# Category Archives: James R. Barrante

Author of best selling supplement, Applied Mathematics for Physical Chemistry, 3rd Edition, James R. Barrante has a new textbook, Physical Chemistry for the Biological Sciences, an iBook for iPads, iPhones, or Macs with OS X Maverick.  To download a sample chapter, go to the iBookstore.  At \$9.99 it makes a great supplement for any physical chemistry course.   http://itunes.apple.com/book/physical-chemistry-for-the-biological-sciences/id900944183?mt=13

August 19, 2014 · 1:15 am

## Carbon Dioxide and Henry’s Law

by James R. Barrante, Ph.D.

There have been a number of unsuccessful attempts to relate the level of carbon dioxide in the atmosphere to the concentration of dissolved CO2 in the oceans using Henry’s Law.  The Henry’s Law equation is a simple linear relationship between the partial pressure of CO2 in the atmosphere to the activity of dissolved CO2 gas

Dissolved carbon dioxide reacts with water to form a weak acid, carbonic acid.  Consequently, the Henry’s Law constant k is not affected significantly by the high ionic strength of seawater.  The activity coefficient of undissociated carbonic acid is close to unity, so the activity of dissolved CO2 can be replaced by its concentration in moles per liter.  This, of course, is not true for the dissociation constants of carbonic acid.  (See Ocean Acidification I and II).   Like all equilibrium constants, the Henry’s Law constant is temperature dependent.  Therefore, an average partial pressure of atmospheric CO2 must be carefully calculated.  That is, it is not acceptable to determine the partial pressure of atmospheric CO2 at various points around the globe and average them together.

One must first recognize that the temperatures of the oceans are not randomly distributed around the globe, but are banded in zones following closely to the latitude lines.  Moreover, the latitude zonal surface seawater areas differ at least two ways.  First, from the equator to each pole the total surface area around the globe is different, and second, each area contains a different amount of land mass.  Consequently, before any calculations can be done, the total relative surface area of each zone must be determined.  This author found that a zonal width of 5˚ was adequate.  Since we are concerned with relative surface area, it is not necessary to use actual values.  The surface area of a zone is easily calculated using the equation

S  =  2 π r h

where is the radius of the globe and h is the height between successive latitude lines.  Table 1. shows the surface area of each zone on a globe of radius 50.0 units.  Notice that the difference in surface area between the first three zones is for all practical purposed zero.  We find that is it not necessary to go much beyond 80˚ north and south.

Table 1.

Latitude Range      Difference in Latitude          Surface Area
(degrees)                                       Lines                          (square units)
0˚ to 5˚                                           4.2                                               1319
5˚ to 10˚                                          4.2                                               1319
10˚ to 15˚                                         4.2                                                1319
15˚ to 20˚                                         4.1                                                1288
20˚ to 25˚                                         4.1                                                1288
25˚ to 30˚                                         3.8                                                1194
30˚ to 35˚                                         3.8                                                1194
35˚ to 40˚                                         3.5                                                1100
40˚ to 45˚                                         3.4                                               1068
45˚ to 50˚                                         3.0                                                 942
50˚ to 55˚                                          2.5                                                  785
55˚ to 60˚                                         2.4                                                  754
60˚ to 65˚                                         2.1                                                  660
65˚ to 70˚                                          1.7                                                  534
70˚ to 75˚                                          1.4                                                  440
75˚ to 80˚                                          0.9                                                  283

Once the surface area of each zone is found, it is necessary to determine the fraction of that area that is water and the average temperature of that water.  Figure 1. shows a grid of the globe designed to do this.   The average temperature of

Figure 1.  Planar grid of globe.

each zone was found by layering the grid shown in Figure 1 on top of a SST map such as those produced by the National Weather Service Environmental Modeling Center for 2014.  Table 2. shows the fraction of each zone that is water and the average temperature of each zone.

Table 2.

Knowing the average sea surface temperature of each zone and using the equation describing the variation of the Henry’s Law constant with respect to temperature, we can determine the partial pressure of CO2 associated with each zone.  To do this, we must assume a level of dissolved CO2.  From the post Ocean Acidification II, a reasonable average value is 1.36 x 10-5M. By multiplying the fraction of ocean by the surface area of a particular zone, we can determine the zone’s contribution of partial pressure CO2 to the average total pressure of CO2.    The results of these calculations are shown in Table 3.

Table 3.

We see that the total weighted average pressure of CO2 is 347.2 ppm, not too bad, considering the uncertainty in the concentration of dissolved CO2.  Keep in mind that this is an equilibrium value that may take a couple hundred years to be established. If the instantaneous level of CO2 is measured to be around 400 ppm, this would indicate that the contribution of CO2 from all other sources, including burning fossil fuels, is about 50 ppm.

It is important to note the silliness in measuring CO2 levels around the globe and assuming that the average has some statistical meaning.  Look at the data in Table 3.  Simply based on ocean temperature around the equator, CO2 levels are well over 400 ppm, while levels in Antarctica are similar to those found during ice ages.  This, obviously, has to be the case to get an average somewhere between these two extremes.  This really puts into question the validity of ice-core data as a proxy for “global” CO2 levels.  It is difficult to understand how samples of ice taken from Antarctica could represent global CO2 levels in violation of Henry’s Law.

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## Thermal Behavior of CO2

Carbon dioxide is a triatomic molecule, but because it is linear, it behaves in many respects like a diatomic molecule.  In fact, the similarity between CO2 and other diatomic molecules is so striking that the molecule can be described as a Hooke’s Law harmonic oscillator, a model that generally works well only for diatomic molecules.

It is well-known that the heat capacity ratio,CP /CV, for most small molecule gases at room temperature or below, calculated using classical equipartition theory, does not agree well with the ratios determined experimentally.  For example, the calculated heat capacity ratio of N2 gas at 25˚C is 1.28, whereas the experimentally determined value is 1.39.  Contrast this to a monatomic gas like argon.  The calculated heat capacity ratio of Ar at 25˚C is 1.67, while the experimentally determined value is 1.62, very good agreement.  The general explanation for the difference between calculated and experimental heat capacity ratios in molecular systems is justified by the difference between the classical and quantum mechanical descriptions of the systems.  This is supported by the fact that as the temperature of N2 gas, for example, is increased, the capacity ratio begins to approach the calculated value of 1.28.

To understand this, consider the following:  according to classical equipartition theory, at any temperature molecules absorb thermal energy through molecular collisions and distribute this energy equally into their various molecular degrees of freedom.  In order for this to work, molecules must be able to absorb energy continuously.  The important thermal molecular degrees of freedom for gases are translational degrees, the motion of the molecule from one point in space to another as specified by its center of mass, rotations of atoms about various axes in the molecule, as specified by moments of inertia, and vibrations of atoms in molecules along certain allowed normal coordinate axes.  Additionally, one must consider that translational motion and rotational motion only involve kinetic energy, whereas vibrational motion involves both kinetic energy and potential energy.  Classically, all these modes of motion absorb energy continuously, and, according to the equipartition theory, are assigned an amount of energy equal to 1/2 kT for each degree, where k is the Boltzmann constant (1.381 x 10-23 J/K) and is the absolute temperature.  As a reminder, remember that monatomic gases possess only translational degrees, and solids possess (for the most part) only vibrational degrees.

To assign the position of the center of mass of a molecule, we need three coordinates.  Consequently, there are three translational degrees of freedom.  Assigning a value of 1/2 kT for each gives us a total of 3/2 kT energy for the translational degrees of freedom.  If a molecule is linear, there is no moment of inertia about the bond axis (assumed to be the x-axis).  Therefore, rotational degrees only involve the y– and z-axes, giving 2/2 kT energy for the rotational degrees of freedom (linear molecule).  Of course, for a non-linear molecule there are three rotational degrees of freedom, giving 3/2 kT energy for the rotational degrees of freedom (non-linear molecule).  We generally determine the vibrational degrees by difference.  Originally, equipartition theory was described in terms of a single point mass, giving a total of 3 degrees of freedom.  To extend this to a molecule, thought of as a collection of point masses, one needs to assign an x-, y, and z-coordinate for each atom in the molecule.  Consequently, a molecule has a total of 3N degrees of freedom, where N represents the total number of atoms in the molecule.

Let us apply this to N2 gas to see how all this works.  Nitrogen gas is a linear molecule and should have 3N  = 3(2) = 6 total degrees of freedom.  But 3 of those degrees are translational degrees and 2 of those are rotational degrees, so 6 – 5 =  1 must be a vibrational mode.  Generalizing these rules:

Translation:  3 degrees of freedom

Rotation:  (linear molecule) 2 degrees of freedom; (non-linear molecule) 3 degrees of freedom

Vibration:  3N – 5  (linear);  3– 6  (non-linear)

We now can calculate the heat capacity ratio of N2:

Translational Energy:    3/2 kT

Rotational Energy:  2/2 kT

Vibrational Energy:  (1 mode)  1/2 kT for kinetic energy; 1/2 kT for potential energy

Total Energy:  7/2 kT

To find the heat capacity at constant volume:

If we assume that N2 is an ideal gas, then  CP  =  CV  +  k  =  9/2 k.  The heat capacity ratio then is

γ  =  CP /CV  =  9/7  =  1.28

Application of Quantum Theory

We know that for all practical purposes translational energy states for all gases are so closely spaced that we can consider them as being continuous.  The same can be said for the rotational energy of most gases, particularly at room temperature.  However, this is not the case for vibrational energy states.  For many diatomic gases, such as nitrogen, the first excited vibrational energy state is separated from the ground state by a value that is too large for the thermal energy around room temperature, approximately 2500 J/mol, to excite nitrogen molecules to higher vibrational states.  Consequently, in the case of nitrogen, thermal energy is partitioned only between translational and rotational degrees.

Translational energy:  3/2 kT

Rotational energy:  2/2 kT

Total energy:  5/2 kT

Therefore:       CV  =  5/2 kCP  =  7/2 k

Ratio  =  7/5  =  1.40, which is very close to the experimentally determined value.

Application of Quantum Statistical Mechanics to Carbon Dioxide

As previously described, carbon dioxide can be treated as a Hooke’s Law diatomic molecule, because it is linear. We find that CO2 has four vibrational modes:  a symmetric stretching mode at 1388 cm-1  , an asymmetric stretching mode at 2349 cm-1  , and a degenerate pair of bending modes at 667 cm-1  .  We can apply the Boltzmann equation to determine the fraction of CO2 molecules in the first excited state for each vibrational mode.  Keep in mind that we are dealing with thermal excitation here, not light excitation.  It does not matter in this analysis whether the modes are infrared active or not.  Recognizing that ∆E  =  hc/λ, where h is Planck’s constant, c is the speed of light, and 1/λ is the wave number (in reciprocal meters), we have at 15˚C (288K)

Symmetric Stretch:  fraction in first excited state  =

=

=     =   9.94 x 10-4

Asymmetric Stretch:  fraction  =  8.22 x 10-6

Each Bending Mode:     fraction  =  3.60 x 10-2

This is a surprising result.  While the asymmetric stretching mode is similar to most diatomic molecules, indicating that thermal energy would not promote CO2 molecules to higher energy states in this mode, this is not the case for the other two modes.  In fact, for the bending modes (3.6 x 2) about 7% of the molecules are already in the first excited state due to thermal collisions, irrespective of the presence of infrared radiation.  To check these numbers for accuracy, let us calculate the heat capacity ratio of CO2 and compare it to the experimental value.  The contributions for translation and rotation are the same as they are in the classical case.  The contributions for vibration are not that straight forward.  Certainly, we can eliminate the contribution from asymmetric stretch.  The contributions from the other modes must be calculated using statistical mechanics.  The appropriate equation to do this is

where R is the gas constant 8.314 J/mol·K.  This will give the heat capacity for a vibrational mode in joules per mole rather than in joules per molecule.  The following contributions are:

Symmetric stretch:    CV  =  0.397 J/mol·K

Bending modes:    CV   =  3.57  J/mol·K

The contributions from translation and rotation (per mole) are 3/2 R + 2/2 R  =  20.79 J/mol·K.  We now add all contributions, keeping in mind that there are two bending modes:

CV(total)  =  20.79  +  0.397  +  3.57  +  3.57  =  28.32 J/mol·K

Assuming CO2 is an idea gas (not a great assumption):    C =  36.64 J/mol·K

Ratio  =  1.294 at 15˚C

The actual published heat capacity ratio of CO2 at 15˚C is 1.304, which is quite close, considering we assumed that the gas was ideal.

Discussion

The band of radiation that is infrared active and important in the greenhouse gas scenario is the bending mode at 667 cm-1 (15 microns).  Symmetric stretch, while able to store a small amount of energy (about 229 J/mol) is not IR active.  The fact that about 7% of the CO2 molecules are already in the first excited state at 15˚C due to molecular collisions will affect their ability to absorb IR radiation.  In fact, even in the absence of a source of IR radiation at 15 microns, CO2 will be radiating a small amount of 15 micron radiation.

If we look at the variation of this effect with respect to temperature, we find that from approximately 250K to 300K the fraction of molecules of CO2 in the first excited vibrational bending mode states changes from about 4.4% to 8.0%, which is relatively small, and most likely would not have a large impact on the molecule’s ability to absorb 15 micron IR radiation.  Nevertheless, this does not justify ignoring it, and it should be included in any calculations dealing with the greenhouse gas behavior of CO2.

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