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### Topic: Calculate the equilibrium concentration of CO2 in beer  (Read 1271 times)

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#### sci0x

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• Mole Snacks: +0/-0 ##### Calculate the equilibrium concentration of CO2 in beer
« on: February 10, 2020, 10:41:55 AM »
Question
At end of fermentation, beer is cooled to 2 degrees C. Height of beer in vessel is 20m, top pressure of 0.1 bar g of co2 is applied. Calculate the equilibrium conc of co2 in the beer in g/l assuming that a reasonable estimate of its average conc can be obtained by considerimg the prevailing pressure at the mid-point of beer in the vessel

Data:
Henrys constant at 2 degrees C = 84.1 x 106Pa (mole fraction) -1
Beer density = 1008kg m-3
Acc due to grav = 9.81 m s-2
Mol weight CO2 = 44
Mol mass beer = 18
Atm pressure = 1.013 bar 1 bar = 105 Pa

My workings:
Absolute pressure = Gauge press + Atmos press
Atmos press = 105 Pa
Co2 press = 0.1 bar g = 10000Pa
Abs press = 10,105 Pa

Hydrostatic pressure = absolute press + (density)(grav)(height at midpoint)
= 10,105 + (1008)(9.81)(10)
10,8989.8 Pa

Calc co2 conc by henrys law
P=KHC
C=P/KH
= 108989.8/84.1x10^6Pa
= 1.29 x 10^-3
Ans x 44 = 0.056 g/l

Can i get some help please, its from a past exam paper
Notes to help solve:
Atmos pressure co2 should be used in calc of absolute pressure from gauge pressure
Hydrostatic pressure should added for the pressure at the mid-point, calculated by multiplying density x acceleration due to grav x liquid height (10m)

Calc of Co2 conc is by henry's law
Mole fraction is converted to g/L

#### mjc123

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• Mole Snacks: +252/-11 ##### Re: Calculate the equilibrium concentration of CO2 in beer
« Reply #1 on: February 10, 2020, 01:04:21 PM »
Atmospheric pressure is not 105 Pa. Did the question say 105 Pa?

#### sci0x

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• Mole Snacks: +0/-0 ##### Re: Calculate the equilibrium concentration of CO2 in beer
« Reply #2 on: February 10, 2020, 01:34:27 PM »
It says 105 Pa, maybe supposed to be 10^5 Pa

Add atmospheric pressure of co2 is 110000

Add hydrostatic pressure for pressure at midpoint
Is 110,000 + (1008)(9.81)(10)
Is 208884.8
This should be the absolute pressure

If this is right i should be able to calc co2 conc by henrys law

So P = KHC
208884.8 Pa = 84.1 x 10^6 Pa (C)
C= 2.48 x 10^-3 moles / L

44g co2 in 1 mole
0.109 g in 2.48x10^-3 mole

Am i on the right track?
« Last Edit: February 10, 2020, 04:14:22 PM by sci0x »

#### peter77

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• Mole Snacks: +0/-0 ##### Calculate the equilibrium concentration of CO2 in beer
« Reply #3 on: August 08, 2020, 06:08:22 AM »
Question
At end of fermentation, beer is cooled to 2 degrees C. Height of beer in vessel is 20m, top pressure of 0.1 bar g of co2 is applied. Calculate the equilibrium conc of co2 in the beer in g/l assuming that a reasonable estimate of its average conc can be obtained by considerimg the prevailing pressure at the mid-point of beer in the vessel

Data:
Henrys constant at 2 degrees C = 84.1 x 106Pa (mole fraction) -1
Beer density = 1008kg m-3
Acc due to grav = 9.81 m s-2
Mol weight CO2 = 44
Mol mass beer = 18
Atm pressure = 1.013 bar 1 bar = 105 Pa

My workings:
Absolute pressure = Gauge press + Atmos press
Atmos press = 105 Pa
Co2 press = 0.1 bar g = 10000Pa
Abs press = 10,105 Pa

Hydrostatic pressure = absolute press + (density)(grav)(height at midpoint)
= 10,105 + (1008)(9.81)(10)
10,8989.8 Pa

Calc co2 conc by henrys law
P=KHC
C=P/KH
= 108989.8/84.1x10^6Pa
= 1.29 x 10^-3
Ans x 44 = 0.056 g/l

Can i get some help please, its from a past exam paper
Notes to help solve:
Atmos pressure co2 should be used in calc of absolute pressure from gauge pressure
Hydrostatic pressure should added for the pressure at the mid-point, calculated by multiplying density x acceleration due to grav x liquid height (10m)

Calc of Co2 conc is by henry's law
Mole fraction is converted to g/L
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#### mjc123

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• Mole Snacks: +252/-11 ##### Re: Calculate the equilibrium concentration of CO2 in beer
« Reply #4 on: August 08, 2020, 11:34:02 AM »
This question has been asked before. Was it by you under a different name? It's the same, down to the spelling mistakes.

#### MNIO

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• Mole Snacks: +12/-3 ##### Re: Calculate the equilibrium concentration of CO2 in beer
« Reply #5 on: August 08, 2020, 02:25:33 PM »
ALWAYS include units!!!

P @10m = P CO2 guage + P hydrostatic
= Patm + PCO2 + ρgh
Patm = 101325 Pa
PCO2 = 0.1 bar * (105 Pa / bar) = 10000 Pa
Phydro = (1008 kg/m3) * (9.80m/s2) * (10m) * (1Pa / 1kg/ms2) = 98784 Pa
Ptotal = 101325Pa + 10000 Pa + 98784 Pa = 210109 Pa = 2.10x105 Pa

then let's find kH... I'm going to use a reference value first to verify the 6.1g/L #
kH* for CO2 in H2O @ 25°C = 29 atm/M * (101325 Pa / atm) = 2.94x106 Pa/M @25°C
* note: that is from the wikipedia henry's law page.
then at 2°C,
kH @2°C = kH@25°C * exp((-dHsol/R)(1/T2 - 1/T1))
= 2.94x106 Pa/M * exp(-2400K)*(1/275K - 1/278K))
= 1.50x106 Pa/M

and finally
[CO2] = 2.10x105 Pa * (1M / 1.50x106 Pa) * (44.01g/mol) = 6.2 g/L

meaning.. we verified the answer of 6.1g/L using an alternate kH

*********
now let's go back to your solution
(1) ALWAYS INCLUDE UNITS !!!
(2) your atmospheric pressure is 105 Pa not 105 Pa.  101325 Pa
(3) take a look at your kH CO2 @ T = 2° value.. that's NOT Pa / M
that value is Pa per MOLE FRACTION of CO2 in the liquid phase < sigh>

**********
ok.. so let's see if we can solve this with YOUR kH value

We're going to simplify this by making 2 assumptions
(1) that beer is essentially water
(2) The mole fraction of CO2 in H2O is  going to be very small
(I can easily show - and you should be able to also - that 6g/L is about 0.14 mol CO2
per 55.5 mol H2O.  So that is a reasonable assumption).

making this equation approximately true.

mol CO2               mol CO2
χ CO2 = ----------------------- ≈ ------------
mol CO2 + mol H2O      mol H2O

and now I can rewrite that kH value as

84.1x106 Pa     84.1x106Pa * (mol H2O)
kH = -------------- ≈ ----------------------------
χ CO2                   (mol CO2)

and now we can finish up.

2.10x105 Pa                 mol CO2                  44.01g CO2     55.5 mol H2O
[CO2] = -------------- * ------------------------------ * -------------- * --------------- = 6.1 g / L
1               (84.1x106 Pa) * (mol H2O)     mol CO2            1 L H2O
« Last Edit: August 08, 2020, 04:59:29 PM by MNIO »

#### billnotgatez

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• Gender:  ##### Re: Calculate the equilibrium concentration of CO2 in beer
« Reply #6 on: August 08, 2020, 11:03:56 PM »
This question has been asked before. Was it by you under a different name? It's the same, down to the spelling mistakes.

Topics merged

#### Borek ##### Re: Calculate the equilibrium concentration of CO2 in beer
« Reply #7 on: August 09, 2020, 03:49:48 AM »
This question has been asked before. Was it by you under a different name? It's the same, down to the spelling mistakes.

Good catch, but IPs from all around the world so I am ready to believe these were different people.
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#### Enthalpy ##### Re: Calculate the equilibrium concentration of CO2 in beer
« Reply #8 on: August 10, 2020, 11:38:23 AM »
Why do you compute a hydrostatic pressure?

Only because the question misleads you?

If CO2 is produced within the vessel and didn't diffuse, you might compute some concentration distribution as limited by the formation of bubbles, but this is not an equilibrium concentration. This equilibrium is set by the gas pressure over the liquid.

#### MNIO

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• Mole Snacks: +12/-3 ##### Re: Calculate the equilibrium concentration of CO2 in beer
« Reply #9 on: August 10, 2020, 03:03:22 PM »
Think about it like this.  Let's say you have a column of water 20m tall in a tank with a small headspace and a pressure of 1 atm air in that headspace.  Now you add 0.1 atm of CO2 to make the total pressure = 1.1 atm.

Then, CO2 starts dissolving in the water and diffusing down into the water.  As the CO2 starts dissolving and diffusing into the water, the pressure in the vapor phase will drop.  So you add more CO2 gas to bring the pressure back to 1.1 atm.  The deeper the CO2 diffuses to in the column of water, the more pressure it experiences (Pobserved = Patm + P hydrostatic).  Since the amount that can dissolve depends on pressure, the deeper you go, the more that can dissolve AND the more CO2 you must add to the vapor phase to bring the pressure back to 1.1 atm

Eventually (through patience or stirring) the system reaches equilibrium.  Where the CO2 in the water is saturated and the vapor phase is stabilized at 1.1 atm.  That's the situation we have here.

You have to assume the solution is saturated with CO2 AND the pressure of the vapor phase is 1.0atm air + 0.1atm CO2.... AT EQUILIBRIUM.

And I suppose if it helps, imagine a perfectly stirred vat of beer and the dissolution of CO2 in the beer is infinitely fast while the addition of CO2 to the tank is slow.  You're going through a valve.  So that as you add CO2 it instantly dissolves.  And when you reach 1.1 atm, the solution is just reaching saturation level.
« Last Edit: August 10, 2020, 04:23:26 PM by MNIO »

#### mjc123

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• Mole Snacks: +252/-11 ##### Re: Calculate the equilibrium concentration of CO2 in beer
« Reply #10 on: August 10, 2020, 05:58:42 PM »
Quote
Since the amount that can dissolve depends on pressure, the deeper you go, the more that can dissolve
The concentration of dissolved gas is proportional to the partial pressure of that gas over the liquid. It does not depend on the total pressure, whether atmospheric or hydrostatic. Thus in this instance the concentration of CO2 is KH*0.1 atm. At equilibrium, this concentration will be the same throughout the liquid, and will not vary with depth.
As you have shown that, on your assumptions, you get the required answer, I can only assume that the person who set the question didn't understand it.

#### MNIO

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• Mole Snacks: +12/-3 ##### Re: Calculate the equilibrium concentration of CO2 in beer
« Reply #11 on: August 11, 2020, 04:12:48 PM »
Mjc123

Quote
Since the amount that can dissolve depends on pressure, the deeper you go, the more that can dissolve
The concentration of dissolved gas is proportional to the partial pressure of that gas over the liquid. It does not depend on the total pressure, whether atmospheric or hydrostatic. Thus in this instance the concentration of CO2 is KH*0.1 atm. At equilibrium, this concentration will be the same throughout the liquid, and will not vary with depth.
As you have shown that, on your assumptions, you get the required answer, I can only assume that the person who set the question didn't understand it.

that is NOT CORRECT.

Think about the headspace of that tank.  The pressure is a result of the molecules of CO2 striking the container walls and reversing direction.  The more molecules per unit volume, the higher the pressure.  The higher the KE of the CO2 molecules (i.e. the higher the temp) the higher the pressure.  Agreed?

Now think about a bubble of CO2 1" below the surface of the water.  The pressure of that bubble will be related to the number of CO2 particles per unit volume in that bubble and their KE.   Just like the headspace, the pressure is related to the collisions of those particles with the container walls.  The walls of that container is the water/gas interface.

Now think about a bubble of CO2 10m deeper in the water.  The pressure will be an additional 1 atm of pressure because the weight of the water is pressing against that gas bubble and the gas bubble is pressing back.  To compensate for the increased external pressure, the gas pressure in that water molecule must increase.  Assuming KE is the same and # gas particles is the same, the volume must decrease to make pressure increase.  And in the end, we get a smaller gas bubble with a higher internal pressure.  Now that gas bubble doesn't "know of" nor have any "knowledge of" or care WHY the pressure is increased, it's simply experiencing a higher pressure. To the gas bubble.... this is NO different that a higher pressure in the headspace of the tank.

at higher pressure, more gas particles are forced through the water/gas interface into the liquid phase and are dissolved by the liquid (this is henry's law).  So, when the internal pressure of the gas bubble is higher, more CO2 dissolves in the surrounding water.

Therefore, CO2 solubility increases as depth increases.

********
now put it all together.
(1) We assume WELL MIXED and AT EQUILIBRIUM.
(2) at depth = x, you have Pressure on that CO2 bubble = PCO2 + Patm + Phydrostatic
(3) the AVERAGE depth ~ the AVERAGE pressure on the AVERAGE CO2 bubble.
« Last Edit: August 11, 2020, 04:24:10 PM by MNIO »

#### Borek ##### Re: Calculate the equilibrium concentration of CO2 in beer
« Reply #12 on: August 11, 2020, 04:52:03 PM »
Now think about a bubble of CO2 10m deeper in the water.

At equilibrium there is no bubble.
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#### MNIO

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• Mole Snacks: +12/-3 ##### Re: Calculate the equilibrium concentration of CO2 in beer
« Reply #13 on: August 11, 2020, 05:02:11 PM »
Now think about a bubble of CO2 10m deeper in the water.

At equilibrium there is no bubble.

do you think solubilty of CO2 at depth depends on the size of the bubble present?

********
now imagine a tank with vigorous mixing
vs
a tank with no mixing

what's the difference in bubbles, solubility and TIME to equilibrium
« Last Edit: August 11, 2020, 05:50:45 PM by MNIO »

#### Borek ##### Re: Calculate the equilibrium concentration of CO2 in beer
« Reply #14 on: August 12, 2020, 03:16:51 AM »
You can't have both equilibrium and concentration gradient at the same time (at least as long as we don't talk about some highly exotic cases).
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