system of reacting gases A & B, and we will add inert gas C.

A <-> 2B

K = p_B²/p_A

p_X - X partial pressure
n_X - number of moles of X
P - total pressure
' marks pressures after C has been added

Before C addition:

p_B = P n_B/(n_B+n_A)
p_A = P n_A/(n_B+n_A)

after C addition

p'_B = P' n_B/(n_B+n_A+n_C)
p'_A = P' n_A/(n_B+n_A+n_C)

in general

p = n RT/V

so

P' = (n_B+n_A+n_C)/(n_B+n_A) P

substituting P' into p' equations:

p'_B = (n_B+n_A+n_C)/(n_B+n_A) P n_B/(n_B+n_A+n_C)
p'_A = (n_B+n_A+n_C)/(n_B+n_A) P n_A/(n_B+n_A+n_C)

canceling:

p'_B = P n_B/(n_B+n_A) = p_B
p'_A = P n_A/(n_B+n_A) = p_A

Partial pressures were not changed, equilibrium is not moving.

--

 

[iRaid]

That's a good answer and makes sense, thanks borek :)