Pressure in shaken Coke can?
Ciaran Brennan
when you shake it?
I've asked many Ph.D.s in physics, chemistry, and materials science this
question over the years. It is the type of question that everyone thinks
they know the answer to, but on closer inspection, finds out that they
don't really understand at all. I've even phoned the Coca-Cola company!
After convincing them that I wasn't pulling their leg, they referred my
question to their research department. But they didn't have the answer
either.
I'm not looking for the obvious (but inadequate) answer - that CO2 in solution
as carbonic acid is released into the gas phase, and more gas in a fixed volume
gives more pressure. I guess the real question is why does shaking make the
carbon dioxide come out of solution?
Think of it this way: when the sealed can is at rest, the CO2 gas is in
equilibrium with the CO2 in solution (i.e. carbonic acid). A certain partial
pressure of the CO2 gas is maintained at which the rate of CO2 absorption into
the liquid equals the rate of CO2 deabsorption into the gas phase.
Now, when the can is shaken, this gas/liquid system is suddenly put into a
non-equilibrium state. The amount of CO2 in the gas phase jumps well above
the equilibrium value, with the accompanying increase in gas pressure. If the
can is left alone for a period of time, the gas pressure slowly declines as the
excess CO2 gas is re-absorbed into solution. Eventually, equilibrium is restored.
Why does shaking disrupt the equilibrium condition in this manner?
A popular answer is that shaking imparts kinetic energy which raises the temperature.
CO2 is less soluble at high temperatures, so the pressure rises. But this can easily be
shown to be untrue. A cold can of Coke from the refrigerator will have a much
greater pressure after a few vigorous shakes than a room temperature can that
has not been shaken. You can measure the temperature - shaking does not cause
much of a change. The cold can is still colder than the warm can.
Another answer is that the mixing of the gas and liquid during shaking induces
more gas to come out of solution. But put some Coke in an ultrasonic cleaner and
see what happens: the gas bubbles come from the bottom of the container, where
no gas/liquid mixing occurs. The gas is released no matter what the volume to
gas surface area of the container is as well.
A case of Coke for the person with the most convincing answer! (I'm the judge.)
Ciaran
cbre...@draper.com
or:
cbre...@mit.edu
David Angelo Pollard
pressure) solution of CO2 in water and the increase in kinetic energy is
enough to cause the 'excess' CO2 to come out of solution, upsetting the
delicate euilibrium. After the agitation ceases, the equilibrium is
re-established and the excess CO2 is reabsorbed back into solution. The
existance of an equibilrium which requires a supersaturated solution is
easy to establish due to the fact that at any temperature, there is a
positive pressure in the container. This also explains why the drink
goes flat after opening as all the CO2 escapes into the atmosphere. I
once worked in the quality control department of a beer company and the
way they checked the amount of carbonation in the beer was to fill a
device which measures the pressure of the gas over the beer and shake it
up to get the maximum pressure. Also the fill machines cap the bottles
within a tenth of a second after filling them under pressure so that
very little gas escapes.
I love Coke and I hope the simplicity of my hypothesis does not
disqualify me from getting that free case.
Thank-you,
David P.
hamil...@ix.wcc.govt.nz
>Why does the pressure go up in a can of Coke (or any carbonated beverage)
>when you shake it?
This should only be "followed up" to sci.physics where this topic has
been beated to death ( I can't/don't know how to set followups )
My guess is that the pressure doesn't increase, but that's until somebody
posts some data using a low-volume pressure transducer _and_ an aerosol
sampler that doesn't significantly increase the volume ( they are just
small valves with a rubber outer seal and a needle to pierce the can -
you pull a high vacuum to the can wall, close the valve which punctures
the can but remains there, sealing the hole, and then open the valve to
vent a small ( about 0.5ml ) sample to the transducer )
>cbre...@draper.com
>or:
>cbre...@mit.edu
Lloyd R. Parker
: Why does the pressure go up in a can of Coke (or any carbonated beverage)
: when you shake it?
: Ciaran
: cbre...@draper.com
: or:
: cbre...@mit.edu
My hunch would be the microscopic dissolved gas bubbles, when
shaken, are given more motion which makes it easier for them to
find other tiny gas bubbles and join up; when these gas bubbles
become big enough, they escape the liquid phase. Sort of analagous
to what happens to a supersaturated solution.
Of course, I could be out in left field!
Joel Polowin
>Why does the pressure go up in a can of Coke (or any carbonated beverage)
>when you shake it?
>
>I've asked many Ph.D.s in physics, chemistry, and materials science this
>question over the years. It is the type of question that everyone thinks
>they know the answer to, but on closer inspection, finds out that they
>don't really understand at all. I've even phoned the Coca-Cola company!
>After convincing them that I wasn't pulling their leg, they referred my
>question to their research department. But they didn't have the answer
>either.
Since I don't usually drink carbonated stuff, I could be off base on this --
but I don't think the pressure actually goes up when you shake the can. You
get more of a mess if you open a can shortly after you shake it, but that's
not because of a pressure change. I expect it's because shaking the can
drops a lot of small bubbles into the liquid, bubbles of CO2 at the relatively
high equilibrium pressure in the sealed can. When you open the can and
decrease the gas pressure above the liquid, the small bubbles expand and
force the foaming mixture out of the can. Those bubbles would also serve
as sites for dissolved gas to come out of solution, and the agitation of
the foaming liquid also would help the stuff to reach the new equilibrium
state quickly, with a lower concentration of dissolved gas and more in the
gas phase, i.e., still more bubbles.
Camacho D R Noguera
I think the important point is the formation of bubbles and their effect on
the interphase surface area:
When the can is at equilibrium, the flux of CO2(g) from the gas into the liquid through the planar interphase is equal to the flux of CO2(g) from the liquid to the gas.
Now, when the bubbles are formed by shaking the can, the interphase is not planar anymore, and if you account for the diffusion layer at each side of the
interphase, it turns out that the effective surface area for the diffusion from the liquid into the gas is slightly larger than the effective surface area for the diffusion from the gas to the liquid. This results in a positive flux from the liquid into the gas, and therefore, an increase in the pressure of the gas phase.
Uhm, I think I'm thirsty now :)
Daniel
Nathan Frei
Ciaran Brennan <cbre...@draper.com> wrote:
>Why does the pressure go up in a can of Coke (or any carbonated beverage)
>when you shake it?
It seems like it's a surface-area-to-volume problem. When the can is in
the fridge, the liquid has a flat, circular surface area, and there's a
dynamic equilibrium (?) across the liquid/gas interface. When shaken,
however, there are suddenly lots and lots of bubbles throughout the
liquid, and the bubbles will present a much larger surface area for the
same volume of liquid to equilibrate over. Hence, dissolved CO/2 leaves
the liquid and enters the gas phase. I seem to remember (it's been a
while since I tried this) that if you shake a can up, increasing the
internal pressure, then leave it alone for a few hours (or days?) the
pressure goes back down, due to reestablishment of the original
equilibrium.
For all the nay-sayers out there who don't think the pressure really goes
up, try this: Take a two-liter bottle of soda, about half full. Cap it
tightly, and notice that it still feels squishy. Shake it a bunch, and
the bottle is a lot less squishy. Pressure rise, natch.
Nathan Frei
Joel Polowin
>For all the nay-sayers out there who don't think the pressure really goes
>up, try this: Take a two-liter bottle of soda, about half full. Cap it
>tightly, and notice that it still feels squishy. Shake it a bunch, and
>the bottle is a lot less squishy. Pressure rise, natch.
Nope. Two different situations. The first case, what the original poster
was talking about, involved a sealed can, under pressure, at equilibrium.
What *you're* talking about, here, is a container which starts out with a
lot of gas in solution and not much pressure above it, which you then
push to equilibrium by violent mixing of the two phases.
Now, if you let *that* settle down, and then give it another shake, is there
a difference in feel before and after you give it that extra shake?
Mariablue
Polowin) writes:
>>For all the nay-sayers out there who don't think the pressure really
goes
>>up, try this: Take a two-liter bottle of soda, about half full. Cap it
>>tightly, and notice that it still feels squishy. Shake it a bunch, and
>>the bottle is a lot less squishy. Pressure rise, natch.
>
>Nope. Two different situations. The first case, what the original
poster
>was talking about, involved a sealed can, under pressure, at equilibrium.
>What *you're* talking about, here, is a container which starts out with a
>lot of gas in solution and not much pressure above it, which you then
>push to equilibrium by violent mixing of the two phases.
I just bought a case of Coke on sale and as I was carying it down some
stairs - the box broke open and every can fell out. The pressure
increased so much that the bottom of each can ballooned out so that it
wouldn't stand up anymore. The pressure goes up.
If I may, I'd like to toss in my 2 cents. Consider delta G for the
formation of a CO2 bubble. The equilibrium is set up between the gas and
the liquid for a given surface area and geometry. So
delta G = -nRT*ln(c/c_e) + gamma * A
where n is the number of moles of CO2 in the gas phase, gamma is the
surface tension (about 70 dynes/cm) and c and c_e are the concentration
and equilibrium concentration w/o surface effects. The first term amounts
to the chemical potential absent any surface effects and is negative for a
supersaturated solution and positive at sub saturation. At equilibrium,
delta G is zero so that for a given pressure we set up equilibrium by
balancing chemical potential and surface effects.
Now (said as Ross Perot might say it) when we shake the can, we create
bubbles (assumed to be spherical) into which dissolved gas can diffuse.
If these bubbles are above a critical threshold but small enough and
numerous enough to increase significantly the gas/liquid surface area in
the can - they will tend to grow because the gas/liquid surface area
times the surface tension is greater. To reestablish equilibrium
(including the bubbles) the number of moles of CO2 in the gas phase must
increase into the same total volume and the pressure goes up.
Once the bubbles (even the very small ones) come to the top, the pressure
remains high until the CO2 can diffuse back into the Coke and the old
equilibrium is established.
Surface effects are everything - try this - take a glass of sprite (it's
clear and colorless) drop in a few grains of salt or sugar. You will see
bubbles nucleating at the corners of the crystals. Why??? Is curvature
important?
Love,
Maria
Christopher V. Sack
some research that appeared in the Journal or Chemical Education (June
'93, v70/6, pg 518).
The authors discuss the situation of what happens to the pressure inside
a soda pop bottle upon shaking. In order to test this, they built a
"barocolameter". It is a pressure guage mounted onto a cola bottle screw
cap that permits measurements of the headspace pressure under various
conditions. Their conclusions are as follows:
1) "If a bottle has been closed for some time (a day or so may be
required), it reaches an equilibrium pressure that is a function only of
the temperature and Herny's law (the solubility of a gas in a liquid is
proportional to the gas pressure.) CO2 in water obeys Henry's law up to
5 atm., and it remains a good approximation up to 10 atm. Shaking has no
effect on the pressure in the headspace of a bottle that has come to
equilibrium."
2) "When a bottle that has reached equilibrium is first opened, the head
pressure immediately falls to one atmosphere. The contents of the bottle
are no longer at equilibrium, and gas slowly leaves the solution until the
beverage 'goes flat'. However, if the bottle is shaken, nucleation sites
in the form of small bubbles are produced which permit CO2 to come out of
solution at a relatively rapid rate. If a thumb (or our barocolameter) is
stoppering the bottle, the pressure at room temperature rises to
approximately three atmospheres, at which point equilbrium is again
reached."
The point of this is that the pressure does not change in the closed
system upon shaking, only the number of nucleations sites. Upon resting,
these bubbles rise back up into the headspace, but the system is always at
its pressure equilibrium. The "gusher" upon opening the recently shaken
bottle appears to the result of increased pressure, but it is just the
CO2 equilibrating with the new lower pressure thru thousands of quite
large nucleation sites.
It could be argued that the temperature of the closed and shaken soda
bottle increases slightly because there is an input of energy, but this
would be very small, not enough to significantly change the Henry's law
pressure.
Sincerely, ___ ___ Christopher V. Sack
Chris / ) | / / ) __ __ | Chemistry Dept.
/ | / (___ __ ) / )| / S.U.N.Y.-E.S.F.
/ | / ) / / / | / Syracuse, NY 13210
(____/* |/* (____/ (__\ (__/ |/ \ <cvs...@mailbox.syr.edu>
JE Brady
Do you have some actual experimental data that shaking does increase the
pressure within the closed can or are you inferring this from the obvious
mess you make after you open the can?
My guess:
Shaking simply sets the stage for very efficient nucleation for the
release of the gas once the can is opened and the CO2 really wants to
escape (you now have a major nonequilibrium state). The key thing with
respect to your question is that there isn't actually a pressure increase
in the closed can. You don't get efficient nucleation after shaking and
allowing the can to stand for a while since the (I hope) small population
of particulates driving the whole process are sitting at the bottom of the
can and are able to nucleate only that much smaller local subvolume of the
package, giving that nice gentle fizzing I so enjoy on a hot day. (Okay,
that last bit is consistent, but possibly lame speculation.)
Jim "I'll take Coke, but prefer Pepsi" Brady
Ciaran Brennan (cbre...@draper.com) wrote:
: Why does the pressure go up in a can of Coke (or any carbonated beverage)
: when you shake it?
: [... some background stuff deleted]
: I'm not looking for the obvious (but inadequate) answer - that CO2 in
JE Brady
>In article <30sh3t$4...@knot.queensu.ca>, pol...@chem.queensu.ca (Joel
>Polowin) writes:
>>>For all the nay-sayers out there who don't think the pressure really
>>>goes up, try this: Take a two-liter bottle of soda, about half full.
>>>Cap it tightly, and notice that it still feels squishy. Shake it a
>>>bunch, and the bottle is a lot less squishy. Pressure rise, natch.
>>
>>Nope. Two different situations. The first case, what the original
>>poster was talking about, involved a sealed can, under pressure, at
>>equilibrium. What *you're* talking about, here, is a container which
>>starts out with a lot of gas in solution and not much pressure above
>>it, which you then push to equilibrium by violent mixing of the two phases.
>I just bought a case of Coke on sale and as I was carying it down some
>stairs - the box broke open and every can fell out. The pressure
>increased so much that the bottom of each can ballooned out so that it
>wouldn't stand up anymore. The pressure goes up.
Again, this is not necessarily the same situation specified in the
original post. When the can hits the floor, or whatever, it deforms. The
pressure jumps (remember, the headspace volume is relatively small, an
apparently small dV for the can may give rise to a relatively large dP for
the headspace) and causes the bottom to balloon out. There's also other
issues that may or may not be pertinent. You just bought the stuff, did
it heat up before you got home (in which case, you may be in a
nonequilibrium state in which shaking may increase the pressure in the can
since the increase in T has changed the CO2 vapor/liquid partition
coefficient). I know, speculation that may not be pertinent to your
example. However, try the following experiment - Shake a can (at room
temp.) as hard as you can. Does the bottom balloon out? I haven't been
able to get this to happen. Maybe I'm not shaking it hard enough.
>If I may, I'd like to toss in my 2 cents. Consider delta G for the
>formation of a CO2 bubble. The equilibrium is set up between the gas and
>the liquid for a given surface area and geometry. So
>delta G = -nRT*ln(c/c_e) + gamma * A
>where n is the number of moles of CO2 in the gas phase, gamma is the
>surface tension (about 70 dynes/cm) and c and c_e are the concentration
>and equilibrium concentration w/o surface effects. The first term amounts
>to the chemical potential absent any surface effects and is negative for a
>supersaturated solution and positive at sub saturation. At equilibrium,
>delta G is zero so that for a given pressure we set up equilibrium by
>balancing chemical potential and surface effects.
The equilibrium set up is the CO2 vapor/liquid partitioning. Surface
effects in this case are minimal.
>Now (said as Ross Perot might say it) when we shake the can, we create
>bubbles (assumed to be spherical) into which dissolved gas can diffuse.
Okay, the equilibrium pressure in the bubbles is greater than for the flat
surface, so gas is partitioned into the bubbles.
>If these bubbles are above a critical threshold but small enough and
>numerous enough to increase significantly the gas/liquid surface area in
>the can - they will tend to grow because the gas/liquid surface area
>times the surface tension is greater. To reestablish equilibrium
>(including the bubbles) the number of moles of CO2 in the gas phase must
>increase into the same total volume and the pressure goes up.
I would also agree that, in principle, the headspace will experience an
increase in pressure since the internal pressure of the small bubbles is
greater than for the headspace (Young-Laplace equation). The question is,
what's the magnitude of the expected increase and timeframe for relaxation
to the initial state? Let's assume that you convert the entire headspace
into 10 micron bubbles. This will give a pressure increase (assuming
memory serves correctly) of the order of 70*10^3 dyne/cm^2 (or
approximately 70*10^3/10^6 atm - roughly 0.1 atm). Now that's not a
whole lot of increase and this does assume all the headspace existed as
10 micron bubbles and there was no relaxation as the bubbles reverted to
the macroscopic headspace.
>Once the bubbles (even the very small ones) come to the top, the pressure
>remains high until the CO2 can diffuse back into the Coke and the old
>equilibrium is established.
But, as the rough calculation above shows, the pressure increase is
extremely small.
>Surface effects are everything - try this - take a glass of sprite (it's
>clear and colorless) drop in a few grains of salt or sugar. You will see
>bubbles nucleating at the corners of the crystals. Why??? Is curvature
>important?
Wrong experiment in a sense. Nucleation is clearly important to the
example. I would say this simply illustates the point that shaking causes
the nucleation to be very efficient. However, with respect to the
original post, all the action occurs after the can is opened and the Coke
very quickly relaxes to the new equilibrium state (partial pressure of CO2
at ambient atmospheric levels).
>Love,
>Maria
Joe Copson
>Ciaran
>cbre...@draper.com
>or:
>cbre...@mit.edu
I can't answer your question but the following may be of interest. Before
pop-top cans existed we discovered that heated Coke cans were quite
stable if after heating the cans were carefully opened. A little gas
comes out but no foam. This was done outside during the summer and
we used cans heated with the sun. Anyway, we would then strike the
can sharply on a curb. The entire contents of the can spewed across
the street, lots of fun. I'd be willing to bet that CO2 cannot disolve
in H2O if the water is mechanicaly stressed with some kind of shock-wave.
Another thing we used to do was shoot a warm can of cola with a small
BB gun. The can explodes rather violently. Small hole in the front
of the can and on the other side the can was ripped from top to bottom.
By the way the gun normally could NOT shoot through both sides of an
empty can.
--
_____________________________________________________________
| Joe Copson 408-353-4914 |
| cop...@netcom.com 415-966-1740 |
|_____________________________________________________________|
Wesley Brzozowski
|> >Why does the pressure go up in a can of Coke (or any carbonated beverage)
|> >when you shake it?
|>
|> >I've asked many Ph.D.s in physics, chemistry, and materials science this
|> >question over the years. It is the type of question that everyone thinks
|> >they know the answer to, but on closer inspection, finds out that they
|> >don't really understand at all. I've even phoned the Coca-Cola company!
|> >After convincing them that I wasn't pulling their leg, they referred my
|> >question to their research department. But they didn't have the answer
|> >either.
|>
|> >I'm not looking for the obvious (but inadequate) answer - that CO2 in solution
|> >as carbonic acid is released into the gas phase, and more gas in a fixed volume
|> >gives more pressure. I guess the real question is why does shaking make the
|> >carbon dioxide come out of solution?
I may be able to provide a bit of rather sketchy data in this area, as I've
actually spent a lot of time shaking up a carbonated beverage in a container
attatched to a pressure gage. For the record, it wasn't in a can but in a 2
liter bottle, and the beverage isn't Coke but homemade soda that I'm in the
process of carbonating. For those familiar with Lindsay Publications, which
sells all sorts of unusual technical books, I've built one of their "soda
carbonators", which consists of a specially machined clamp on the 2 liter bottle,
connected with a beverage hose to a CO2 cylinder and regulator that's set to
about 40 PSI. This is normally done to a bottle that's only about 2/3 full,
so there's lots of room to slosh the liquid around.
Usually I'll turn on the gas and set the pressure right on 40, then shake the
bottle to dissolve the gas. Initially, the gas flows fairly freely from the
cylinder and dissolves into the liquid I'm carbonating. As the thing nears
equilibrium, which I'm guessing from the fact that I can turn the cylinder
valve off, shake like crazy, and the pressure does not go down, I may stop and
rest for a moment, as I've just given my arms and upper body a fairly vigorous
workout. In fact, I may set the thing down and leave it for 1/2 hour or more
before remembering to get back to it.
At this point, the valve to the cylinder is still off, the pressure is still up
around 40, and I'm assuming that it's fairly close to equilibrium because another
shake does not lower the pressure, from more gas being dissolved. Here's the
point that might be relevant to this discussion: the pressure NEVER goes up
upon being shaken. Perhaps to be more precise, it never goes up enough that I
can see it on my pressure gage, so the suggested increase can't be greater than
2-3%, if it exists. Based on this experience, I'd like to suggest that the
original poster's question as to why the pressure goes up may be flawed. The
fizzing over that we see upon opening a shaken bottle may be due to something
other than the pressure going up.
The beverage hose in my carbonator also contains an additional valve that is
used to keep the pressure up in the bottle while I disconnect it from the
cylinder. That way, I can bring the bottle to the sink, release the pressure
slowly, and any fizz-over goes into the sink. One thing I've noticed is that
the shaking process whips the liquid up almost into a foam, and the smaller
bubbles take a long time to go away. As long as there are bubbles, a slight
release in pressure causes a mound of foam to form in the bottle. If I can't
see bubbles, the amount of foam is much smaller. As another interesting
sideline, if I give the bottle a good solid rap with a wooden spoon, the small
bubbles seem to consolidate and vanish very quickly. After such a rap, I can
release the pressure immediately with very little foam forming.
Okay, so my data is qualitative rather than quantitative, and it's mostly based
on recollection, rather than deliberate observation. Still, what I've seen
seems to suggest that the shaking introduces small bubbles below the surface
of the liquid, and these little suckers seem to correlate with the presence
of foam when the pressure is released. Based on this, I'd suspect that the
fizz-over is caused by one of two things, or a combination of both.
1). The bubbles beneath the surface provide nucleation sites for the gas
dissolved in the liquid, causing the formation of foam.
2). The bubbles beneath the surface expand upon the release of pressure, forming
the foam, and pushing the liquid above them upward.
However, no pressure increase is observed upon shaking. Should someone suggest
some useful experiments to try to provide better data, I'll be glad to go make
some soda (purely in the interests of science, of course) and I'll even be
willing to selflessly attempt to persuade my kids to find a practical way to
dispose of the soda, during this hot weather. :-)
Wes B.
--
*******************************************************************************
* Wes Brzozowski * Nothing is a waste that makes a memory *
* we...@vnet.ibm.com * - Ned Rorem *
*******************************************************************************
Max Muir
A case of Coke for the person with the most convincing answer! (I'm the judge.)
Ciaran
Fascinating! I suggest that there is some kind of ordering effect present
in the solution and that shaking it disrupts this ordering which is (I suggest)
important for solvation. I think the CO2 - H2O bonds will be relatively
weak hydrogen bonds which can be easily broken by even modest shearing forces,
so I suppose (i) raising the temperature (ii) shaking the solution will
break these forces, causing the CO2 to leave solution. Once agitation of
the solution has ceased I believe the ordering can be re-established,
leading to re-solvation of the CO2.
H
\
O--H - - O=C=O - - H--O
\
H
It may be that the formation of the solvation sphere around the CO2 is
the only ordering factor involved but it may be possible that domains
of ordered solute within the solution are also important (warning, bull-
shit in progress . . ).
I want to spend some time looking at solvation spheres, so if anyone
has any bright ideas, I'd like to here them.
Yours,
Max
--
*******************************************************************************
* DISCLAIMER: Unless indicated otherwise, everything in this note is *
* personal opinion, not an official statement of Biosym Technologies, Inc. *
*******************************************************************************
Keith Warren Rickert
>I can't answer your question but the following may be of interest. Before
>pop-top cans existed we discovered that heated Coke cans were quite
>stable if after heating the cans were carefully opened. A little gas
>comes out but no foam. This was done outside during the summer and
>we used cans heated with the sun. Anyway, we would then strike the
>can sharply on a curb. The entire contents of the can spewed across
>the street, lots of fun. I'd be willing to bet that CO2 cannot disolve
>in H2O if the water is mechanicaly stressed with some kind of shock-wave.
>Another thing we used to do was shoot a warm can of cola with a small
>BB gun. The can explodes rather violently. Small hole in the front
>of the can and on the other side the can was ripped from top to bottom.
>By the way the gun normally could NOT shoot through both sides of an
>empty can.
This may not have much to do with the CO2 in the can at all.
The hydrostatic shock involved in shooting the can full of coke
can do an impressive amount of damage.
Shooting identical cans with carbonated and non-carbonated
liquids inside would be a more controlled experiment
(even better if the non-carbonated one is pressurized to the same
level with something that doesnt dissolve as well.
Keith
--
Keith Rickert | "Aye, though we hunted high and low, and hunted
ke...@imppig.caltech.edu | everywhere, of the three men's fate, we found no
ric...@cco.caltech.edu | trace, in any time, in any place, but a door ajar,
| and an untouched meal, and an overtoppled chair."
James E. Baumgardner II
Max Muir <m...@biosym.com> wrote:
> Why does shaking disrupt the equilibrium condition in this manner? [..]
>
> A case of Coke for the person with the most convincing answer! (I'm
> the judge.)
>
> Ciaran
>
>Fascinating! I suggest that there is some kind of ordering effect present
>in the solution and that shaking it disrupts this ordering which is (I suggest)
>important for solvation. I think the CO2 - H2O bonds will be relatively
>weak hydrogen bonds which can be easily broken by even modest shearing forces,
>so I suppose (i) raising the temperature (ii) shaking the solution will
>break these forces, causing the CO2 to leave solution. Once agitation of
>the solution has ceased I believe the ordering can be re-established,
>leading to re-solvation of the CO2.
Hydrogen bonds between CO2 (non-polar) and water? Where does the
negative charge come from?
It makes more sense if taken in terms of pure equilibrium. Soda
goes 'flat' since (Carbonated beverages are simply H2CO3, sugar, and
additives) in the reaction H2CO3 --> H2O + CO2, the carbon
dioxide escapes to the surrounding air and no longer participates in the
forward reaction (Hence H2CO3 --> H2O + CO2 K>1). This reaction takes a
long time to occur when undisturbed... the surface area of the air-cola
interface is relatively small. However, when the can is shaken, (open or
not), the surface area of the interface becomes quite large, increasing
the reaction rate. If the can is open, then the cola will simply fizz
more rapidly. If the can is closed, pressure will build up inside the
can for a short while. If you leave the can closed, through equilibrium
(average molecular motion), carbon dioxide will slowly diffuse back into
the cola. (Note that when shaken, then left undisturbed, the can returns
to normal). But of course, if opened, the pressure from the CO2 will
cause it to shoot out of the can, bringing CO2 still in the can as well
as the rest of the cola.
--
Jim Baumgardner (jbau...@bronze.coil.com)
"Would you _please_ stop clicking my head?"
murphy kenneth p
James E. Baumgardner II <jbau...@bronze.coil.com> wrote:
>In article <1994Jul26.2...@biosym.com>,
>Max Muir <m...@biosym.com> wrote:
>> Why does shaking disrupt the equilibrium condition in this manner? [..]
>>
>> A case of Coke for the person with the most convincing answer! (I'm
>> the judge.)
>>
>> Ciaran
>>
>>Fascinating! I suggest that there is some kind of ordering effect present
>>in the solution and that shaking it disrupts this ordering which is (I suggest
Well, maybe. The problem should, of course, be considered in terms of the
equilibria, but the above discription doesn't quite do this. The change in
surface area in the closed can doesn't affect the equilibrium and only affects
the kinetics in the open can. However, shaking the can is adding energy to it
and this energy is partially absorbed in the tranfer of CO2 from the solution to
the gas phase (remember that gases are less soluble at higher temperature). The
thermal energy added to the system by shaking is, in time, lost to the surround-
ings and the gas redissolves.
Kip
--
Dr. Kenneth P. Murphy e-mail: k-mu...@uiowa.edu
Department of Biochemistry office: (319)335-8910
Univeristy of Iowa lab: (319)335-7936
Iowa City, IA 52242 FAX: (319)335-9570
mah...@esvx12.es.dupont.com
gas in the headspace escapes. The CO2 dissolved in the water escapes only
slowly, as it takes substantial energy to initiate bubbles. If the can is
shaken just before opening there are bubbles in the liquid. When it is opened
the headspace gas escapes as before, but the bubbles that are in the liquid
allow additional gas to escape rapidly.
There is no difference in pressure between the shaken and
unshaken coke as some posters have stated and demonstrated. There is a
difference in kinetic approach toward equilibrium after the can has been
opened, depending on whether it has been nucleated with bubbles or not.
The release of gas from the body of the liquid just after
opening the shaken can carries liquid out with the flow of gas. In
addition there will be some droplets in the gas phase of the shaken can. These
greatly increase the perception of gas flow.
Walter Mahler
Max Muir
Hydrogen bonds between CO2 (non-polar) and water? Where does the
negative charge come from?
I must admit I am a little hazy on H bonding. I had not realised that H bonding
can only occur between charged species. For example, don't water molecules H
bond with each other?
H H
\ /
O---H - - O
\
H
I have hitherto thought that neutral molecules can H bond to one another, but I
may be wrong. I also had not realised that the presence of a permanent electric
diple in both molecules is necessary either. Doesn't this say, oxalic acid
would not be able to H bond in water?
When you say CO2 is non-polar do you mean it has no molecular dipole? I agree
with that statement. Like 1,4 dichlorobenzene, the symmetry means that there
is no resultant static electric dipole, however, I would say that it is a quite
polar molecule myself :-) and MOPAC agrees with me, tho' amusing results are
obtained from Merz and Besler ESP subroutines for CO2 with MOPAC 6.0. Here's
my MOPAC output. ab Initio atomic charges are usually greater than this, btw.
FINAL GEOMETRY OBTAINED CHARGE
GRAPH NODIIS MMOK
INSIGHT generated AMPAC/MOPAC input file
DATE: Wed Jul 27 19:35:20 1994
C 0.0000000 0 0.000000 0 0.000000 0 0 0 0 0.4480
O 1.1861261 1 0.000000 0 0.000000 0 1 0 0 -0.2238
O 1.1864401 1 179.963312 1 0.000000 0 1 2 0 -0.2241
As you can see the O atoms are negative wrt the C atom, i.e.,
d- d+ d-
O=====C======O
I propose that the mechanism of solvation involves the formation of H bonds
between the negatively charged O atom and the positively charged H atom of
a water molecule, within the usual angle constraints that exist for H bonds
(that is the O--H- - -O angle must be 180 > x > ~170 degrees.
d- d+ d- d+ d-
O=====C======O - - - H-----O
\
\
H
I had assumed that this accounts for the greater solubility (in water) of
CO2 wrt other non-polar gases which do not have polar bonds, e.g., N2, O2.
2-
Presumably, by your reasoning, the carbonate ion CO is also a non-polar
molecule, since it has no permanent dipole 3 because of its D
symmetry. 3h
I will attend to the other part of your article when I return from
Pittsburgh next week. Does carbonic acid really exist? Can you give
me a reference for its structural parameters?
Regards,
James E. Baumgardner II
murphy kenneth p <kpmu...@blue.weeg.uiowa.edu> wrote:
>
>Well, maybe. The problem should, of course, be considered in terms of the
>equilibria, but the above discription doesn't quite do this. The change in
>surface area in the closed can doesn't affect the equilibrium and only affects
>the kinetics in the open can. However, shaking the can is adding energy to it
>and this energy is partially absorbed in the tranfer of CO2 from the solution to
>the gas phase (remember that gases are less soluble at higher temperature). The
>thermal energy added to the system by shaking is, in time, lost to the surround-
>ings and the gas redissolves.
>
Really, I meant a combination of the two (equilibrium determining
whether or not more gas is released, and the kinetics determining how
quickly)... it's been a long day. <grin>.
However, kinetic energy is not a viable explanation. Using Maxwellian
speed distribution for CO2 (0.040 kg/mol), calculation yields that the
average speed is 400 m/s at 300K. Shaking the can does not provide
enough kinetic energy to change this speed significantly. (I remember
hearing somewhere that stirring your coffee raises the temperature on the
order of 10^-1 K as the rotational energy is converted to thermal
agitation, but don't bet any money on it -- terrible memory <grin>).
ch_...@titan.kingston.ac.uk
I was particularly taken with Wesley Brzozowski's comments (quote):
"As another interesting sideline, if I give the bottle [of soda] a good solid
rap with a wooden spoon, the small bubbles seem to consolidate and vanish very
quickly. After such a rap, I can release the pressure immediately with very
little foam forming."
Years ago, a mate of mine showed me that it was possible to open a can of beer
without getting sprayed with the contents by first flicking the side of the can
hard with one's fingernail. Never fails, and now it's become a habit.
In recent years, a large number of british beers have been sold with a "widget"
in the bottom of the can (yes, one brewer even calls it a widget in their
adverts), the purpose of which is to cause excessive foaming when the can is
opened in order to put a good foamy head on the beer. With one or two
exceptions, the effect is to cause the contents to come flying out the second
the can is opened, so a large quantity is lost (usually over one's clothing)
and you only get two thirds of the contents in a glass (it certainly stops
uncouth persons drinking straight from the can). The instructions on the can
tell me that to avoid this effect, I should chill the can before opening it.
What do they think I am? An american?
Oops, sorry chaps, no offence meant.
Phil Miller Tate
Kingston University, UK
Ivan D. Reid
ch_...@titan.kingston.ac.uk writes...
> The instructions on the can
>tell me that to avoid this effect, I should chill the can before opening it.
>What do they think I am? An american?
I had an engineer once who would contend that you are probably a
physicist! When all else fails, read the instructions...
Ivan Reid, Paul Scherrer Institute, CH. iv...@cvax.psi.ch
James E. Baumgardner II
Max Muir <m...@biosym.com> wrote:
> Jim Baumgardner w
> Hydrogen bonds between CO2 (non-polar) and water? Where does the
> negative charge come from?
>
Well, hydrogen bonding does occur regardless of permanent dipole (i.e.
London dispersion forces), but the likelihood of CO2 with no net dipole
hydrogen bonding to water simply defies logic at "room" temperature.
> d- d+ d-
> O=====C======O
>
>I propose that the mechanism of solvation involves the formation of H bonds
>between the negatively charged O atom and the positively charged H atom of
>a water molecule, within the usual angle constraints that exist for H bonds
>(that is the O--H- - -O angle must be 180 > x > ~170 degrees.
But then again, as you can see, these dipoles are incredibly weak, and
would not withstand the motion from thermal energy. (As I said in a
previous post, using the Maxwellian speed distribution, the average speed
(8RT/pi*M)^.5 for CO2 at 300K is 400m/s). Certainly, these bonds may
form, but they will be broken before they can be formed. CO2 is normally
"bound" to water in H2CO3, and not by hydrogen bonds. These bonds simply
cannot explain the pressure...
>
> d- d+ d- d+ d-
> O=====C======O - - - H-----O
> \
> \
> H
>
> 2-
>Presumably, by your reasoning, the carbonate ion CO is also a non-polar
>molecule, since it has no permanent dipole 3 because of its D
>symmetry. 3h
[CO3]2- has a permanent dipole. Note the unshared electron pair on the
"top" of the tetrahedron.
>
>
>I will attend to the other part of your article when I return from
>Pittsburgh next week. Does carbonic acid really exist? Can you give
>me a reference for its structural parameters?
Carbonated water = Carbonic acid. The body's pH management is determined
by H2O + CO2 <--> H2CO3 <--> H+ + [CO3]2-. (Has to do with selective
permittivity and dissolved CO2, I believe).
O-H
/
O=C
\
O-H
H2CO3
Keith Warren Rickert
>In article <1994Jul28.0...@biosym.com>,
>Max Muir <m...@biosym.com> wrote:
>> Jim Baumgardner w
>> Hydrogen bonds between CO2 (non-polar) and water? Where does the
>> negative charge come from?
>>
>Well, hydrogen bonding does occur regardless of permanent dipole (i.e.
>London dispersion forces), but the likelihood of CO2 with no net dipole
>hydrogen bonding to water simply defies logic at "room" temperature.
>> d- d+ d-
>> O=====C======O
>>
>>I propose that the mechanism of solvation involves the formation of H bonds
>>between the negatively charged O atom and the positively charged H atom of
>>a water molecule, within the usual angle constraints that exist for H bonds
>>(that is the O--H- - -O angle must be 180 > x > ~170 degrees.
>But then again, as you can see, these dipoles are incredibly weak, and
>would not withstand the motion from thermal energy. (As I said in a
>previous post, using the Maxwellian speed distribution, the average speed
>(8RT/pi*M)^.5 for CO2 at 300K is 400m/s). Certainly, these bonds may
>form, but they will be broken before they can be formed. CO2 is normally
>"bound" to water in H2CO3, and not by hydrogen bonds. These bonds simply
>cannot explain the pressure...
Those dipoles are far from incredibly weak, as they are permanent
dipoles. Its just that the geometry leaves the molecule with no
total dipole, but the individual parts do have quite a strong dipole.
Hydrogen bonding of water to CO2 is probably a first step in the
mechanism of solvation of CO2 to H2CO3.
>>Presumably, by your reasoning, the carbonate ion CO is also a non-polar
>>molecule, since it has no permanent dipole 3 because of its D
>>symmetry. 3h
>[CO3]2- has a permanent dipole. Note the unshared electron pair on the
>"top" of the tetrahedron.
What tetrahedron? its a planar molecule, there is a set of delocalized
pi orbitals between the carbon and oxygen atoms.
Max Muir
>> Why does shaking disrupt the equilibrium condition in this manner? [..]
>>
>> A case of Coke for the person with the most convincing answer! (I'm
>> the judge.)
>>
>> Ciaran
>>
>>Fascinating! I suggest that there is some kind of ordering effect present
>>in the solution and that shaking it disrupts this ordering which is (I suggest
>>weak hydrogen bonds which can be easily broken by even modest shearing forces,
>>so I suppose (i) raising the temperature (ii) shaking the solution will
>>break these forces, causing the CO2 to leave solution. Once agitation of
>>the solution has ceased I believe the ordering can be re-established,
>>leading to re-solvation of the CO2.
>
>negative charge come from?
>
>goes 'flat' since (Carbonated beverages are simply H2CO3, sugar, and
>additives) in the reaction H2CO3 --> H2O + CO2, the carbon
>dioxide escapes to the surrounding air and no longer participates in the
>forward reaction (Hence H2CO3 --> H2O + CO2 K>1). This reaction takes a
>long time to occur when undisturbed... the surface area of the air-cola
>interface is relatively small. However, when the can is shaken, (open or
>not), the surface area of the interface becomes quite large, increasing
>the reaction rate. If the can is open, then the cola will simply fizz
>more rapidly. If the can is closed, pressure will build up inside the
>can for a short while. If you leave the can closed, through equilibrium
>(average molecular motion), carbon dioxide will slowly diffuse back into
>the cola. (Note that when shaken, then left undisturbed, the can returns
>to normal). But of course, if opened, the pressure from the CO2 will
>cause it to shoot out of the can, bringing CO2 still in the can as well
>as the rest of the cola.
>--
>Jim Baumgardner (jbau...@bronze.coil.com)
>"Would you _please_ stop clicking my head?"
Well, Jim, I agree with the direction you are going, but I don't think you have
said it clearly enough. Leaving aside the question of the existence of the
"carbonic acid" for a moment, the original poster expressed interest in the role
of the bubbles, and I was speculating about the chemical environment of the
dissolved gas (if there is one), not the fizzikal chemistry involved, which is
dull.
CO ----> CO
2 (aq.) <---- 2 (g)
It seems likely that the Coke factory fills the cans at atmospheric pressure.
When they are sealed, CO2(g) comes off the solution until the partial pressure
of the CO2 above the solution is large enough so that the rate of dissolving of
gaseous CO2 in the solution is equal to the rate of CO2 being given off by the
solution, i.e., equilibrium is established. I think that this means that the
pressure inside the can is slightly greater than atmospheric: this would account
for the slight hiss of pressure one often hears when a can is opened.
When one opens a can of Coke one is increasing the volume of the system to well,
effectively infinity, and CO2 is given off by the solution until equilibrium is
re-established. Since the volume of the atmosphere is very large and the partial
pressure of atmospheric CO2 relatively low the final concentration of CO2 is v.
low: the drink has gone flat.
As you state, the surface area of the solution is vastly increased by the
entrainment of bubbles caused by shaking and it seems obvious that, in some
way these bubbles cause the CO2 to come off the solution more quickly than
usual. It may be as you suggest (I think), that it is merely an increase
in surface area. I tried simply blowing through a straw into my Coke (this
causing a certain amount of annoyance to my nearby collegues): that causes
effervesence, and actually, merely inserting the straw causes bubbles to
form on the surface of the straw, but nothing like the explosive release of
Coke when you shake the can immediately prior to opening it.
I have never been inside a Coke can when it is shaken but I suppose that many
small bubbles are formed,rather than a few big bubbles as I obtain when blowing
through a straw into the Coke. Many small bubbles would of course provide a
larger surface area than a few big ones. I suppose the extra force of the
bubbles from the shaking comes from the fact they are formed under pressure and
therefore have a higher internal pressure than bubbles formed at atmospheric
pressure, like those from blowing through a straw.
To return to the question of the chemical state of the CO2 in solution. I
haven't seen any evidence to support the existence of carbonic acid. If you
could point me at some refs I'd be grateful, since it is an interesting
example of a geminal dihydroxide, compounds which are quite rare, in fact the
only one I can think of is chloral hydrate.
H--O
\
C==O
/
H--O
Max Muir
>Jim Baumgardner w
>Max Muir <m...@biosym.com> wrote:
>> Jim Baumgardner w
>> Hydrogen bonds between CO2 (non-polar) and water? Where does the
>> negative charge come from?
>>
>
>Well, hydrogen bonding does occur regardless of permanent dipole (i.e.
>London dispersion forces), but the likelihood of CO2 with no net dipole
>hydrogen bonding to water simply defies logic at "room" temperature.
Hmm, I think London forces are not the same as H-bonds. Moore says that
London forces are attractive forces due to the polarization of a molecule
due to the instantaneous dipole arising from asymmetry of the electron
clouds in a neighboring molecule. Please correct me if I am wrong but
aren't H bonds directional, _mostly_ ionic, and permanent?
Also I think they occur only between only the most electronegative atoms,
viz., O, N, F, whereas Moore illustrates London forces by discussing the
attraction of two Ar atoms. I am not yet convinced that the molecular
dipole is needed for a fully-fledged H bond. Consider oxalic acid which
has no large dipole (is it zero?) yet with a melting point of 190 degrees
and a MW of only 90 is surely hydrogen bonded. On the other hand chemistry
is vast and I suspect I could find an example of something to justify
anything if I looked long enough :-)
H--O O . . H--O O . . H--O O--H
\ // \ // \ /
C---C C---C C---C
// \ // \ // \\
O O--H . . O O--H . . O O
Why does this remind me of Eqyptians?
>> d- d+ d-
>> O=====C======O
>>
>>I propose that the mechanism of solvation involves the formation of H bonds
>>between the negatively charged O atom and the positively charged H atom of
>>a water molecule, within the usual angle constraints that exist for H bonds
>>(that is the O--H- - -O angle must be 180 > x > ~170 degrees.
>
>But then again, as you can see, these dipoles are incredibly weak, and
>would not withstand the motion from thermal energy. (As I said in a
>previous post, using the Maxwellian speed distribution, the average speed
>(8RT/pi*M)^.5 for CO2 at 300K is 400m/s). Certainly, these bonds may
>form, but they will be broken before they can be formed. CO2 is normally
>"bound" to water in H2CO3, and not by hydrogen bonds. These bonds simply
>cannot explain the pressure...
Cotton and Wilkinson. Chemistry of the Main Group Elements, Hydrogen,
section 6-10: Oxo Acids.
They don't have any structural parameters or any analytical data but they say
"Carbonic acid is exceptional in that the directly measured pk1, 6.38, does
not refer to the process
+ -
H CO = H + HCO
2 3 3
since carbon dioxide in solution is only partly in the form of H CO but
2 3
largely present as more loosely hydrated species CO (aq)."
2
They say substantially the same thing again in two places in the book i.e.,
that carbonic acid is present only as a minor component of the solvated species
and that the bulk of the solute exists as solvated CO as I previously
suggested. 2
>>
>> d- d+ d- d+ d-
>> O=====C======O - - - H-----O etc.
>> \
>> \
>> H
If these loose hydration forces are not hydrogen bonds, what are they?
If they are only dispersion forces, why is CO2 so much more soluble in water
than O2, N2 and benzene (which also have polarizable electron clouds)?
C & W also spend time justifying the anomalously high first pKa of H2CO3,
apparently due to the large portion of the CO2 present as dissolved gas
CO2(aq).
+ _
[ H ] [ HCO ]
3
__________________ = 4.16 x 10^-7
[ H CO ]
2 3
4.16 x 10^-7 isn't much of an acid is it?
It seems to me H2CO3 should be stronger than that (if it existed). How about
this? There is no need to invoke the H2CO3 at all.
d- d+ d- d- d+ d-
O=====C======O O=====C======O
` ' _ ` '
` ' OH(aq) ` '
` ' ----> _` ' +
O------H <---- O------H + H
/ +
/ H |
H (aq) | fast
V
_
O
\
C===O --> etc, boring 2nd dissociation
/ consistent with expectations
H---O of supervisor.
>>Presumably, by your reasoning, the carbonate ion CO is also a non-polar
>>molecule, since it has no permanent dipole 3 because of its D
>>symmetry. 3h
>
>[CO3]2- has a permanent dipole. Note the unshared electron pair on the
>"top" of the tetrahedron.
I believe [CO3]2- is trigonal and planar. See 'Basic Inorganic Chemistry',
Cotton and Wilkinson, 'Oxacids of Carbon', p 111. "The ions are planar. In
[CO3]2- because of delocalized pi-bonding the bond lengths are equal and 120
degrees." I don't think it can have a dipole if this is correct as the charge
distribution is symmetrical, and therefore [CO3]2- _is_ non-polar in the same
sense as CO2.
_ _
O O O
\\ _ \ \ _
C---O <--> C===O <--> C---O
_ / _ / //
O O O
>>Does carbonic acid really exist? Can you give
>>me a reference for its structural parameters?
>
>Carbonated water = Carbonic acid. The body's pH management is determined
>by H2O + CO2 <--> H2CO3 <--> H+ + [CO3]2-. (Has to do with selective
>permittivity and dissolved CO2, I believe).
>
> O-H
> /
> O=C
> \
> O-H
>
> H2CO3
That is what we learnt, isn't it? The thing that bothers me is that no-one
has ever isolated this important biological molecule and I have seen no
NMR data: clearly it is an easy thing to do, one need only look at the 13C
spectrum of carbonated water, or look at the 1H spectrum of a D2O solution
of CO2 after a little H2O has been added. There should be two shifts for
the C atom and two broad peaks for the protons at the usual shift for DOH
and another for the deuterated carbonic acid at ? 10 delta? Nothing with
covalent bonds should be so elusive: after all Chapman isolated C4H4 30
years ago.
I've had a look at HCA and I don't remember the formation of H2CO3 being
part of the mechanism. C&W give the mechanism of the reaction and don't
invoke it. I don't think it exists.
In any case I think we've answered the original poster's question many times
only: as you suggest the increased surface area due to the many bubbles
formed by shaking caused the system to equilibrate rapidly, even explosively!
>--
>Jim Baumgardner (jbau...@bronze.coil.com)
>"Would you _please_ stop clicking my head?"
That's a very funny .sig
James E. Baumgardner II
Max Muir <m...@biosym.com> wrote:
>
>Hmm, I think London forces are not the same as H-bonds. Moore says that
>London forces are attractive forces due to the polarization of a molecule
>due to the instantaneous dipole arising from asymmetry of the electron
>clouds in a neighboring molecule. Please correct me if I am wrong but
>aren't H bonds directional, _mostly_ ionic, and permanent?
My thoughts on that were that the CO2 --> H2O H-bonding could not occur
because of the tremendous speeds of the CO2 molecule at room
temperature. I suggested that perhaps CO2's London dispersion forces
could account for solvation at near 0 K. However, I did a bit of
research on my own, and discovered that CO2 solvation is a bit trickier
than I thought.
>
> H--O O . . H--O O . . H--O O--H
> \ // \ // \ /
> C---C C---C C---C
> // \ // \ // \\
> O O--H . . O O--H . . O O
>
>
>Why does this remind me of Eqyptians?
Sorry, I just like this picture. <grin>.
>since carbon dioxide in solution is only partly in the form of H CO but
> 2 3
>largely present as more loosely hydrated species CO (aq)."
> 2
Well, of course, since there are three equilibria occurring.
H O (l) + CO (g) <--> H CO (aq)
2 2 2 3
+ -
H CO (aq) <--> H (aq) + HCO (aq)
2 3 3
- + 2-
HCO (aq) <--> H (aq) + CO (aq)
3 3
Now that I've had a chance to look at it, the dipoles in CO2 ARE
significant. Mea culpa, mea culpa, mea maxima culpa. <grin>. I had
always thought that CO2 dissolved in H2O was equivalent to H2CO3 (a
virtually immediate reaction), though I hadn't stopped to think of the
mechanism of CO2 solvation in water. Now that I've re-looked at the
figures, it looks possible for the dipoles to hold in the CO2. Of
course, we can't forget the importance of the H+ and HCO3-, because it
makes for a nice "fudge factor" in calculations...
> O O O
> \\ _ \ \ _
> C---O <--> C===O <--> C---O
> _ / _ / //
> O O O
>>Carbonated water = Carbonic acid. The body's pH management is determined
>>by H2O + CO2 <--> H2CO3 <--> H+ + [CO3]2-. (Has to do with selective
>>permittivity and dissolved CO2, I believe).
>That is what we learnt, isn't it? The thing that bothers me is that no-one
>has ever isolated this important biological molecule and I have seen no
>NMR data: clearly it is an easy thing to do, one need only look at the 13C
Again, I had always associated H2O + CO2 --> H2CO3 as being the fast step
in the proton loss, but I hadn't stopped to think of the mechanism of
solvation.
>In any case I think we've answered the original poster's question many times
>only: as you suggest the increased surface area due to the many bubbles
>formed by shaking caused the system to equilibrate rapidly, even explosively!
Agreed. So whose name goes first on this theory? Heh heh heh...
>
>>Jim Baumgardner (jbau...@bronze.coil.com)
>>"Would you _please_ stop clicking my head?"
>
>That's a very funny .sig
My sister's husband made an application for the Atari Falcon030
that had a .TIF of my head, combined with a digitized "Hey, quit it."
that I unknowingly recorded for him. He made it so that when you clicked
my head, it played the sound. Easy, eh? But then it got uploaded to the
Internet (atari.archive.umich.edu), and was distributed throughout the
world with its source code. Ah well... I guess I'm famous now. <grin>.
Max Muir
>they are only dispersion forces, why is CO2 so much more soluble in water
>than O2, N2 and benzene (which also have polarizable electron clouds)?
from what I just learned last year in High School chem... O2 and N2 are
sollube to a noticeable extent due to London forces. However, CO2 is
somewat more soluble because it reacts with water to some extent to form
carbonc acid.
Aldwin Bolante
Ah, another devotee of carbonic acid. Can you provide some data that
might help a chemist characterize this mysterious substance? How about
some IR data, or maybe a 13C spectrum. Combustion data? Melting point?
How about a 1H nmr spectrum?
Thanks in advance.