Alt.survival removed form newsgroups.
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On Mar 14, 2:50 pm, "stuart.g…@comcast.net"
<stuart.g…@comcast.net> wrote:
> fredfigh…@spamcop.net wrote:
> > Isn’t the CO2 concentration in the ocean well
> > below saturation? Equilibrium and saturation
> > are the same thing.
> Equilibrium:
> K = {X]/[Y] = constant, for [X] and [Y] that can vary. Change [X], and
> [Y] seeks a new value until the ratio of [X] to [Y] is K.>
> Saturation:
> [Y] = constant, regardless of the value of [X], or visa versa. K is NOT
> constant.
Wrong. K remains constant. Keeping in mind that our
system is gas over water, if [Y] is kept constant then
[X] changes until [X]/[Y] = K. [X] changes by dissolution
of X from the atmosphere above, or the release of X from
the solution.
The equilibrium constant doesn’t change with concentration,
that is why it is called a CONSTANT. For gas and water
K is inversely proportionate to temperature and directly
proportionate to partial pressure of the gas and for the range
of temperatures and partial pressures of interest here can be
assumed to have the simple relationship of K = k p/T,
where k is the product of the two constants of proportionality
for the separate temperature and partial pressure relationships.
NET dissolution, that is net flow of X from the atmosphere
to the solution, occurs when [X]/[Y] < K. Net evolution,
that is net flow of X from the solution to the atmosphere
occurs if [X]/[Y] > K.
> > You seem to think that the measured concentrations
> > [of CO2 in ocean water, FF]
> > represent an equilibrium condition. Why?
> Probably he thinks that because it is the definition of equilibrium, if
> given enough time for equilibrium to be reached.
I’ll wait to see if Mr Hughes concurs.
> Mind you, I used numbers that were known to be at equilibrium, and I
> reached the same conclusion that the ocean warming is the cause of the
> increased CO2 concentrations in the atmosphere.
IIUC what you did yes, you did use equilibrium numbers. You
used concentration at saturation from the CRC handbook.
You did NOT use the concentrations observed in ocean water,
right?
> >>As various processes remove CO2 from solution in the ocean
> >>(primarily by the formation of carbonates), more CO2 is dissolved at the
> >>liquid/gas interface.
> >>If the temperature of the water increases, the equilibrium will shift by
> >>releasing CO2 from solution.
> > Only if the increase in temperature causes a shift
> > to the other side (supersaturation) of equilibrium.
> Super saturation?!
Yes. Unless the solution is supersaturated at the temperature and
partial pressure at the air/water interface, the solute will not
evolve.
> > Equilibrium occurs at saturation. That is the
> > definition of saturation,
> This is pointless. It would take a month of chemistry lessons to bring
> you up to speed, and I’m pretty sure that you would resist every single
> step to reach a common language.
Splorf. In order to support your calculation you had to redefine
the equilibrium constant to be a variable that depends on the
concentration of the solute.
Think about that.
–
FF
- Hide quoted text — Show quoted text -
fredfigh…@spamcop.net wrote:
> Alt.survival removed form newsgroups.
> Alt.chem added to newsgroups.
> On Mar 14, 2:50 pm, "stuart.g…@comcast.net"
> <stuart.g…@comcast.net> wrote:
>>fredfigh…@spamcop.net wrote:
>>>Isn’t the CO2 concentration in the ocean well
>>>below saturation? Equilibrium and saturation
>>>are the same thing.
>>Equilibrium:
>>K = {X]/[Y] = constant, for [X] and [Y] that can vary. Change [X], and
>>[Y] seeks a new value until the ratio of [X] to [Y] is K.>
>>Saturation:
>>[Y] = constant, regardless of the value of [X], or visa versa. K is NOT
>>constant.
> Wrong. K remains constant. Keeping in mind that our
> system is gas over water, if [Y] is kept constant then
> [X] changes until [X]/[Y] = K. [X] changes by dissolution
> of X from the atmosphere above, or the release of X from
> the solution.
> The equilibrium constant doesn’t change with concentration,
> that is why it is called a CONSTANT. For gas and water
> K is inversely proportionate to temperature and directly
> proportionate to partial pressure of the gas and for the range
> of temperatures and partial pressures of interest here can be
> assumed to have the simple relationship of K = k p/T,
> where k is the product of the two constants of proportionality
> for the separate temperature and partial pressure relationships.
> NET dissolution, that is net flow of X from the atmosphere
> to the solution, occurs when [X]/[Y] < K. Net evolution,
> that is net flow of X from the solution to the atmosphere
> occurs if [X]/[Y] > K.
I’m sorry, you’re just being stupid. I explained to you the difference
between equilibrium and saturation, my whole point being that they mean
two different things, and you just ignored that was the point and
pretended that I meant them to be the same.
Talking to you is like talking to a rock – You don’t grasp the concept.
On Mar 14, 8:15 pm, "stuart.g…@comcast.net"
- Hide quoted text — Show quoted text -
<stuart.g…@comcast.net> wrote:
> fredfigh…@spamcop.net wrote:
> > Alt.survival removed form newsgroups.
> > Alt.chem added to newsgroups.
> > On Mar 14, 2:50 pm, "stuart.g…@comcast.net"
> > <stuart.g…@comcast.net> wrote:
> >>fredfigh…@spamcop.net wrote:
> >>>Isn’t the CO2 concentration in the ocean well
> >>>below saturation? Equilibrium and saturation
> >>>are the same thing.
> >>Equilibrium:
> >>K = {X]/[Y] = constant, for [X] and [Y] that can vary. Change [X], and
> >>[Y] seeks a new value until the ratio of [X] to [Y] is K.>
> >>Saturation:
> >>[Y] = constant, regardless of the value of [X], or visa versa. K is NOT
> >>constant.
> > Wrong. K remains constant. Keeping in mind that our
> > system is gas over water, if [Y] is kept constant then
> > [X] changes until [X]/[Y] = K. [X] changes by dissolution
> > of X from the atmosphere above, or the release of X from
> > the solution.
> > The equilibrium constant doesn’t change with concentration,
> > that is why it is called a CONSTANT. For gas and water
> > K is inversely proportionate to temperature and directly
> > proportionate to partial pressure of the gas and for the range
> > of temperatures and partial pressures of interest here can be
> > assumed to have the simple relationship of K = k p/T,
> > where k is the product of the two constants of proportionality
> > for the separate temperature and partial pressure relationships.
> > NET dissolution, that is net flow of X from the atmosphere
> > to the solution, occurs when [X]/[Y] < K. Net evolution,
> > that is net flow of X from the solution to the atmosphere
> > occurs if [X]/[Y] > K.
> I’m sorry, you’re just being stupid. I explained to you the difference
> between equilibrium and saturation, my whole point being that they mean
> two different things, and you just ignored that was the point and
> pretended that I meant them to be the same.
> Talking to you is like talking to a rock – You don’t grasp the concept.
How about if you work through an example for me?
Suppose I put a small open beaker of pure water (no
dissolved gases) into a large room with a pure CO2
atmosphere at one atmosphere pressure. I say large
and small so that the dissolution of CO2 and the
evaporation of water do not significantly affect the
partial pressure of the CO2 in the room.
Allow the system to equilibrate at 25 degrees C.
At equilibrium, what is the concentration of the
CO2 in the water?
What would be the concentration if the water
were saturated with the CO2, instead of being
equilibrium?
You can find the necessary data here:
http://dwb.unl.edu/Teacher/NSF/C09/C09Links/www.chem.ualberta.ca/cour…
Table: Molar Henry’s Law Constants for Aqueous Solutions at 25oC
Gas Constant Constant
(Pa/(mol/dm3)) (atm/(mol/dm3))
…
CO2 2.937 x 10+6 29.76
…
–
FF