EAS 3030 9/16/08 LAD

Weathering reactions

We can consider many weathering reactions as acid-base reactions. Acidity in soils can be generated from several sources. The main ones are:

• Hydration of CO2 (yields carbonic acid): CO2 + H2O  H2CO3 • Organic decomposition to yield organic acids (R-COOH  R-COO- + H+)

• Sulfide oxidation to yield sulfuric acid (2FeS2+4H2O+7.5O2  4H2SO4+Fe2O3) • Atmospheric deposition of mineral acids such as H2SO4 and HNO3.

The hydrogen ion released by these acids can react with carbonate minerals (e.g.

CaCO3) or alumino-silicate minerals (feldspars, micas, clays ….). Carbonate and silicate weathering reactions can be considered separately:

For carbonates, we can write:

++ - 1) CaCO3 + H2 CO3  Ca + 2HCO3

In this reaction carbonic acid (from CO2 and water) and calcium carbonate react to form bicarbonate ion and calcium ion. The net reaction yields no H+, so the carbonic acid has been neutralized by reaction with carbonate. It also yields no solid phases – the carbonate mineral undergoes complete dissolution. This sort of reaction is know as “congruent” dissolution.

An example silicate reaction would be the reaction of potassium feldspar (a primary mineral) to form kaolinite, a secondary mineral . The reaction of a primary mineral to form a secondary mineral plus solutes is called “incongruent” dissolution.

+ + 2) 2KAlSi3O8(Kspar) + 2H + 9H2O  Al2Si2O5(OH)4(kaol) + 4H4SiO4(aq) +2K

If the source of acidity is CO2, we can write equivalently:

+ - 3) 2KAlSi3O8(Kspar) + 2CO2 + 11H2O  Al2Si2O5(OH)4(kaol) + 4H4SiO4(aq) +2K + 2HCO3

In the breakdown of K-feldspar, acidity is consumed. We can see that H+ is used up and K+ is released to solution, a kind of “ion exchange” reaction. Also released to solution is dissolved silica or silicic acid, written H4SiO4.

As a general observation, we can say that acids are neutralized (H+ consumed) by weathering minerals. The reaction products are “base cations” (Na+, K+, Ca++, Mg++), and - if CO2 is the acid another product will be bicarbonate ion (HCO3 ). In the case of alumino-silicate weathering, silicic acid (a.k.a. “dissolved silica”) is released as well. Secondary alumino-silicate minerals are typically clays, and have increased hydroxyl and/or water content, relative to their primary precursors.

1 EAS 3030 9/16/08 LAD Because of its low solubility in typical environments, aluminum is often largely retained in the secondary mineral. It is sometimes a useful approximation to treat Al as a “conservative” element, i.e. one that does not enter or leave the system of interest. We can then normalize the other elements in a mineral to the Al content to examine net mass transfer. For example, in reaction 2, we can write:

Table 1 Si/Al K/Al H/Al K-feldspar 3 1 0 kaolinite 1 0 2

Since Al is neither gained nor lost in this reaction, it serves as an index element for the others. The secondary kaolinite is depleted in Si and K relative to the primary mineral, but enriched in H (as hydroxyl). Typically secondary clay minerals have increased molar volumes (lower densities) than primary alumino-silicates.

An equilibrium constant for reaction #2 can be written 4):

2 + 4 [K ] [H4 SiO4 ] K'= 2 [H + ]

by taking the square root of both sides we can reduce it to 5) ! + [K ] 2 K = [H4 SiO4 ] [H + ]

Taking logarithms, we can write

" [K + ]% 6) logK = log$ ' + 2log[H4 SiO4 ] # [H + ]&

+ + We can make a plot with log[H4SiO4] for the “x” axis and log([K ]/[H ]) for the “y” axis. Then equation 6 has the form of a straight line y = mx + b with a slope m of -2 and a “y” int!e rcept of logK:

" [K + ]% 7) log$ ' = (2log[H4 SiO4 ]+ logK # [H + ]&

The value of logK can be found in some tables of geochemical data, or calculated from thermodynamic data for the Gibbs free energy of formation (∆G˚f) of the reactants and pr!oduc ts:

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8) ∆G˚reaction = ∑∆G˚f(products) - ∑∆G˚f(reactants)

!"G 0 9) log K = r 2.303RT

-1 -1 -1 The value of R = 8.3143 J mol K , so if the ∆G˚f data are given in kilojoules per mole (kJ/mole) and the temperature is 25˚C (298.15 K), equation 9 can be simplified to:

-1 10) logK = - ∆G˚reaction/5.708 kJ mol .

Gibbs free energy of formation data for some of the phases we will work with are in the table below:

Table 2 0 phase formula ∆Gf , kJ/mole gibbsite Al(OH)3 -1151 kaolinite Al2Si2O5(OH)4 -3800 pyrophyllite Al2Si4O10(OH)2 -5275 muscovite KAl3Si3O10(OH)2 5606

K-feldspar KAlSi3O8 -3767 albite NaAlSi3O8 -3715 analcite NaAlSi2O6•H2O -3090

Na-beidellite Na0.33Al2.33Si3.67O10(OH)2 -5382 K+ -283.27 Na+ -261.91

H4SiO4 -1316.6

H2O -237.13 Ca2+ -553.58 H+ 0

Data from J. I. Drever, 1997, The Geochemistry of Natural Waters, Prentice Hall

For reaction 2 (Kspar -> kaolinite) the ∆G˚reaction is given by the “products minus reactants” (with proper stoichiometric weighting), and will be:

11) ∆G˚reaction = [-3800 + 2(-283.27) + 4(-1316.6)] - [2(-3767) + 2(0) + 9(237.13)] = 35.23 kJ mol-1.

Using equation 9:

-1 -1 -1 12) log K’Kspar-kaol = -35.23 kJ mol /(2.303 × 8.3143 kJ mol K × 298.15 K) = -6.17

3 EAS 3030 9/16/08 LAD Since K = √K’, from the properties of logarithms we have that

13) log K = log K’/2 i.e. log K = -6.17/2 = -3.09

Note that, although we found it convenient to reduce the order of the equation 4 to equation 5, we must still calculate ∆G˚reaction according to the stoichiometry we have written for equation 2. But, note also we could have equally well written equation 2 in an alternative but still balanced fashion:

+ + 14) KAlSi3O8 + H + 4.5H2O  0.5Al2Si2O5(OH)4 + K + 2 H4SiO4 which would lead directly to the definition of K in equation 5. The two approaches are equivalent and thermodynamically consistent.

Now we have numerical values for both the slope of this reaction line (-2) and it’s + + y-intercept (-3.09) on a plot of log([K ]/[H ]) vs. log([H4SiO4]). What does this look like? See Figure 1.

Follow the accompanying Powerpoint graphics with the equations below as we build the stability relations for the system K-Al-Si-H2O.

In order to complete the system, we will consider the following phases and the reactions between them: K-feldspar, kaolinite, muscovite and gibbsite. We will also include opal (amorphous silica).

The next reaction we write is kaolinite to gibbsite:

15) Al2Si2O5(OH)4 (kaol) + 5H2O  2Al(OH)3 (gibb) + 2 H4SiO4

Unlike the K-spar reaction, this is not an acid-base reaction. Gibbsite is an “end-product” of weathering, completely stripped of base cations, and silica, leaving only the highly insoluble aluminum hydroxide. It is typically found in highly weathered tropical soils.

The equilibrium constant for this reaction is simply;

2 16) K’(kaol-gibb) = [H4SiO4]

From the data in Table 2, we can write -1 17) ∆G˚reaction = [-3800 + 5(-237.13)] - [2(-1151) + 2(-1316.6)] = 50.45 kJ mol .

From equation 10 we obtain

-1 -1 18) log K’(kaol-gibb) = -50.45 kJ mol /(5.708 kJ mol ) = -8.8

19) K = √K’= [H4SiO4]

4 EAS 3030 9/16/08 LAD 20) log K = logK’/2 = -4.4 (i.e has a numerical value of 10-4.4).

+ + On the plot of log([K ]/[H ]) vs. log[H4SiO4], the equilibrium boundary appears as a straight vertical line (i.e. undefined slope) with an “x-intercept” of –4.4 (Figure 2).

The next reaction we add is between muscovite and K-feldspar. Muscovite is a mica, and can be a primary mineral. The mineral illite, a fine-grained clay is chemically and structurally very similar to muscovite, and is commonly a secondary mineral. Here we use the thermodynamic data for muscovite.

+ + 21) 3KAlSi3O8(Kspar) + 2H + 12H2O  Kal3Si3O10(OH)2 (muscl) + 4H4SiO4(aq) +2K

The equilibrium constant is

[K + ]2 22) K'= [H SiO ]6 [H + ]2 4 4

+ [K ] 3 23) KKspar"musc = + [H4 SiO4 ] ! [H ]

Using data from Table 2 we find that ∆Grxn = 74.42 kJ mol-1, logK’ = -13.0, and logKkspar-musc = -6.5. As for equation 7, we can write: ! + + 24) log([K ]/[H ]) = -3log[H4SiO4] + logKkspar-musc

Thus, on our chosen axes, the reaction boundary between K-spar and muscovite will have a slope of –3 and a y-intercept of –6.5 (Figure 3). Because they have different slopes, the K-spar - muscovite reaction boundary intercepts the K-spar - kaolinite line. Thus, depending on both initial conditions and the reaction path, K-spar could react to form either muscovite or kaolinite. As we complete the diagram we will be better able to see which reaction is relevant for a given set of conditions.

Muscovite may also react to form gibbsite:

+ + 25) KAl3Si3O10(OH)2 (muscl) + H + 9H2O  3Al(OH)3 (gibb) + K + 3H4SiO4

The equilibrium constant is

[K + ] 26) K = [H SiO ]3 musc"gibb [H + ] 4 4 and we obtain directly

! + + 27) log([K ]/[H ]) = -3log[H4SiO4] + logKmusc-gibb

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From table 2, the value of ∆Grxn = 74.42 kJ mol-1, and logKmusc-gibb = -9.5. The muscovite-gibbsite reaction boundary will plot as a line with a slope of –3 and an intercept of –9.5, and is parallel to the K-spar-muscovite line (figure 4).

Muscovite may also react to form kaolinite:

+ + 28) 2KAl3Si3O10(OH)2 (muscl) + 2H + 3H2O  3Al2Si2O5(OH)3 (gibb) + 2K

The equilibrium constant is:

[K + ]2 29) K'= [H + ]2

[K + ] 30) Kmusc"kaol = + ! [H ]

Following the usual procedure with data from Table 2, we obtain logKmusc-kaol = 3.8. Since this reaction does not produce dissolved silica, the equilibrium line will have a s!lope of zero and a y-intercept of 3.8 (Figure 5).

The preceding are all the reactions among the alumino-silicate phases we will consider in this K-Al-Si-H2O system, but there is one additional phase that we will add to bound our diagram. The precipitation or dissolution of amorphous silica (opal) is often rapid enough that it provides an “upper limit” for silica solubility in natural systems. The reaction may be written:

31) SiO2(amorph) + 2H2O  H4SiO4

32) Kopal = [H4SiO4]

Because of variation in both the composition and structure of this phase there is some uncertainty in the thermodynamic data, or at least how a given laboratory experiment can be applied to real systems. Recent experimental data give a logKopal value near –2.7, so reaction 31 will have an undefined (vertical) slope and an “x-intercept” of –2.7 on our diagram (Figure 6). Ordinarily, we do not expect to see dissolved silica (H4SiO4) concentrations above 10-2.7 M at 25˚ C (equivalent to ca. 2000 µmolar), as they should be limited by the solubility of amorphous silica.

Now we have a “complete” diagram, but it is useful to resize the axes to focus on the environmentally relevant portion (Figure 7). [K+]/[H+] ratios in natural waters are usually > 1, so we set the lower limit of the y-axis to zero (log(1)). The upper limit for silicic acid is given by opal saturation, so we set the x-axis maximum to –2. The lower limit of the x-axis and upper limit of the y-axis are simply chosen to cover almost all likely compositional ranges.

6 EAS 3030 9/16/08 LAD The next task is to sort out which are the stable phases and reaction boundaries (Figure 8). Starting in the upper right hand corner, the stable phase is K-feldspar. The stability field of K-feldpsar is bounded by the reaction with muscovite when log([K+]/[H+]) > 3.8, and with kaolinite below this value. Assuming log([K+]/[H+]) > 3.8, at the boundary K- feldspar will react to form muscovite. Moving further to the right into the muscovite field implies that K-spar is no longer stable, so the extension of the K-spar-kaolinite reaction line into the muscovite stability field is called a “metastable” boundary. Since neither kaolinite nor K-spar are stable in the muscovite field, the reaction between them is not favored, and could only occur if some kinetic inhibition prevented muscovite from forming.

At lower values of log([K+][H+]) < 3.8, muscovite is not stable. The reaction muscovite – kaolinite occurs at log([K+][H+]) = 3.8. Once muscovite or kaolinite react to form gibbsite, the extension of the muscovite-kaolinite line to the left of the diagram into the gibbsite field is no longer stable. Similarly, beginning from the lower left gibbsite field, + + for log([K ][H ]) < 3.8, increasing [H4SiO4] causes the reaction to form kaolinite before equilibrium with muscovite is reached. So, the extension of the muscovite-gibbsite line into the kaolinite field is another metastable boundary, sometimes termed a “metastable extension”. To reiterate, metastable boundaries are not the stable or equilibrium boundaries we predict. However, in the absence of the true stable phase (typically because it is too slow to form), they may be relevant.

A final “cleaned up” diagram is presented in Figure 9. We can use this diagram to understand equilibrium sates and reaction paths.

Returning to the question of kinetic inhibition of the formation of a phase, we can consider a reaction very similar to that between opal and water: quartz and water. reaction may be written:

33) SiO2(qtz) + 2H2O  H4SiO4

34) Kqtz = [H4SiO4]

Quartz is a nearly ubiquitous phase, but is very slow to precipitate below about 70˚ C, so determining its solubility at low temperature has been surprisingly difficult. Recent work gives a logKqtz value of –3.8, so, like opal, reaction 33 will have an undefined (vertical) slope but an “x-intercept” of –3.8 on our diagram (Figure 10). If weathering solutions were in true equilibrium with quartz, we would not expect to see dissolved silica (H4SiO4) concentrations above 10-3.8 M at 25˚ C (equivalent to ca. 180 µmolar). However, it is not uncommon to see measured values well above 180 µM in low temperature solutions (i.e. supersaturation with respect to quartz), because the rate of quartz precipitation is so slow. Nevertheless can be helpful to draw the quartz saturation line on our diagram, because it is a useful reference point.

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