Decarbonating Soda Water Prac Help (1 Viewer)

[jksdhjkfsjdkf]

hey there ... i've jsut completed the decarbonation soda water prac and have begun writing it up... i NEED HELP

Q! - wat are soem sources of error ( i heated my bottle of soda water up in a beaker, unopened ofcoures)

Q2 - explain in terms of the equilibrium reaction of CO2 dissolved in water, why a stream of small bubbles can be seen rising in a bottle of soda water when the lid is loosened. (is this jsut because it has been depressurized and lower pressure means lower solubility of carbon dioxide)

some general infomation on what i coukld put as answers THNX

 

[pLuvia]

http://community.boredofstudies.org/240/acidic-environment/90108/decarbonation-soft-drink.html

Take a look at this thread

 

[richz]

hey there ... i've jsut completed the decarbonation soda water prac and have begun writing it up... i NEED HELP

Q! - wat are soem sources of error ( i heated my bottle of soda water up in a beaker, unopened ofcoures)

Q2 - explain in terms of the equilibrium reaction of CO2 dissolved in water, why a stream of small bubbles can be seen rising in a bottle of soda water when the lid is loosened. (is this jsut because it has been depressurized and lower pressure means lower solubility of carbon dioxide)

some general infomation on what i coukld put as answers THNX

wow thats a very dangerous exp. are u sure u actually did it and the teacher approved of it, becuz if u did i hope u didnt spalsh soda water everywhere

 

[Dreamerish*~]

Sources of error include:

• Slight inaccuracy when weighing, as changes in atmospheric pressure can affect sensitive scales
• It isn't possible to get rid of all the CO₂ that's present in the soft drink
• Liquid may have been lost due to careless handling
• Water may have been lost to the environment when heating

As for question two, the solubility of carbon dioxide reaction is as follows:

CO_2(g) + H₂O_(l)

H₂CO_3(aq)

When you open the lid of the soft drink bottle, you are decreasing the pressure and letting out some of the CO₂ - a reactant. Le Chatelier's principle states that when a system at equilibrium is disturbed, the system will adjust itself to minimise the disturbance. When the concentration of one of the reactants is decreased, the equilibrium will shift to the left to counteract this decrease, producing more CO₂ and H₂O from H₂CO₃. That is, more CO₂ is coming out from solution. That would explain the bubbles rising to the surface.

 

[Riviet]

Something interesting to also note:

If you leave the bottle sitting there without it's lid on, the CO₂ floating around inside the bottle will continue to exit the bottle neck and by Le Chat's principle, the system at equilibrium before will shift to the left further to replace the loss of CO₂ that was floating around inside. This continues like a chain reaction until there is no more CO₂ in solution. Then you are just left with a sugary solution, which tastes weird.

 

[jksdhjkfsjdkf]

thanks heaps for the info...but let me clear up my original post...i put my opened bottle of soda water in a beaker full of water...the beaker was then heated creating a "hot bath"

 

[Dreamerish*~]

This continues like a chain reaction until there is no more CO₂ in solution.

Technically, that's not true.

If you leave the drink out long enough, the solubility of CO₂ falls 1.2 x 10^-5 mol/L, which is the solubility at atmospheric pressure and temperature.

So there's still some CO₂ in solution after all.

 

[richz]

thanks heaps for the info...but let me clear up my original post...i put my opened bottle of soda water in a beaker full of water...the beaker was then heated creating a "hot bath"

o ok u opened the lid. If u closed the lid. I wudnt want to know the chaos u created

 

[Riviet]

Technically, that's not true.

If you leave the drink out long enough, the solubility of CO₂ falls 1.2 x 10^-5 mol/L, which is the solubility at atmospheric pressure and temperature.

So there's still some CO₂ in solution after all.

Yes I am aware of that, but that's what theoretically would happen, but it doesn't, mainly because of the conditions that the bottle is in. But thanks for clarifying that.

 

[OzV]

I am doing this prac with my chemistry class tomorrow (hmm - this morning ...) and instead of leaving it in the open to work by itself I am going to get them to place the soft drink can in a bell jar and suck out enough air to lower the pressure enough to make the carbon dioxide bubble out but not enough to make the water bubble out. Given that the boiling point of carbon dioxide is -78.5 deg C and that of water is 100 deg. C this isn't too hard to achieve and if done slowly you don't make a mess. This way you have min evap. of water too.

It is an interesting proposition to investigate the errors in this experiment. According to the ideal gas law, PV=nRT, the molar volume of a gas at 25 deg C and 100kPa is V=nRT/P = 1mole x 8.314J/K/L x 298.15K/100kpa = 24.79L

You can use the difference in mass to calculate the moles and plug this into Volume = moles x molar volume to find the volume but this assumes SLC. In reality SLC are prob not achieved so the above equation will have inherent errors. You can also use PV=nRT to get exactly the same answer using 298.15K and 100kPa, BUT if you also measure the actual pressure and actual temperature on the day you can plug these figures into PV=nRT to get the more accurate answer and the closer the answer is to the one using V=molar volume/moles then the closer your conditions were to SLC. Now you can convert your answer to what you would expect to get at SLC with your actual data using Boyle's Law; P₁V₁=P₂V₂ where P=pressure and V = volume if you assume that the carbon dioxide is behaving like an ideal gas (which it isn't at lower pressures because the particles have lower KE and therefore there is some attraction between particles - more error!!!) - so you can check and verify your results pretty easily and get an idea of just how big your errors are.

 

[NightShadow]

what i'm concerned with the that an equilibrium exists in a closed system only...

so opening the cap is like opening the system, and no longer qualifies as a traditional equilibrium?

but i guess you say that the room now is part of the closed equilibrium

but how closed...is a closed equilibrium then?

 

[Dreamerish*~]

what i'm concerned with the that an equilibrium exists in a closed system only...

so opening the cap is like opening the system, and no longer qualifies as a traditional equilibrium?

but i guess you say that the room now is part of the closed equilibrium

but how closed...is a closed equilibrium then?

Yes, the equilibrium must be in a closed system. This means no matter (gas, liquid, solid, whatever) can get in or out. A school classroom wouldn't be considered a closed system because there are gaps between doors, windows are open, aircon vents, etc.

It has to be closed, very closed.

 

[Riviet]

Yes, the equilibrium must be in a closed system. This means no matter (gas, liquid, solid, whatever) can get in or out. A school classroom wouldn't be considered a closed system because there are gaps between doors, windows are open, aircon vents, etc.

It has to be closed, very closed.

That's why you can stay in a classroom for hours and still be breathing, because air is still able to get into the room!
A system can only be closed or open, it can't be half closed/half open. Other examples of closed systems include a vacuum and an unopened can or bottle of soft drink. Also note that after opening the lid of a carbonated drink, it is a open system from then on; even if you screw the lid back on, you are not sealing all the air from exiting/entering, although you are slowing down the process of the CO₂ coming out of solution and into the air and out of the bottle and vice versa.

 

[Dreamerish*~]

Also note that after opening the lid of a carbonated drink, it is a open system from then on; even if you screw the lid back on, you are not sealing all the air from exiting/entering

A closed bottle is a closed system. Once you screw the lid back onto an opened bottle of soft drink, it becomes closed again. The system will again reach equilibrium, however, this new equilibrium would contain different concentrations of reactants and products from the old one.

 

[Riviet]

I thought it wouldn't be closed, just because the gap between the inside of the lid and the top of the bottle opening would leak very small amounts of CO₂ and other gases.

 

[Mountain.Dew]

I thought it wouldn't be closed, just because the gap between the inside of the lid and the top of the bottle opening would leak very small amounts of CO₂ and other gases.

mmmmmmm i think for the HSC, this wouldnt matter too much. i wouldnt be too concerned with it. stick with the theoretical principles.

 

[Dreamerish*~]

I thought it wouldn't be closed, just because the gap between the inside of the lid and the top of the bottle opening would leak very small amounts of CO₂ and other gases.

Negligible amounts, I would imagine. That probably wouldn't matter.

Anyhow, an unopened bottle would have had such a gap too, wouldn't it?

 

[OzV]

what i'm concerned with the that an equilibrium exists in a closed system only...

so opening the cap is like opening the system, and no longer qualifies as a traditional equilibrium?

but i guess you say that the room now is part of the closed equilibrium

but how closed...is a closed equilibrium then?

However, this is not really the point of the experiment. You are right in saying that the system will only be in equilibrium when the top is on and sealed tight to prevent any leakage. But if a reaction system that IS in equilibrium is disturbed, by say... opening the lid, it will attempt to reverse the effects of the disturbance to re-establish a state of equilibrium. Of course this is impossible in an open system when the reaction system involves gases which is why it IS possible to exploit this tendency to drive the reaction system in a particular direction to achieve the aim by manipulating either the pressure (partial or total), temperature, or concentration of the products/reactants. In this case the aim was to decarbonate a carbonated solution by exploiting the principles of chemical equilibria and in doing so calculate the volume of gas dissolved in the solution at SLC. Once it's initial mass has been determined (as a closed system) you can't actually achieve the aim if it is maintained as a closed system (and could be potentially dangerous if you tried it ). If it were me I would be focusing my concerns on being able to explain things like;

<ul><li>why does a drop in pressure affect the solubility of CO₂ in water ? </li>
<li>how did you exploit this to ensure the CO₂ was driven off completelty?</li>
<li>how did you minimise errors or how did your experimental design introduce errors?</li>
<li>why does increasing the temperature of the reaction system cause a shift in equilibrium to the left (using the terms exothermic and endothermic to explain your answer)?</li>
<li>Increasing the temperature of a reaction system involving solids and liquids increases the solubility of reactants in the solvent so why doesn't this occur when gases are involved?</li>
<li>Why does heating a solution lower the total pressure above the solution in an open system...</li></ul>

etc

 

[OversizeSweater]

hey!
i need to plan an experiment to show the factors that affect a system at equilibrium, and i can only think of this one... i need to know how change in temp, volume, concentration and pressure affect THIS equilibrium. and how do i conduct an experiment to investigate this..

can anyone suggest anything?