Background and Basic Chemical Principles: Elements, Ions ...€¦ · in environmental geochemistry require under-standing of the relationships among aqueous solutions, geological

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Background and Basic Chemical Principles:Elements, Ions, Bonding, Reactions

1.1 AN OVERVIEW OFENVIRONMENTAL GEOCHEMISTRY –

HISTORY, SCOPE, QUESTIONS,APPROACHES, CHALLENGES FORTHE FUTURE

“The best way to have a good idea is to have a lotof ideas.” (Linus Pauling)

“All my life through, the new sights of naturemade me rejoice like a child.” (Marie Curie)

Environmental geochemistry encompassesresearch at the intersection of geology, environ-mental studies, chemistry and biology, and at itsmost basic level, is designed to answer questionsabout the behavior of natural and anthropogenicsubstances at or near the surface of Earth.The scope includes topics as diverse as trace metalpollution, soil formation, acid rain and sequestra-tion of atmospheric carbon, and most problemsin environmental geochemistry require under-standing of the relationships among aqueoussolutions, geological processes, minerals, organiccompounds, gases, thermodynamics, kinetics,

and microbial influences, to name a few. A goodexample is the fate and transport of lead in theenvironment. In some areas much of the lead inthe Earth surface environment is (was) derivedfrom combustion of leaded gasoline, because evenafter it had been banned, Pb tends to persist in soil.The original distribution of Pb was controlled atleast in part by atmospheric processes rangingfrom advection to condensation and precipitation.The fate and transport of Pb deposited on the

land surface is controlled by the interactionsand relationships among lead atoms, solidscompounds (e.g. inorganic minerals or organicmatter), potential for uptake into plants or otherorganisms, and the composition of water in soils,lakes or streams (including dissolved gases like O2

and CO2). In cases where Pb falls on soils bearingthe carbonate anion (CO3

–2), the formation of leadcarbonate (PbCO3) can result in sequestration oflead in a solid state where it is largely unavailablefor uptake by organisms. If the PbCO3 is thermo-dynamically stable, the lead can remain seques-tered (i.e. stored), but changes in chemicalregime can destabilize carbonates. For example,

Environmental and Low Temperature Geochemistry, First Edition. Peter Crowley Ryan.© 2014 Peter Crowley Ryan. Published 2014 by John Wiley & Sons, Ltd.Companion Website: www.wiley.com/go/ryan/geochemistry

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acidic precipitation that lowers the pH of soil cancause dissolution of PbCO3, but how muchof the carbonate will dissolve? How rapidly? Muchlike the melting of ice at 10 �C, geochemicalprocesses are kinetically controlled (some morethan others), so even in cases where phases existout of equilibrium with their surroundings, wemust know something about rate laws in orderto predict how fast reactions (e.g. dissolution ofPbCO3) will occur.If Pb is dissolved into an aqueous form, addi-

tional questions of fate and transport must beaddressed – will the Pb remain in solution, thusfacilitating its uptake by plants? Or will it becarried in solution into a nearby surface waterbody, where it could be consumed by a fish oramphibian? Or will other soil solids play a rolein its fate? Will it become adsorbed to the surfaceof a silicate clay or organic matter, transporteddownstream until it ultimately desorbs in lakesediments? We also need to consider the possibil-ity that the PbCO3 does not dissolve, but rather isphysically eroded into a stream or lake, where itmight dissolve or remain a solid, possibly becom-ing consumed by a bottom feeder, from whichpoint it could biomagnify up the food chain.Environmental geochemistry has its origins in

ground-breaking advances in chemistry and geol-ogy ushered in by the scientific breakthroughs ofthe late 19th and early 20th centuries, particularlyadvances in instrumental analysis. The NorwegianVictor M. Goldschmidt is considered by many to bethe founder of geochemistry, a reputation earnedby his pioneering studies of mineral structuresand compositions by X-ray diffraction and opticalspectrograph studies. These studies led Gold-schmidt to recognize the importance and preva-lence of isomorphous substitution in crystals, aprocess where ions of similar radii and chargescan substitute for each other in crystal lattices.Goldschmidt’s peer, the Russian Vladimir I.

Vernadsky, had come to realize that mineralsform as the result of chemical reactions, andfurthermore, that reactions at the Earth surface

are strongly mediated by biological processes.The application of geochemistry applied to envi-ronmental analysis mainly arose in the 1960sand 1970s with growing concern about contam-ination of water, air and soil. The early 1960s sawpublication of Rachel Carson’s Silent Spring, andearly research on acid rain at Hubbard Brook inNew England (F.H. Bormann, G.E. Likens, N.M.Johnson, and R.S. Pierce) emphasized the inter-disciplinary thinking required for problems thatspanned atmospheric, hydrologic, soil, bioticand geologic realms.Current research in environmental geochemis-

try encompasses problems ranging from microm-eter-scale (e.g. interactions between minerals andbacteria or X-ray absorption analysis of tracemetal speciation), local-scale (e.g. acidmine drain-age, leaking fuel tanks, groundwater composition,behavior ofminerals in nuclearwaste repositories)to regional (acid rain, mercury deposition, datingof glacier retreat and advance) and global (climatechange, ocean chemistry, ozone depletion). Mod-ern environmental geochemistry employs analyt-ical approaches ranging from field mapping andspatial analysis to spectrometry and diffraction,geochemistry of radioactive and stable isotopes,and analysis of organic compounds and toxic tracemetals. While the explosion of activity in this fieldmakes it impossible to present all developmentsand to acknowledge the research of all investiga-tors, numerous published articles will be cited andhighlighted in throughout the text, and a casestudy that integrates many concepts is presentedin Appendix I.

1.2 THE NATURALLY OCCURRINGELEMENTS – ORIGINS ANDABUNDANCES

The chemical elements that comprise Earthwere mainly produced during the Big Bangapproximately 12−15 billion years ago. The earlyuniverse was extremely hot (billions of degrees)

2 Background and Basic Chemical Principles: Elements, Ions, Bonding, Reactions

and for the first few seconds was comprised only ofmatter in its most basic form, quarks. After about15 s, the atomic building blocks known asneutrons, protons, electrons, positrons, photonsand neutrinos began to form from quarks, andwithinmoments after the Big Bang, the first actualatoms formed. Protons combined with neutronsandelectrons to formhydrogen (1

1H) and its isotopedeuterium (1

2H, or D), which rapidly began to formhelium (He) through fusion, a process inwhich thenuclei of smaller atoms are joined to create larger,heavier atoms: 2H ! He + energy, or to be moreprecise:

11H + 2

1H= 32He+ γ +E ð1:1Þ

where γ is the symbol for gammaradiation emittedduring nuclear fusion. (note: basic principles ofatomic theory are presented in Section 1.3).Small amounts of lithium were probably also

produced in the first few minutes or hours afterthe Big Bang, also by fusion (in a simple senserepresented as H + He = Li). From a graph of theabundances of elements in the solar system(Fig. 1.1), it is clear that H and He are the mostabundant, and that, for the most part, elementabundance decreases exponentially with increas-ing atomic number up to atomic number 50, andbeyond that elements are quite rare.

It is also clear that some elements (e.g. Li, Be,B, Sc) appear to be anomalously uncommon ascompared to their neighbors, whereas othersseem to be present in anomalously high concen-trations (e.g. Fe, Ni, Pb). It is also interesting tonote the sawtooth pattern produced by alterna-tion of relatively abundant even-numberedelements as compared to neighboring odd-numbered elements (a phenomenon describedby the Oddo–Harkins rule). How do theseobserved trends relate to the processes thatformed the elements? The answer lies in basicprinciples of nuclear fusion.

Elements larger than He were formed byfusion in stars in the first few million to hundredsof millions of years after the Big Bang by a processgenerally referred to as stellar nucleosynthesis.Gravitational forces had produced a contracting,spinning disc-like mass of primordial H and Heknown as the solar nebula that contained theenergy necessary to form heavier elements byfusion as follows:

42He + 4

2He = 84Be+ γ ð1:2Þ

42He + 8

4Be = 612C+ γ ð1:3Þ

612C + 6

12C = 2010Ne + 4

2He ð1:4Þ

This process skipped over 3Li and 5B and the 48Be

that formed was very unstable and was eitherrapidly transformed to 12C before it decayed (seemiddle reaction above) or was destroyed byradioactive decay.

Fusion was able to form elements up to iron(2656Fe), but beyond Fe no heat is produced duringfusion – that is, the process is no longer exother-mic when fusing nuclei heavier than Fe. In fact,the iron nucleus is so stable that fusion reactionsinvolving iron actually consume energy (it is anendothermic process), so without the heat neededto fuel fusion reactions, another process hadto take over to form the heavier elements. Thisprocess is known as neutron capture and canbe represented like this:

Fe

Pb

U

Ca

C

N

O

F

Sc

Au

HHe

Abu

ndan

ce

1010

108

106

104

102

100

10–2

0 10 20 30 40 50 60 70 80 90Atomic number

Si

Th

Fig. 1.1 Abundance of elements in the solar systemnormalized to Si = 106 on a logarithmic y-axis – thisis a standard means of normalizing and plottingvalues for this type of data set.

The Naturally Occurring Elements – Origins and Abundances 3

5626Fe + 1

0n = 5726Fe ð1:5Þ

5726Fe + 1

0n = 5826Fe ð1:6Þ

5826Fe + 1

0n = 5926Fe ð1:7Þ

The Fe59 atom is unstable and undergoesspontaneous radioactive decay by beta emissionas follows:

5926Fe=

5927Co+ β – ð1:8Þ

In this case, the negatively charged beta particle(sometimes written as –1

0e or e–) effectively repre-sents the transformation of a neutron (0

1n) to aproton (1

1p). Neutron capture combined withradioactive decay then formed progressivelyheavier elements up to the heaviest naturallyoccurring element, uranium. Some elements suchas 56Ni and 56Co are unstable and undergo radi-oactive decay to form stable 56Fe, helping toexplain the relative abundance of Fe as comparedto elements with similar atomic number.Neutron capture (equations 1.5–1.7) takes

place via two main mechanisms. The r-process(“r” is for rapid) takes place in core-collapse super-novae, where there is a high flux of neutrons andextremely high temperatures (e.g. > 109 K), andnucleosynthesis involves a rapid series of neutroncapture reactions starting (typically) with 56Fe –

the r-process explains the origin of ~50% ofatoms heavier than Fe. The other main meansby which heavy elements can be produced isknown as the s-process (“s” is for secondary),in which nucleosynthesis occurs by means ofslow neutron capture. The difference is thats-process neutron-capture nucleosynthesis occursin asymptotic giant branch (AGB) stars, whichhave lower temperatures (e.g. 103–104 K) thansupernovae, and thus the s-process requirespre-existing (hence the “secondary” nature)heavy isotopes that can function as seed nuclei.Getting back to fusion, it is clear that progres-

sive fusion reactions involving atoms with even

numbers of protons will lead to the sawtoothpattern in Fig. 1.1, but there is also anothercontributing factor to this pattern. The Oddo–Harkins rule states that atoms with an evennumber of protons in their nuclei are more stablethan their odd-numbered counterparts. This isbecause, during nucleosynthesis, nuclei with anunpaired proton were more likely to capture anadditional proton, producing a more stable protonarrangement in the nucleus of atoms with evennumbers of protons. For additional informationon nucleosynthesis and the origin of the elements,the reader is referred to the accessible and more-detailed presentation in Gunter Faure’s textPrinciples and Applications of Geochemistry.As the universe continued to cool, galaxies and

solar systems began to form. The solar nebula thatwas to form our solar system cooled and began tosolidify into small masses known as chondritesand eventually larger masses known as planetesi-mals (on the order of tens of km in diameter).Those bodies closest to the early sun were moreenriched in heavier elements (especially Si, Al,Mg, Fe, Ca, Na, K), in part because centrifugalforces effectively flung lighter elements (H, He)preferentially to the farther reaches of the solarsystem (other important influences include tem-perature, pressure, redox conditions and nebulardensity, but these factors that will not be coveredhere). The end result is that the inner planets areterrestrial and rocky (Mercury throughMars) andenriched in heavier elements, whereas the outerplanets are gaseous (Jupiter and beyond) andenriched in lighter elements (think of the possibil-ity of methane oceans or methane ice on Jupiter’smoon Titan).Planets ultimately formed when gravitational

forces caused accretion of planetesimals. Theaccretion of what was to become Earth producedheat that left the proto-planet in a molten orsemi-molten state and allowed relatively denseFe and nickel (Ni) to sink to the core of the Earth,whereas relatively light silicon (Si), aluminum (Al),magnesium (Mg), calcium (Ca), sodium (Na) and


potassium (K) floated to the top to form Earth’scrust, leaving the Fe-Mg-Ni-Cr-Si mantle inbetween core and crust. While this is a broadgeneralization, the result is a differentiated Earth(Table 1.1), one where average continental crust(~ between 25–60 km thick) has a felsic composi-tion much like that of granite, whereas oceaniccrust (~ between 5 and 10 km thick) is composi-tionally mafic (or basic) and comprised mainly ofbasalt, rock that is less silica-rich and relativelyenriched in Fe, Mg and Ca relative to continentalcrust. The mantle is comprised of rock types like

peridotite and dunite and has an ultramafic(or ultrabasic) composition.

One example of how compositional differentia-tion of the Earth is important in an environmentalsense is related to soils formed by weathering ofrocks derived from the mantle – these ultramaficsoils tend to be depleted in plant nutrients such asCa and K, and enriched in the trace metals Niand Cr (and in some places, arsenic). Finally, theprimordial atmosphere of Earth was comprisedofCO2,H2OandNgasesderived fromvolcanicerup-tions. Oxygen in its form as O2 gas is a relatively

Table 1.1 Chemical differentiation of Earth withmajor elements and selected trace elements. Major elements (SiO2

through P2O5) are presented in units of wt% oxides and the trace elements are presented in concentrations of partsper million (ppm, or mg/kg). Data sources are as follows: granite is the United States Geological Survey granitestandard “G-2”; basalt and ultramafic data are averages from Turekian and Wedepohl (1961) and Vinogradov(1962); sandstone and carbonate rock data are unpublished analyses of early Paleozoic sedimentary rocks fromnorthwestern Vermont performed by the author; and shale is the North American Shale Composite (Gromet et al.,1984). Trace element values with asterisks are averages from Turekian and Wedepohl (1961) and Vinogradov(1962). Additional resource include the text The Continental Crust: Its Composition and Evolution by Taylor andMcLennan (1985) and the chapter by Rudnick and Gao (2003) in Treatise on Geochemistry.

Granite Basalt Ultramafic Sandstone Shale Carbonate

SiO2 69.4 49.3 42.1 71.3 64.8 8.34TiO2 0.48 1.86 0.05 0.70 0.80 0.12Al2O3 15.4 15.3 2.25 10.6 17.0 1.52Fe2O3

∗ 2.66 12.0 13.7 5.03 5.70 1.07MnO 0.03 0.22 0.20 0.10 0.25 0.07MgO 0.75 7.39 38.2 2.52 2.83 21.2CaO 1.96 9.9 2.22 3.03 3.51 66.7Na2O 4.08 2.47 0.66 1.56 1.13 0.21K2O 4.48 0.98 0.02 4.61 3.97 0.55P2O5 0.14 0.25 0.04 0.16 0.15 0.07SUM 99.4 99.7 99.4 99.7 100.1 99.8As 0.25 2.2 0.8 1.0 28 1.6Ba 1880 315 0.7 595 636 99Cr 10∗ 185 1800 20 125 10Co 4.6 47 175 2.3 20∗ 3.6Cu 11 94 15 2.5 45∗ 1.3Ni 10∗ 150 2000 10.1 58 3.3Pb 30 7 0.5 8.3 20∗ 2.8Th 13∗ 3.5 0.005 4.4 12.3 0.54U 3∗ 0.75 0.002 1.3 2.7 1.6Zn 86 120 40 50 95∗ 13

The Naturally Occurring Elements – Origins and Abundances 5

recent addition to Earth’s atmosphere, havingbegun to accumulate slowly and in a step-wisemanner in the atmosphere after the appearanceof photosynthetic algae 3 billion years ago.

1.3 ATOMS AND ISOTOPES:A BRIEF REVIEW

A schematic sketch of a carbon atom, consisting of acentral positively charged nucleus surrounded by anegatively charged “cloud” of electrons, ispresented in Fig. 1.2. All atoms consist of a nucleusthat contains positively charged protons andneutral neutrons, subatomic particles with a massof 1 atomic mass unit (1 amu, or 1 Dalton [Da]).The actual mass of a proton is 1.6726 × 10–24

g, so it is more convenient to say that themass of aproton is one amu or Da. The +1 charge on aproton = 1.602 × 10−19 coulomb. The numberof protons in the nucleus is what distinguishesatoms of one element from another – hydrogenhas 1 proton, helium has 2, carbon has 6, ura-nium has 92 protons, and so on. If there is anucleus with some other number of protons than

92, it is not uranium. The number of protons iscommonly referred to as the atomic number (Z).The mass of a neutron is effectively the same as

the mass of a proton (1 amu; 1.6749 × 10–24 g),and sum of protons and neutrons is the atomicmass or mass number or atomic weight –

all are essentially synonymous terms, (see backinside cover for atomic weights chart). Virtuallyall of the mass of an atom is contained in itsnucleus. The carbon atom in Fig. 1.2 contains anucleus with 6 protons and 6 neutrons, a config-uration represented in this text and inmany otherplaces as 6

12C. So, while the number of neutrons isnot explicitly given in this notation, it is implied,and in the case of the most common form ofuranium, 92

238U, there must be 146 neutrons inthe nucleus along with 92 protons to producethe atomic mass of 238.The carbon atom in Fig. 1.2 contains 6 protons

and likely also contains 6 neutrons, but it couldcontain 7 or 8 neutrons. Differences in the numberof neutrons (6, 7 or 8) give rise to the three iso-topes of carbon. Carbon-12, or 12C, is the mostabundant isotope of carbon and it contains 6 neu-trons; 13C contains 7 neutrons and it is much rarerthan 12C (Chapter 10). 12C and 13C are both stable

e–e–

e– e–

e– e–6 P+6 N

e–

e–

e–

e–

e–

e–

Fig. 1.2 Schematic sketches of Bohr models of a carbon atom (12C) showing, on the left, 6 protons and6 neutrons in the nucleus with 6 electrons in 2 separate orbitals, and on the right, an atom that attempts toshow the actual size of the nucleus compared to the electron cloud, yet even here the nucleus is ~100 timeslarger than an actual nucleus. The example on the right also introduces the idea of separate orbitals in theouter electron “shell”.


isotopes of carbon and the ratios of these twoisotopes in plants, rocks, waters and sedimentshave proven very useful in environmental analysis.Carbon also has a radioactive isotope, 14C,atoms of which are produced in the atmospherein the presence of cosmic rays and ultimatelyundergo radioactive decay to nitrogen. This topicand many others in the field of isotope geochemis-try are covered in Chapters 10 and 11.Balancing the positive charge of the protons are

electrons located in specified positions, or energylevels, outside of the nucleus. The charge on anelectron is exactly opposite that of a proton – itis –1, or –1.602 × 10–19 coulomb, and neutrallycharged atoms contain equal numbers of protonsand electrons. Furthermore, while the “electroncloud” surrounding the nucleus is of a virtuallynegligible mass, it occupies a volume that is ordersof magnitude larger than the volume of thenucleus (atoms are mostly open space).For the purpose of this text and virtually all

environmental geochemistry, the main concernto investigators is the outermost shell of electrons(the valence shell), because that is the part ofthe atom most intimately involved in bondingand ionization. Neutral atoms with atomic num-ber (Z) < 20 contain somewhere between 1 and 7electrons in their valence (outermost) shell; thenoble gases Ne and Ar contain 8 valence-shellelectrons. The most stable valence shell electronconfiguration for elements with Z< 20 is one thatconsists of 8 electrons, so all atoms with Z < 20seek to produce ions or form bonds that result ineight valence electrons. This is done either bylosing, gaining or sharing electrons (Section 1.6).In elements with Z> 20, the presence of d and forbitals makes the octet rule inapplicable; none-theless, heavier elements (Fe+2 and Fe+3; Ni+2;U+4 and U+6) lose electrons in predictable mannersto form more-stable configurations, and the formsof these ions are commonly known and enableprediction of their behaviors.One convenient way to depict valence electrons

is with Lewis electron dot diagrams (Fig. 1.3).

Sodium, with one valence electron, and oxygen,with six, can be depicted as such:

Sodium satisfies the octet rule by losing an elec-tron, becoming Na+1. The neutral oxygen atomwith six valence electrons can most easilysatisfy the octet rule by gaining two electrons,producing O–2, the most common form of oxygenin nature.

The Aufbau principle describes how electronorbitals are populated in a systematic manner. Inbrief, electrons occupy orbitals of fixed energylevels and tend to occupy the lowest energy levelspossible to create a stable atom. Orbitals are filledin a relatively predictable, sequential manner,starting with the lowest quantum number(n = 1), for which there is only one orbital (thes-orbital) which can be occupied by two electrons(of opposite spin).

For a neutral hydrogen atom with onlyone electron the symbol is 1s1. For neutral He with2 electrons, the notation is 1s2. Quantum number2 (n=2) contains both s- and p- orbitals, where thes-orbital again can be occupied by one electron pairand the p-orbital can host up to 3 electron pairs.For neutral lithium with 3 electrons, the notationis 1s2 2s1. For neutral oxygen, the notation is 1s2

2s2 2p4. (to satisfy the octet rule, oxygen gains twoelectrons, producing O–2 with an electron configu-ration like that of neon: 1s2 2s2 2p6). The thirdquantum level (n=3) contains s-, p- and d-orbitals,where the s- and p-orbitals can contain 1 and 3electron pairs, respectively, and the d-orbital canhost 5 electron pairs (10e–s). However, the 3dorbital exists at a higher energy level than the 4s orbital, so the 4 s orbital is filled before the 3d.

ONa ONa

Na

Fig. 1.3 Lewis electron dot diagrams showing valenceelectrons of sodium and oxygen in their ground states(left) and in the compound Na2O (right), where eachsodium has lost an electron to oxygen, resulting in twoNa+ and one O–2.

Atoms and Isotopes: A Brief Review 7

The sequence in which these orbitals are filled is asfollows:

1s!2s!2p!3s!3p!4s!3d!4p!5s!4d!5p!6s!4f!5d!6p!7s!6d

Two examples of larger atoms are calcium andiron. The notation for calcium is:1s2 2s2 2p6 3s2 3p6 4s2. Note that the second and

third quantum levels contain eight electrons, butthat the fourth contains only two. What must cal-cium do to satisfy the octet rule when forming anion? It loses two electrons, taking on an electronconfiguration like that of argon (1s2 2s2 2p6 3s2

3p6) and becoming Ca+2. The notation for iron is:1s22s2 2p6 3s2 3p63d64s2 (26 electrons balance

the 26 protons to produce a neutral Fe atoms). Atthis point the octet rule cannot be used to predictcommon oxidation states – iron tends to occur asFe+2 or Fe+3 depending on redox conditions (moreto come on redox later in this chapter).

1.4 MEASURINGCONCENTRATIONS

Concentrations of elements or compounds aremeasured in a few common ways. Only in rarecases do scientists directly measure individualatoms/compounds, for even microscopic crystalsof minerals at the micrometer scale such ashydroxides and silicate clays typically containmillions or billions of atoms.

1.4.1 Mass-based concentrations

One commonmeans of measuring concentrationsis by units such as milligrams per kilogram(mg/kg) for solids and milligrams per liter(mg/L) for liquids, or sometimes in units of micro-grams (μg/kg or μg/L) or nanograms (ng/kg orng/L) for trace elements. Note that mg/L = μg/mL. Weight percent is a common mass-based

approach for expressing element concentration incases where elements are in high concentrations(e.g. Table 1.1), a good example being Si, whichcomprises approximately 28% by weight (or mass)of the continental crust (or, as wt% oxide, approx-imately 59% of the crust – this conversion is pre-sented below).A few other useful facts to know about units of

concentrations are:For solids:

mg/kg is also known as parts per million (ppm),where mg is a milligram (10–3g)

μg/kg is also known as parts per billion (ppb),where μg is a microgram (10–6g)

ng/kg is also known as parts per trillion (ppt),where ng is a nanogram (10–9g)For liquids:

mg/L is also known as parts per million (ppm)μg/L is also known as parts per billion (ppb)ng/L is also known as parts per trillion (ppt)

Weight percent oxide is a common means ofexpressing major (abundant) elements in soilsand rocks. Note in Table 1.1 that Si, Al, Fe andthe other major elements are presented in unitsof SiO2, Al2O3, Fe2O3 and so on. This is donepartly by convention (or habit) and also becausethe major elements tend to occur in silicateminerals bonded to oxygen. The conversionfactors for wt% element to wt% oxide for thecommon oxides are presented here in Table 1.2.In the case of converting Ca to CaO, the

conversion is determined as the molar massof CaO divided by the molar mass of Ca, i.e.56.08 � 40.08 = 1.399.In the case of Al2O3, the conversion factor

is determined as the molar mass of Al2O3

divided by the mass of the equivalent amount ofAl in the oxide, i.e. Al2O3 � (2 ∗Al) = 101.957 �(2 ∗ 26.98) = 1.889. For Na to Na2O, the conver-sion is themolarmass of Na2O divided by themolarmass of 2 ∗Na= 61.98 � (2 ∗ 22.99) = 1.348.For elements that can occur in more than one

oxidation state (e.g. Fe+2, Fe+3), values may be


presented as either one of the oxidation states (i.e.either as FeO or Fe2O3), or as a combination of thetwo if the relative abundances of Fe+2 and Fe+3

are known. In many cases where iron oxidationstate is not known, all iron is presented in termsof Fe2O3.However, mass-based measurements like

μg/L or wt% are not always the best way toexpress concentrations. Consider for examplegroundwater with 98.7 μg/L nickel (59Ni) and98.7 μg/L of uranium (238U); i.e. both elementsare present in equal concentrations of 98.7 ppb(evaporating the liter of water would leave98.7 μg each of Ni and U). However, given thatU atoms (238 g/mol) are ~4 times heavier thanNi (59 g/mol), there must be more Ni atoms –

in fact, there are approximately four times asmany Ni atoms as there are U atoms.

1.4.2 Molar concentrations

Quantifying concentrations on a molar basis hasits roots in the work of Italian scientist AmadeoAvogrado, who in 1811 realized that equalvolumes of gases at identical pressures and tem-peratures contain equal numbers of atoms (ormolecules in the case of gases like N2 and O2),

even though their atomic masses differed. Theterm mole (abbreviated mol) describes the num-ber of atoms of a given element required to form amass equal to the atomic mass of the substance, ingrams. For C, this mass is 12.011 grams. For U,this mass is 238.03 g, and so on. For all elements,the number of atoms required to form the atomicmass in grams is 6.0221 × 1023 atoms, a valueknown as Avogadro’s number. 238.03 g ofuranium (1mol of U) contains 6.0221 × 1023

atoms; 4.002 g of helium (1mol of He) contains6.0221 × 1023 atoms.

Themole is a very useful concept in chemistry –most equations are expressed in terms of moles ofreactants and products. In order to express Niand U concentrations in terms of moles per liter(mol/L), or for trace elements like these, μmol/L,the mass concentration must be multiplied bythe inverse of the molar mass (and a conversionfor g to μg) as follows:

ForNi : 98:7μg=L ∗ 1mol=58:693g ∗1g=106μg=1:68×10 – 6mol=L

ForU : 98:7μg=L ∗1mol=238:03g ∗1g=106μg=0:415×10 – 6mol=L

It is often helpful to express units in easy tocommunicate terms, so in this case mol/L wouldprobably be converted to micromoles per liter(μmol/L), by multiplying mol/L by 106 μmol/mol:

1:68×10 –6mol=L ∗106μmol=mol =1:68μmol=Lof U in the groundwater

0:415×10 –6mol=L ∗106μmol=mol =0:415μmol=Lof Ni in the groundwater

In a solid (e.g. sediment or rock) with 1mol/kgeach of Fe (55.85 g/mol) and Al (26.98 g/mol):(1) there is an equal number of atoms of Fe andAl in the soil; and (2) Fe comprises a greater massof the soil than Al (whether expressed as wt%,g/kg or mg/kg). Given that mass units are acommon way of expressing concentration, itmay be necessary to convert from molar unitsto mass units. A few algebraic calculations allowconversion frommol/kg to 3 common units, g/kg,mg/kg and wt%.

Table 1.2 Conversion factors for wt% elementto wt% oxide.

Element C.F. Oxide

Al 1.889 Al2O3

Ca 1.399 CaOFe 1.286 FeOFe 1.430 Fe2O3

K 1.205 K2OMg 1.658 MgOMn 1.291 MnONa 1.348 Na2OP 2.291 P2O5

Si 2.139 SiO2

Ti 1.668 TiO2

Measuring Concentrations 9

Given that 1 mol of Fe = 55.85 g:55.85 g/mol ∗ 1mol/kg = 55.85 g/kg Fe…or 55.85 g/kg ∗ 1000mg/g = 55850mg/kg Fe… or 55.85 g/kg ∗ 0.1 = 5.855% Fe (by weight)

1 mol of Al = 27 g; 27 g/mol ∗ 1 mol/kg =27 g/kg Al

… or 26.98 g/kg ∗ 1000mg/g = 26980mg/kg Al… or 26.98 g/kg ∗ 0.1 = 2.698% Al (by weight)

1.4.3 Concentrations of gases

Atmospheric gas concentrations are typicallyexpressed as the proportion of the total volumeaccounted for by a given gas. For example, thecurrent atmospheric concentration of CO2 is�400 ppmv, indicating that 400 out of everyone million molecules of gas in Earth’s atmos-phere is CO2. At the onset of the industrialrevolution atmospheric CO2 was 280 ppmv.Less-abundant gases are often expressed interms of ppbv or pptv (parts per billion or tril-lion, volumetrically), and the major componentsof the atmosphere like the fixed gases N2, O2 andAr, are expressed in terms of percent (by vol):N2 = 78.1%, O2 = 20.9% and Ar = 0.9%(the amounts vary depending on the amountof H2O vapor in the air, which can range from0 to 4% by volume). Expressed in this way, CO2

comprises approximately 0.0400% percent ofthe atmosphere (vol%), but clearly units ofppmv are more useful for a gas like CO2. Gasesalso dissolve in liquids, and units of concentr-ation in these cases are commonly mg/L ormmol/L.

1.4.4 Notes on precision and accuracy,significant figures and scientific notation

A few important topics related to data analysisand presentation of results are encompassed bythe concepts of precision and accuracy. Simply

stated, accuracy describes how closely a meas-ured value agrees with the actual value. Theaccuracy of chemical analyses can be tested byanalyzing standards of known concentration.Consider a certified standard solution that con-tains 250mg/L of aluminum (Al), and fiveanalyses of this standard on your spectrometerproduces results of 237, 271, 244, 262, and240mg/L. The mean of those five values is 251mg/L – the average value is very close to the cer-tified value of 250mg/L. One way to express theaccuracy of this test is as a percent difference fromthe certified value:

251mg=L –250mg=Lð Þ�250mg=L½ � ∗ 100=0:4%

However, the five results are somewhat lackingin precision, which is basically a measure ofthe reproducibility of results – how closely domeasured results agree with each other? It isconceivable to produce results with a high degreeof precision that are lacking in accuracy. Forexample, after recalibrating the spectrometerand re-analyzing the Al standard, values noware 277, 281, 274, 278 and 275mg/L. The meanvalue of 277mg/L is farther from the certifiedvalue of 250 mg/L (the difference from the certi-fied value is 10.8%), but the results are definitelymore precise.One way to calculate uncertainty is to

determine the standard deviation of the data.The formula for standard deviation and detailson its appropriate use can be found in a statisticstextbook, but using Microsoft Excel’s standarddeviation formula produces σ values as follows:For the values 237, 271, 244, 262 and 240mg/L,σ = 14.9

For the values 277, 281, 274, 278 and 275mg/L,σ = 2.74

Which are better, analyses with greater accuracybut less precision (251 + 14.9 mg/L), or greaterprecision but less accuracy (277 + 2.74 mg/L)?Precise values are easier to correct because thereis less uncertainty than if you have to deal with


accurate values plagued by low precision.This emphasizes one of the reasons why it isimportant to run standards when making analyt-ical measurements. Of course, the ideal situationis to use a well-calibrated instrument for whichaccuracy and precision are both high, butregardless, it is imperative that researchers seekto quantify both parameters when makingmeasurements.In addition to quantifying uncertainty, it is

very important to present numerical results in amanner that relates to the sensitivity of the meas-urement, or the degree of confidence associatedwith that measurement, while also trying to avoidpropagation of error. Every measurement hassome limited number of significant digits(or significant figures). Measurements of pHmade using litmus paper can only be reportedto 1 significant figure (e.g. pH = 6), whereas mea-surements made with well-calibrated probes maybe reported to 3 significant figures (e.g.pH = 5.87).As a rule, calculations should be carried out

using all figures with each of their representativesignificant figures, and then the final result shouldbe rounded at the end of the calculation. Usingan example where the average concentrationof uranium in groundwater in an aquifer is14.3 μg/L, the aquifer volume is 1.241 × 108m3

and average porosity (and thus %water in the sat-urated zone) is 18%, the resulting mass of U in theaquifer is:

14:3μg=L ∗ 1:241×108m3 ∗1000L=m3 ∗0:018=3:2×1011μgU

The final result is limited by the two significantfigures in 18% (note that conversions involvingliters to cubic meters, cm3 to mL, etc. do not limitsig figs).Also important to understand is the signifi-

cance of zeros. Consider the following values:7200, 700043, 0.0436, 0.043600. How manysignificant figures are reported for each, and why?

• 7200 is assumed to have 2 significant figuresbecause any zero at the end of a number andbefore a decimal point is assumed to not besignificant. How many significant figures does847 000 possess? Three. 7200.0 contains 5significant figures.• 700 043 has 6 sig figs. Any zeros within anumber are significant.• 0.0436 has 3 sig figs because any zeros after adecimal point and before the first non-zero digitare not significant.• 0.043600 has 5 significant figures becausezeros that follow a non-zero digit after a decimalpoint are considered significant (as in 7200.0example above).Some of these examples serve to illustratewhy scientists tend to express values in terms ofscientific notation. Using scientific notation,the values above become:7.2 × 103 (2 sig figs)7.2000 × 103 (if we needed to report 7200 to

5 sig figs, this is how it would look)7.00043 × 105 (6 sig figs)4.36 × 10–2 (3 sig figs)4.3600 × 10–2 (5 sig figs)As a closing thought, one has to wonder aboutthe precision of the road sign in southwesternEcuador shown in Fig. 1.4. Being literal aboutsignificant figures, the distance to Engunga isprecisely indicated as being between 14995 and15005m. (note: it is a very small town, so theprecision may be justified).

1.5 PERIODIC TABLE

The Periodic Table of the Elements (front insidebook cover) is one of the most useful tools toresearchers and students of chemistry, geochem-istry and biochemistry. First fully developed by theRussian chemist Dmitri Mendeleev in 1869 andrefined ever since, it lists elements in order ofatomic mass (from left to right in each row, andalso from top to bottom in each column) and

Periodic Table 11

by similarities in chemical properties. Row 3 ofthe periodic table begins with sodium (atomicnumber Z = 11) and progresses to the right withincreasing Z all the way up to Ar (Z= 18).The atomic mass for each element represents a

weighted average of themass of the isotopes of thatelement, normalized to 12C – observe that theatomic mass of sulfur (S, with Z= 16) is 32.065,reflecting weighted average of 32S and heavierisotopes. The S atomic mass of 32.065 is derivedfrom the abundances of the four stable S isotopesas is demonstrated in the following mass-balancecalculation (data from Faure, 1986):

32S=95:02% 33S =0:75% 34S =4:21%36S=0:02%

32S :0:9502 ∗ 31:97=30:37789433S :0:0075 ∗ 32:97= 0:24727534S :0:0421 ∗ 33:97= 1:43013736S :0:0002 ∗ 35:97= 0:007194

Sum=32:062500

Atomic masses are sometimes regarded as dimen-sionlessnumbers, but are also expressed in termsofamu (Daltons) and also in terms of grammolecularweights (i.e. the grammolecularweight, ormass ofone gram of sulfur, is 32.065 g). All atomicmassesare normalized to the mass of 12C, and this is whythe mass of 32S is expressed as 31.97 g/mol.Columns (or groups) of elements generally

contain elements with similar valence electron

configurations, that is, the outermost shell ofelectrons tends to behave in a similar manner forelements of a given row. Good examples are col-umn 1, the alkali metals, all of which lose one elec-tronwhen they form chemical bonds, resulting in aseries of +1 charged ions including Na+ and K+;column 2, the alkaline-earth metals that formdivalent cations (e.g. Ca+2); and column 8, thenoble gases, including elements like argon andneonwith complete valence shells (as a result, theydo not form chemical bonds in nature). Elements incolumn 7, the halogens, tend to gain one electronwhen forming chemical bonds, resulting in halideanions with a –1 charge (e.g. Cl–).It may not come as a surprise that alkali metals

tend to form bonds with halides, such as:

Na+ +Cl – !NaCl ð1:9Þ

The point to understand here is that the periodictable presents information in a systematic way thatcan help to predict the behavior of elements inenvironmental systems. Like Na+, K+ also can formbonds with chloride (Cl–) to form a different salt,KCl (a substitute for NaCl in reduced-sodium diets).In fact, K+ and Na+ substitute for one another inmanyminerals. The periodic table also implies thatarsenic (As) might substitute for phosphorous (P)in mineral structures, which it does.In 2003, the environmental geochemist

Bruce Railsback from the University of Georgia

Fig. 1.4 Road sign insouthwestern Ecuador.


developed an innovative new periodic table forgeologists known as An Earth Scientist's PeriodicTable of the Elements and Their Ions, one whereelements are organized according to theiroccurrences in geological environments (http://www.gly.uga.edu/railsback/PT.html). This newformulation is designed to predict how elementsand ions behave in the environment. Unlike theconventional periodic table originally envisionedby Mendeleev, the Earth scientist’s periodic tableorganizes elements by charge, so it shows manyelements multiple times because many elementshave numerous valence states. It also containsabundant information on abundance of different

elements in soils, seawater, mantle vs. crust, ionicradii in crystals, and much more. For example,Fig. 1.5 shows the relative solubilities of oxidesof various ions, and from this inset it is apparentthat Al and Ti form insoluble oxides (corundum,Al2O3; rutile, TiO2) but that Na, K, N and S donot (they are more likely to occur as soluble ionssuch as Na+, K+, NO3

– and SO4–2).

The Earth scientist’s periodic table also showsthe numerous species of nitrogen that exist in nat-ural systems (e.g. valence states and commonmolecules of N), predicted behaviors, attributesand affinities of the abundant and trace cationsin natural systems, and more (Railsback, 2003).

Be+4 N+5

Na+ Mg+2 Al+3 Si+4 P+5

K+ V+5 Cr+6Ca+2 Sc+3 Ti+4

Rb+ Y+3 Mo+6

La+3 Hf+4 Ta+5

–7.4

–2.4

4.4

9.9 –8.1 –3.9

–9.7

–1.37

2.77

–7.6

28.9

14.0 1.4

4.3

6.7

Sr+2

Th+4

Li+ B+3Be+2

–9.7

Rutile

Baddeleylite

Thorianite

Molybdite

Lime

Corundum Quartz

Rutile Oxide mineral name

Log activity of cation speciesin distilled water at 25 ° C.

S+6

Periclase

Nb+5Zr+4

Fig. 1.5 Solubilities of ions as afunction of ionic charge, from Inset4 of An Earth Scientist's PeriodicTable of the Elements and Their Ions(Railsback, 2003). Oxides of theelements are used as reference(e.g. lime, periclase, etc.). Note thatlow-charge base cations/alkalimetals and cations of the alkalineEarth metals are relatively soluble,as are high-charge cations suchas S+6 and N+5, which formpolytatomic anions (SO4

–2, NO3–).

Cations with +3 and +4 charges(e.g. Al+3, Ti+4, Zr+4) are insolublein most surficial environments, withactivities in H2O at 25 �C of ~10–8

to 10–10. (Railsback 2003.)

Periodic Table 13

1.6 IONS, MOLECULES, VALENCE,BONDING, CHEMICAL REACTIONS

It is relatively rare to find elements in their native,neutral state in nature. A few examples includethe diatomic gases hydrogen (H2), oxygen (O2)and nitrogen (N2), solids like gold (Au) and graph-ite and diamond (both forms of C), and the noblegases (He, Ne, Ar, etc.). Most elements occur asions, molecules or compounds in liquids (petro-leum, alcohols, H2O and the dissolved species itcontains), solids (minerals, proteins, humus),and gases (CO2, CH4, NO2). Ions are chargedatoms, atoms that have either gained or lostone or more electrons from their neutral state.Examples of ions include the cation Na+1 andthe anion S–2. Most of the elements form cations,because metals generally lose electrons to formcations, and most of the elements are metals.A few common metal cations are Mg+2, Al+3,Cr+3,+6, Fe+2,+3 – note that some elements havemore than one oxidation state. Generally onlythose elements in the upper right of the periodictable form anions (e.g. O–2, N–3, F–1, etc). Oxida-tion state is a term for the charge on an ion: “theoxidation state of iron in swamps is generally +2(divalent), whereas in streams iron tends to occurin the +3 (trivalent) oxidation state”.Molecules are formed when two or more

atoms are joined by a chemical bond, and com-pounds are a specific type of molecule formedwhen two or more atoms of different elementsare joined by chemical bonds. The gases H2, O2

and N2 are comprised of molecules but they arenot compounds (they are diatomic gases). Exam-ples of compounds include NaCl, H2O and C8H18

(octane). Most substances in nature, other thansome gases and a few metals, are compounds.There are two main types of chemical bonds

responsible for forming molecules and com-pounds, ionic bonds and covalent bonds. Thesetwo bond types represent polar ends of the bond-ing spectrum, but it is useful to consider each typeseparately, and then examine intermediate cases.

We will also examine metallic bonds and Van deWaals bonds, but first will begin with ionic bonds.Ionic bonding occurs between atoms of ele-

ments with very different valence electron config-urations. Coulomb forces describe interactionsbetween charged atoms or molecules (i.e. ions),and when the charges are opposite (i.e. involvingcations and anions), the result is attraction(conversely, the Coulombic interaction betweenlike-charged particles, e.g. 2 cations, causesrepulsion).The classic example of Coulombic attraction is

the bond between an alkali metal cation (e.g.Na+) and a halogen anion (e.g. Cl–). The cationand anion are electrostatically attracted to eachother and the result is formation of an ionic bondproducing a solid, in this case cubic vitreouscrystals of halite (NaCl).Loss of an electron by sodium and its incorpo-

ration into the valence shell of chlorine can beviewed in the sense of a Lewis electron-dotdiagram (Fig. 1.6):The way this type of bond commonly forms in

the natural world is when sodium and chlorideions are dissolved in water and ultimately becomeattracted to each other when their concentrationsbecome sufficiently high to allow formation ofsolid crystals. (More details about the controlson aqueous processes are covered in detail inChapters 4, 5, and 9).

1.6.1 Ionic bond strength

The strength of an ionic bond is largely controlledby the charges on the ions and by the ionic radii ofthe ions involved in the bonding – this is a conceptthat should be intuitively apparent because ionswith higher (opposite) charges will be attracted

Na Cl

Fig. 1.6 Schematic representation of formation of anionic bond by transfer of the Na valence electron to Cl.


more, and the closer the spacing of the ions,generally the stronger the bond.The attraction of two ions can be quantified by

Coulomb’s Law:

Fc = k∗ q1∗ q2ð Þ=ε∗r ð1:10Þ

where k is a constant (described below), q1 and q2are the values of ionic charges on the ions, ε is thedielectric constant and r is the distance betweenthe nuclei of the two ions joined by the bond.When Fc is negative the ions are attracted; the

negative sign indicates that the system (the twoions) have shifted to a lower (more stable) energystate than is the case when the ions are separated.Given that all other terms (k, ε, r) are positive,negative Fc results from two ions with oppositecharge, i.e. a cation and an anion.The constant k is expressed as:

k=1= 4πεoð Þ ð1:11Þ

where εo, the permittivity constant (sometimesknown as Po or D), equals 8.854 × 10–12 C2 J–1

m–1 and results in a value for the constant k=8.998 × 109 J �m/C2, where J = Joules and C =1.602 × 10–19 Coulombs, a measure of charge.The dielectric constant ε expresses the effect of

the ambient environment on the strength of thebond, the best example being the difference in ionicbond strength in dry air as compared to water. At20 �C and 1 atm, the dry air value for ε= 1.0,whereas the value for ε in water at 20 �C = 88.Therefore, all other factors being equal, an ionicbond is 88 times weaker in water than in dry air.For Na+ and Cl– ions, q1 and q2 are 1 and –1. The

ionic radius of Na+ in a crystal of NaCl is 1.16 Åand the ionic radius for Cl– is 1.67 Å (where anangstrom is 10–10m). The reason that the Cl–

anion is larger than the Na+ cation is that Cl–

contains an additional shell of electrons. (In gen-eral, anions have larger radii than cations becauseanions gain electrons when they form ions.)The bond distance between nuclei will be:

1:16×10 – 10m+1:67Å×10 – 10m=2:83×10 –10m

so the attractive energy of the NaCl bond in air(ε = 1.0) is:

Fc = 8:998×109J�m=C2 ∗ –1:602×10 –19� �

∗ +1:602×10 –19� ��2:83×10 – 10m

= –8:15×10 – 19J

Note that units of charge in the numerator (C2,Coulombs squared) cancel with C2 units from theconstant, and also that m (meters, from r) in thedenominator cancel with m in the numerator ofk, leaving us with units of J (joules), which expressenergy lost (if J < 0) or gained (if J > 0), wherenegative values indicate transition to amore stablestate.

This calculation applies to a single Na–Cl atompair. If we wished to compute this for a mole wewould multiply the result by Avogadro’s number(one mole of NaCl would contain 6.022 × 1023atoms each of Na and Cl):

−8:15×10 –19J=atom ∗6:022×1023atoms=mol= −4:91×105 J=mol = −491kJ=mol

If we compare this to the bond between K+1 (ionicradius = 1.52 Å) and Cl–1, a bond that results information of KCl (sylvite or potassium chloride),we find that:

Fc = 8:998×109J�m=C2 ∗ −1:602×10 –19� �

∗ +1:602×10 –19� ��3:19×10 – 10

m= −7:23×10 – 19J

and

–7:23×10 –19J=atom ∗ 6:022×1023atoms=mol= −4:35×105J=mol = –435kJ=mol

Considering the attraction of cation and anion,NaCl forms a stronger bond than KCl becausethe smaller ionic radius of Na allows the Na andCl nuclei to be held closer.

Ions, Molecules, Valence, Bonding, Chemical Reactions 15

In the case of magnesium chloride (MgCl2), theionic charge on Mg is +2, and its ionic radius inMgCl2 is 0.86 Å. So in this case,

Fc = 8:998×109J �m=C2 ∗ –1:602×10 –19� �

∗ +2∗1:602×10 – 19� ��2:53×10 –10m

= –1:82×10 – 18J

and

–1:82×10 –18 J=atom ∗ 6:022×1023atoms=mol= 1:10×106 J=mol = –1100 kJ=mol

Based on Coulombic attraction, the higher chargeand smaller radius of Mg+2 compared to K+ andNa+make theMgCl2 lattice energy approximatelytwice that of KCl and NaCl.For ionic solids in water (ε = 88 in the denom-

inator), the lattice energies of NaCl, KCl andMgCl2 are 5.58 kJ/mol, 4.95 kJ/mol and 12.5kJ/mol, respectively, which is a quantitativeway of saying that ionic bonds are approximatelytwo orders of magnitude weaker in water than indry air.It is important to emphasize that the treatment

presented above only considers the attractionbetween two ions and does not consider other fac-tors associated with the strength of ionic bonds,two of which include (1) the effect of other ionsand the geometry of the lattice structure (thiscan be assessed using the Madelung constant,which is not covered here), and (2) temperature;for example, the melting points for NaCl, KCland MgCl2, respectively, are 801, 770 and714 �C, differences which are not explained byCoulomb’s Law.

1.6.2 Covalent bonds

Covalent bonds involve overlap of electronorbitals, i.e. they are bonds that form as a result ofsharing of electrons between atoms. Unlike ionicbonding, where atoms have very different attrac-tions to valence electrons, atoms that formcovalent

bondshave similarvalence electron configurations,and because their valence electrons are attracted totheir nucleiwith similar strength, it is impossible forone atom to lose its valence electron(s) to anotheratom to form a bond.One classic example of covalent bonding

involves the attraction involved with formationof the diatomic gases H2, N2, O2. When two oxy-gen atoms combine to form O2, there is clearly nodifference in the valence electron configuration orin the attraction of each oxygen atom for its elec-trons. The atoms are effectively equal in structure,so the solution to the octet rule for O2 involvesreorganization of electrons and overlappingto produce valence shells that each contain8 electrons (viewed as an electron dot diagramin Fig. 1.7).Both O atoms have satisfied the octet rule the

same way, by overlapping electron orbitals to pro-duce a stable configuration. So, while eachO atom has six valence electrons, two of themare shared with the adjacent O atom by orbitaloverlap to produce a strong chemical bond. In thiscase the bond between the oxygen atoms is adouble bond because it involves 2 pairs of elec-trons, as contrasted to a typical single bond thatinvolves only one pair of electrons (like H-O bondsin water molecules).Covalent bonds also occur between elements

with similar electronegativities (Fig. 1.8), includ-ing H and O in H2O, C and O in CO2, and Siand O in SiO2. Chemical bonds that are a mix ofionic and covalent (e.g. SiO2) are termed polarcovalent bonds.

O O

O=O

Fig. 1.7 Lewis electron dot diagram of an oxygenmolecule (O2) shown using two different notations.Note double bond between oxygen atoms consisting oftwo electron pairs depicted by the double dashed linesin the lower example.


1.6.3 Electronegativity

Electronegativity is a measure of the attractionof a nucleus for its valence electrons. Na andK have low electronegativities because they read-ily lose their valence electron to satisfy the octetrule, resulting in +1-charged cations. Conversely,fluorine and chlorine have strong attractions tothe electrons in their valence shell and further-more have the ability to pull valence electronsaway from nearby atoms to form –1-chargedanions and satisfy the octet rule – these elementshave high electronegativity values. Electronega-tivity is important because it helps to predict thetype of chemical bond that will form between ele-ments – elements with large differences in electro-negativity form dominantly ionic bonds, whereaselements with low differences in electronegativityform covalent or polar covalent bonds.Figure 1.8 presents electronegativities of the

elements and Fig. 1.9 shows the % ionic characterof some selected chemical bonds expressed bydifference in electronegativity. Nearly all bondsinvolving atoms of two elements (i.e. not includ-ing diatomic gases like H2 or O2 native elementssuch as Au or Ag) involve some aspect of electrontransfer (ionic character) and orbital overlap

(covalent character) – the greater the differencein electronegativity, the greater the ionic charac-ter (and usually, the greater the tendency to dis-sociate and dissolve in water).

The electronegativity difference (Δχ) for NaClis 3.0 (Na) – 0.9 (Cl) = 2.1, resulting in a pre-dominantly ionically bonded compound. For a

H

2.1

Li1.0

Na0.9

K0.8

Rb0.8

Cs0.8

Fr0.7

Ra0.9

Ba0.9

Sr1.0

Ca1.0

Mg1.3

Be1.6

Sc1.4

Y1.2

La-Lu1.1–1.2

Ac1.1

Th1.3

Hf1.3

Zr1.3

Ti1.5

V1.6

Nb1.6

Ta1.5

Pa1.5

U1.4

Mo2.2

1.6

Mo2.2

1.6MnCr

Tc1.9

Re1.9

Fe1.8

Ru2.2

Os2.2

Co1.9

Ir2.2

Ni1.9

Pd2.2

Cu1.9

Ag1.9

Cd1.7

Zn1.7

B

2.0

C

2.5

N

3.0

O

3.4

F

4.0

Al

1.5

Si

1.9

P

2.2

S

2.6

Cl

3.2

Ga

1.8

In

1.8

Tl

1.6

Ge

2.0

Sn

2.0

Pb

2.3

As2.2

Sb

2.1

Bi

2.0

Se

2.6

Te

2.1

Po

2.0

Br

3.0

I

2.7

At

2.2

Electronegavity (χ) of elements

If ∆χ< 1, bond is dominantly covalentIf ∆χ> 2, bond is dominantly ionic

Rh2.3

Hg2.0

W2.4

Au2.5

Pt2.3

Fig. 1.8 Electronegativity of elements based on Linus Pauling’s early 20th-century research.

0 1.0 2.0 3.0

100

75

50

25

0

% Io

nic

cont

ent o

f bon

d

Difference in electronegativity (χ)

MgO

SiO2

KF

LiFKCl

NaCl

HCl

LiBrAl2O3

HBr

H2SN2

NH3 H2O

CaO

CO2Covalent

Ionic

Polarcovalent

Polarcovalent

Fig. 1.9 Plot of relationship between difference inelectronegativity of two atoms in a bond and thepercent ionic character of the bond. Note that nobond is 100% ionic; however, diatomic gases(e.g. N2, Cl2, O2) and the C–C bonds in diamondare 100% covalent.. Based on early and middle20th-century work of Linus Pauling and morerecent research by Lu et al. (2006).

Ions, Molecules, Valence, Bonding, Chemical Reactions 17

purely covalent molecule like N2, O2 Cl2 ordiamond (pure C), Δχ is zero. Methane (CH4)and hydrogen sulfide (H2S) are strongly covalent(Δχ = 0.4 and 0.5, respectively). SiO2 is a mix ofcovalent and ionic (Fig. 1.9) and the term polarcovalent bond (or polar bond) is used todescribe this type.CaCO3 (calcium carbonate, the mineral

calcite) contains bonds between Ca–O and C–O.The Ca–O bonds are predominantly (~70%)ionic (Δχ = 3.5–1.0 = 2.5), whereas the C–Obond is predominantly (~80%) covalent (Δχ =3.5–2.5 = 1.0). Understanding this concept isimportant when predicting solubilities of minerals;for example, in water, calcite dissolves to produceCa+2 ions and the polyatomic CO3

–2 anion insolution (see Chapter 5 for details of carbonategeochemistry). The covalent character preservesthe C–O bond and as a result the carbonate anion(CO3

–2) is a common constituent of natural waters.Consider also the examples of NaCl (very ionic)

and SiO2 (which contains polar covalent bonds).Ionically bonded compounds like halite are highlysoluble in water, but quartz is very insoluble,which helps to explain the common occurrenceof white quartz sand beaches – if the covalentbonds in quartz were readily destroyed by waterwe would have no sandy beaches. In fact, weprobably wouldn’t have mountains or canyonseither, for the strong bonds in rock-formingminerals are primarily polar covalent. In theabsence of water, ionic bonds and covalent bondsare generally both strong, and appreciablystronger than the other types of chemical bondsin nature such as metallic bonds and dipole bonds(e.g. hydrogen bonds, Van der Waals bonds).

1.6.4 Metallic bonds, hydrogen bondsand van der Waals forces

Metallic bonds occur among metals such as Cr,Cu, Fe, Ni and Zn in solid compounds, but unlikeionic or covalent bonds, once valence electronsare released by a metal atom, they are not fixed

with a specific atom, but rather migrate throughthe crystal structure. This type of bond occurs insulfide minerals such as pyrite, and also in nativeelements such as Cu, gold (Au) and silver (Ag).These bonds are weaker than covalent and ionicbonds and are part of the reason why metal-bearing sulfide minerals are relatively unstableat the Earth surface.Dipole bonds, also known as van der Waals

bonds (and sometimes called van der Waalsforces), exist between electrically neutral mole-cules or compounds with some unequal distribu-tion of charge. A great example of a dipole bondexists between water molecules. Water is adipolar compound, with a positively charged poleand a negatively charged pole (remember that themolecule as a whole is neutral). Fig. 1.10 presentsa schematic sketch of three adjacent watermolecules attracted by dipolar bonds – in thiscase, the bond occurs between hydrogen atomsat the positively charged pole (δ+) of water mole-cules and the negatively charged pole (δ–) pro-duced by valence electrons from the highlyelectronegative oxygen atom in the adjacentwater molecule (Fig. 1.10).This specific type of dipolar bond is known as a

hydrogen bond, and although shown betweenadjacent H2O molecules in this case, hydrogenbonds can also occur between H and O (or N)within amolecule. Note that the dots surroundingthe O atoms represents electrons – oxygen satis-fies the octet rule by sharing an electron withan electron from each of the H atoms, and the

(δ−)

(δ−)

(δ+)

(δ+)

(δ−)

(δ+)

H

H HO

OH

HO

H

Fig. 1.10 Diagram of three water moleculesshowing polarity of the water molecule (δ+, δ–)as well as hydrogen bonds (dashed lines) betweenH and O in adjacent molecules.


H–O bonds within the H2O molecules (within theovals) are single bonds (dominantly covalent)involving one pair of electrons each.Water has a permanent dipole, but van der

Waals bonds also exist between electrically neutralmolecules or compounds where the electrostaticattractions are temporary, commonly existing astransitory states involving alternating positiveand negative charge distributions that minimizesrepulsion. The example in Fig. 1.11 representstwo different transitory states of adjacent atomswhere the dots are a schematic representation ofan electron cloud. The delta symbol (δ–) representsthe side of the atom or molecule with the greaterconcentration of negative charge (Fig. 1.11):While only schematic, this diagram indicates

how alternating electron distributions can leadto attraction between adjacent atoms or nonpolarcompounds. Both A and B are transitory states,and in fact, both are unstable states – as soonas the electrons take on one configuration theyare driven in the opposite direction by repulsiveforces, causing the other configuration. Thisthen drives the electron cloud back towards con-figuration A, and this alternation of transitorystates facilitates atomic or molecular attraction.Although these bonds are far weaker than cova-lent, ionic or metallic bonds, they are importantto understand because they control the meltingand boiling points of many compounds, particu-larly nonpolar organic compounds includingpesticides, fuels and solvents.

1.7 ACID–BASE EQUILIBRIA,pH, K VALUES

Chemical reactions in natural systems commonlyoccur in the presence of water, and understandingacid–base chemistry is crucial to understanding alarge proportion of issues in geochemistry, includ-ing the solubility of minerals and trace metals,chemical weathering, the decomposition oforganic matter, speciation of chemical elementsand reactions in the atmosphere.

Following the convention established by theSwedish chemist Svante Arrhenius, an acid is acompound that releases hydrogen ions (H+1) insolution (i.e. when dissolved in water). A fewexamples of classic acids are the inorganic acidshydrochloric acid (HCl), sulfuric acid (H2SO4)and nitric acid (HNO3), and organic acids suchas formic acid (HCOOH, also known as methanoicacid, it is what ants sting with) and acetic acid(HCH2OOH, also known as ethanoic acid, or morecommonly as vinegar).

Using nitric acid as an example, acids behave insolution as follows:

HNO3 ! H+1aqð Þ + NO3

–1aqð Þ ð1:12Þ

In water, HNO3 dissociates to release hydrogenions (and an equal amount of nitrate anions) intosolution. The subscript (aq) is used here to indi-cate that the ions are dissolved in water (i.e. areaqueous). The hydrogen ion actually does notexist alone in aqueous solutions, but rather bondswith water molecules and exists as the hydro-nium ion (H3O

+), but it is commonly representedin chemical equations as H+.

TheArrheniusdefinitionofabase is a compoundthat, when dissolved in water, releases hydroxylanions (OH–1) in solution. A few classic examplesof bases are sodium hydroxide (NaOH), calciumhydroxide (Ca[OH]2) and ammonium hydroxide(NH4OH). In solutions, bases behave as follows:

NaOH=Na+1 aqð Þ +OH –1aqð Þ ð1:13Þ

δ– δ– δ– δ–

A B A B

Fig. 1.11 Van der Vaals bonds between adjacentatoms. A and B represent nuclei of adjacent atomsand dots are electrons. In the example on the left,electron clouds are shifted to the left, causingattraction of the atoms. In the example on the right,electron clouds are shifted to the right, alsoenhancing attraction of the atoms. If this type ofalternating motion is synchronized, the atomswill be weakly attracted.

Acid–Base Equilibria, pH, K Values 19

Strong acids and bases dissociate completely, ornearly so. A strong base like NaOH may almostcompletely dissolve in solution, producing a highconcentration of hydroxyl anions and resultingin a very basic solution. A weak base like NH4OH(ammonium hydroxide) is much less soluble, so insolution it producesamuch lowerconcentrationofOH–. Similarly, the mineral acids listed above arestrong acids, while the organic acids (formic andacetic acid) are relativelyweak and producemuchlower concentrations of H+ in solution.The pH scale is the conventional means of

expressing the acidity or alkalinity of a solution.It is defined by the concentration1 of H+ ions insolution according to the equation:

pH= – log H+½ � ð1:14Þ

where [H+] is the concentration of hydrogen ions insolution. Inahighlyacidic solutionwith10–2 mol/Lof H+, the pH = 2. In an alkaline solution with 10–11mol/L of H+, the pH= 11. In aqueous solutions,the product of the concentrations of H+ and OH– is10–14 ([H+] ∗ [OH−] = 10–14).What this translatesto is that if the concentration of H+ = 10–4, the con-centration of OH–= 10–10. In this case, pH= 4 andqualitatively, it makes sense that the solution isacidic because there is far more H+ than OH–. Inanalkaline solutionwith [OH–] = 10–2, the concen-tration of [H+] = 10–12 and the pH= 12.Consider now the dissociation of two acids, one

a strong acid (nitric acid, HNO3) and one a rela-tively weak acid (carbonic acid, H2CO3).

1ð ÞHNO3 !H+ +NO –3 ð1:15Þ

2ð ÞH2CO3 !H+ +HCO –3 ð1:16Þ

The dissociation of nitric acid can be representedaccording to the Law of Mass Action as follows:

KaHNO3 = H+½ �∗ NO –3

� �= HNO3½ �=2:4×101

ð1:17Þ

The concentrations of H+ and NO3– are equal

when HNO3 dissolves and if we assume a HNO3

concentration of 1, [H+] = [NO3–] = √24 = 4.9.

The same treatment for carbonic acid produces:

KaH2CO3 = H+½ �∗ HCO –3

� �= H2CO3½ �=10 –6:37

ð1:18Þ

If, as was done with HNO3, we assume an H2CO3

concentration of 1, [H+] = [HCO3–] = √10–6.37 =

6.5 × 10–4. In other words, the concentration ofH+ produced by nitric acid in water is approxi-mately ten thousand (104) times greater thanH+ produced by an equivalent amount of carbonicacid in water.Ka values for some relatively common acids are

presented in Table 1.3.Two additional definitions of acids and bases

include Brønsted and Lewis classifications.A Brønsted acid is a substance that can

donate a proton (i.e. H+) to another substance,and a Brønsted base is a substance that canaccept a proton from another substance. The fol-lowing chemical reaction illustrates thisrelationship:

HCl +NH3 =Cl – +NH+4 ð1:19Þ

where HCl (the Brønsted acid) donates a protonthat is accepted by the Brønsted base NH3

(ammonia). The result is formation of a chlorideanion and ammonium, where Cl– can be termeda Brønsted base (it can accept a proton) andNH4

+ is considered a Brønsted acid. Mostminerals can be viewed as Brønsted basesbecause they consume H+ when they undergochemical weathering in soils. A good example isthe weathering of gibbsite, Al(OH)3, in acidic soils:

1 Note: this section discusses aqueous species (e.g. H+, NO3–) in

terms of concentration (e.g. [NO3–]), yet often species in solu-

tion are represented by activity (e.g. a NO3–), a term that takes

into account effects of other ions in solution on the effectiveconcentration (often activities are less than actual concentra-tions). In this introductory section and in many otherresources, concentrations are used; in Chapter 4, the conceptof activity is introduced.


Al OHð Þ3 + 3H+ !Al+3 +3H2O ð1:20Þ

Consider also that NaOH, described as anArrhenius base above because it yields dissolvedOH– in solution, is also considered a base by theBrønsted definition:

NaOH+H+ !Na+ +H2O ð1:21Þ

Lewis acids are substances that can accept elec-tron pairs when forming bonds (H+ is a good exam-ple),whereasLewisbasesare electron-pairdonors(OH– is a good example of this type of substance).

1.8 FUNDAMENTALS OF REDOXCHEMISTRY AND CHEMICALREACTIONS

Reduction–oxidation (redox) chemistry refers toprocesses that take place when atoms gain or loseelectrons, andoften involve reactionswhereoxygenis transferred. Electron transfer facilitates exchange

of energy that is crucial to processes across thechemical spectrum, from aquifer and soil dynamicsto photosynthesis and degradation of toxic organiccompounds. Innature, redoxreactionsoften involvechanges to the oxidation state of elements like car-bon, nitrogen, oxygen, sulfur, manganese and ironthat can exist in different oxidation states (e.g. car-bon exists in many oxidation states, including C–4,C0, C+2, C+4), where the change from one oxidationstate to another involves gain or loss of electrons.

Oxidation refers to the loss of electrons by anatom. Two common examples are the oxidation ofiron from its ferrous state (Fe+2) to its ferric state(Fe+3) by loss of one electron, and of nitrogen fromN+3 to N+5 by loss of two electrons. These tworeactions can be represented as follows:

Fe+2 = Fe+3 + e – ð1:22ÞN+3 =N+5 +2e – ð1:23Þ

Where do those liberated electrons go? Theyprobably were pulled away from the oxidized atom

Table 1.3 Selected acids and their Ka values, pKa values and conjugate bases.

Acid Conjugate Base

Name Formula Ka pKa Formula Name

Hydroiodic acid HI 3.2 ∗ 109 9.5 I− IodideHydrobromic acid HBr 1.0 ∗ 109 9.0 Br− BromideHydrochloric acid HCl 1.3 ∗ 106 5.1 Cl− ChlorideSulfuric acid H2SO4 1.0 ∗ 103 3.0 HSO4

− Hydrogen sulfate anionNitric acid HNO3 2.4 ∗ 101 1.4 NO3

− NitrateHydrogen sulfate ion HSO4

− 1.0 ∗ 10−2 −2.0 SO4−2 Sulfate anion

Sulfurous acid H2SO3 1.3 ∗ 10−2 −1.9 HSO3− Hydrogen sulfite anion

Phosphoric acid H3PO4 7.1 ∗ 10−3 −2.2 H2PO4− Dihydrogen phosphate anion

Dihydrogen phosphate ion H2PO4− 6.3 ∗ 10−8 −7.2 HPO4

−2 Hydrogen phosphate anionHydrogen phosphate ion HPO4

−2 4.2 ∗ 10−13 PO4−3 Phosphate anion

Nitrous acid HNO2 7.2 ∗ 10−4 −3.1 NO3− Nitrite anion

Hydrofluoric acid HF 6.8 ∗ 10−4 −3.2 F− FluorideMethanoic (formic) acid HCOOH 1.8 ∗ 10−4 −3.7 HCOO− Methanoate (formate) anionBenzoic acid C6H5COOH 6.3 ∗ 10−5 −4.2 C6H5COO

− Benzoate anionEthanoic acid CH3COOH 1.8 ∗ 10−5 −4.7 CH3COO- Ethanoate (acetate) anionCarbonic acid H2CO3 4.4 ∗ 10−7 −6.4 HCO3

− Bicarbonate anionHydrogen carbonate ion HCO3

− 4.7 ∗ 10−11 −10.3 CO3−2 Carbonate anion

Hydrogen sulfide H2S 1.1 ∗ 10−7 −7.0 HS− Hydrogen sulfide anionAmmonium ion NH4

+ 5.8 ∗ 10−10 −9.2 NH3 Ammonia

Fundamentals of Redox Chemistry and Chemical Reactions 21

because a neighboring atom had a greater attrac-tion for those valence electrons. The oxidation ofone atom cannot occur without a correspondingchange to another atom. This change is reductionand it takes place when an atom gains electrons.Two common examples are the reduction of

oxygen gas (where oxygen is O0) to oxygen anions(O–2), and of C+4 (e.g. the C in CO2) to molecularcarbon, C0 (e.g. the C in some organic matter).

O2 + 4e – =2O –2 ð1:24ÞC+4 +4e – =C0 ð1:25Þ

While it is useful to examine individual examplesof reduction or oxidation, loss of electrons fromone atom results in gain of electrons for anotheratom. A simple example that you can probablyvisualize is the oxidation of ironmetal (Fe0) to ironoxide (Fe2O3) where Fe occurs in its trivalent orferric state (Fe+3).

4Fe0 +3O2 =2Fe2O3 ð1:26Þ

It helps to examine individual reduction and oxi-dation pairs in order to see where and how theexchange of electrons takes place:

Fe0 = Fe+3 +3e – ð1:27Þ½O2 +2e – =O –2 ð1:28Þ

All elements in their pure state, like the Fe atomand O atom (represented as ½O2) shown above,have an oxidation state of zero.The oxidation of iron metal (Fe0) by oxygen gas

(O2) involves the loss of three electrons from aneutral iron atom and gain of two electrons bya neutral oxygen atom. Clearly, one oxygen atomcannot cause the oxidation of one iron atom fromFe0 to Fe+3, and this brings up the need to balanceredox reactions, as follows:

2 ∗ Fe! Fe+3 +3e –� �

3 ∗ ½ O2 +2e – !O –2� �

This results in a balanced pair of reactions (withrespect to electrons) where two iron atoms lose

six electrons and three oxygen atoms gain 6electrons:

2Fe!2Fe+3 +6e –

1:5O2 +6e – !3O –2

and the paired redox reaction can be expressed as:

2Fe +1:5O2 !2Fe+3 +3O –2 ð1:29Þ

The electrons are excluded from this redoxreaction because there are 6e– on both sides(products and reactants) and thus cancel eachother out, yet electron flux is implied by thechange in oxidation states of Fe andO atoms. Terminologically, iron is the reducingagent that donates electrons to oxygen, causingoxygen to become reduced (and iron to be oxi-dized); oxygen is the oxidizing agent that pullselectrons from iron, which results in reductionof oxygen and oxidation of iron. In reality, the oxi-dation of iron ends up producing iron oxide,shown here as the mineral hematite:

2Fe +1:5O2 ! Fe2O3 ð1:30Þ

Hematite consists of Fe in its most oxidized state(Fe+3, ferric iron) and oxygen in the form that ittakes in virtually all compounds, O–2.

The terms “oxidizing” and “oxidized” areeffectively synonymous. An oxidized soilwill likely contain abundant available O2

as well as minerals that are stable in anoxidized (or oxidizing) environment, e.g.iron oxides. This soil will be oxidizingbecause if an organic compound or sulfidewere to be transported into the soil, anoxidation reaction would likely lead todestruction of the reduced substance.Similarly, reduced and reducing environ-ments are effectively synonymous.


Redox chemistry comprises some of the mostimportant reactions in the realm of geochemistryand biochemistry. Microbial activity often playsan important role in redox chemistry becauseelectron transfer is an energy source – a classicexample is the microbially mediated decomposi-tion of fuels and solvents in soils, where the oxida-tion of organic carbon provides energy to themicrobe and results in the transformation ofleaked fuel into H2O and CO2. More-detailed infor-mation on redox reactions is presented inChapter 4 on aqueous geochemistry.

1.9 CHEMICAL REACTIONS

Reactions among elements and compounds havebeen presented in a few ways in this introductionthus far, including reference to chemical bonding,the formation of elements and redox chemistry.Chemical equations are algebraic expressions thatrepresent quantities and charges of constituentsinvolved in chemical reactions, and in some waysare the language of geochemistry, or at least areone of the languages. Accordingly, the followingsection will present a few fundamental conceptsabout chemical reactions, what they represent,how to balance them and how to interpret them.First, a few general rules. Chemical reactions

commonly take place in the presence of water, butif water is not produced or is not consumed by thereaction, it is not listed in the reaction. Given thatthe 1st Law of Thermodynamics states that mattercan neither be created nor destroyed, but rathercan only change forms, chemical reactions shouldnot give the illusion that the 1st law is being violated–whatthismeans is thatall chemical reactionsmustbe balanced. If there are four oxygen atomsexpressed on the reactants side of the equation, thenthere also must be four oxygen atoms on the pro-ducts side. Similarly, if the reaction is of the redoxvariety, the charges (sum of + and –) should beequal on both sides. If the reaction involves nuclearfusion or fission, then energy and mass should beequal on both sides of the equation.

1.10 EQUILIBRIUM,THERMODYNAMICS AND DRIVINGFORCES FOR REACTIONS:SYSTEMS, GIBBS ENERGIES,ENTHALPY AND HEAT CAPACITY,ENTROPY, VOLUME

This section begins with the example of pyrite(FeS2 or iron disulfide), a mineral that forms inO2-poor environments that include deep crustallevels and anoxic surface environments likeswampy muds (technically iron sulfide that formsat Earth surface temperatures is a poorly orderedform of ~ FeS, e.g. mackinawite). Iron, in theFe+2 state, and sulfur, in a combination of S–2

and S0 states (average = S–1), both occur in chem-ically reduced forms in the mineral pyrite – theseoxidation states are stable in reducing/anoxic/anaerobic/O2-poor environments.

Given the conditions under which pyrite forms,it is possible to predict its fate in oxidizing (O2-rich)environments, common examples being soilsexposed to the oxygen-rich, water vapor-bearingatmosphere, or an O2-rich bubbling stream,where the stable forms of iron and sulfur areFe+3 and S+6. Under oxidizing conditions likethese pyrite undergoes chemical oxidation, pro-ducing iron hydroxide with its characteristicrusty orange stains typical of many rock outcropsand soils. The reaction of pyrite, water and oxy-gen to produce iron hydroxide, sulfuric acid andfree electrons (oxidation!) can be expressed asfollows:

2FeS2 +7H2O+7:5O2 !2Fe OHð Þ3 + 4H2SO4

ð1:31Þ

(In reality this reaction occurs in two or moresteps, often in the presence of sulfur-oxidizingbacteria). Viewing this reaction in terms of theextent of iron oxidation would likely result in acharacteristic pattern observed for many reac-tions in nature (Fig. 1.12).

In Fig. 1.12, the initial system (pyrite in contactwith atmosphere) is out of equilibrium – both the

Equilibrium, Thermodynamics and Driving Forces for Reactions 23

Fe and S are unstable in reduced forms whenexposed to O2-rich air (e.g. by a landslide thatexposes fresh un-oxidized rock) and the pyritebegins to react. This reaction is expressed asFe(II) ! Fe(III) in the graph, and note that ini-tially the reaction occurs rapidly but as the reac-tion progresses the rate steadily decreases andultimately ceases altogether. There are two prob-able explanations:1 the system has reached equilibrium, a condi-tion where net concentrations of products andreactants do not change. Dynamic equilibriumsometimes occurs in natural systems, when therate of formation of products equals the rate offormation of reactants. If the Fe(OH)3 and H2SO4

produced by pyrite oxidation are not leachedout of the system, a dynamic equilibrium may beestablished where the rate of the forward reactionshown above is equal to the rate of the reversereaction – i.e. FeS2, O2 and H2O are producedat the same rate as are Fe(OH)3 and H2SO4. Atdynamic equilibrium, reactions are taking place,namely “products” are dissolving to produce“reactants” at the same rate that “reactants” pro-duce “products”, but the net concentrations ofproducts and reactants do not change with time.However, any change to a variable involved in a

dynamic equilibrium (e.g. concentration of reac-tant or product, volume, pressure, temperature)will cause a shift to counter the change. For exam-ple, removing a reactant will shift the reactiontoward the direction of reactants – the rate willslow, or reactants will be produced at the expenseof products.Or, if products are lost from the system,the reactionwill continue to form products, whichleads to the other possibility:2 The reaction has run to completion. In somenatural systems, where products are lost dueto leaching (e.g. of H2SO4 in the case above)or degasification (e.g. of CO2 with decompositionof hydrocarbons), a dynamic equilibrium cannotbe established. If the pyrite reaction above stoppedbecause all Fe(II) had been consumed to produceFe(III) (i.e. if the reactionwere to run to completionbecause products are being lost from the system),that system will not reach dynamic equilibrium.The natural environment differs from the labora-tory in that reactants are often lost from soils, rocksand groundwaters, and dynamic equilibrium maynot apply. In other cases, reactions in nature maynot reach equilibrium because reaction rates arevery slow – the reaction never proceeds past theearly convex part of the disequilibrium curve. Thisoften occurs in soils, where igneous minerals suchas amphiboles and pyroxenes that are stable inhigh-temperature, low-O2 environments persist ina state of disequilibrium because the rates at whichthey decompose in weathering environments arerelatively slow.

1.10.1 Systems, species, phases andcomponents

In spite of certain limitations, examining environ-mental systems through the lens of equilibriumthermodynamics can be very useful. It canhelp determine the direction in which chemicalchanges will take place (e.g. pyrite is unstableand will oxidize in contact with the atmosphere)and also to infer rates because the farther a system

Time

Exte

nt of F

e(I

I)F

e(I

II)

Disequilibrium Equilibrium

Fig. 1.12 Typical progression of a geochemicalsystem towards equilibrium. A high degree ofdisequilibrium in the early stages causes rapid ratesof change, but as the system approaches equilibrium,rates decrease logarithmically.


is from equilibrium, the faster it will react toreach equilibrium. A common term used inthermodynamics is system, which is a somewhatarbitrary definition of the components we wish toconsider. Depending on the question, a systemmight be an aquifer, or a pore within an aquifer,or the entire hydrologic cycle; it could be an entiregranitic pluton, or it might be a micrometer-sizedfluid inclusion in a quartz crystal. It reallydepends on the scale of study. If the question isclimate change, the entire troposphere might beconsidered as the system, or a smaller systemcomprised only of a landfill might be consideredif the main concern is a single source of carbon(e.g. CH4).Systems can be open (e.g. a stream, the

atmosphere, a leaking landfill), where materialis added or lost, or closed, where flow of materialis restricted (e.g. tiny pores within impermeablefine-grained sediments). In some cases, systemsare closed with respect to solids but open withrespect to gases or heat, and in other cases sys-tems can be defined as closed to physical and ther-mal flux, in which case they are isolated.Systems are comprised of components, phasesand species.In geochemistry, species are microscopic

entities, commonly ions or gases such as Ca,CO3

–2, SO4–2, CO2 or H2S, whereas phases are

defined as physically separable parts of a system,typically minerals, liquids (e.g. H2O) and distinctgases (dissolved CO2 and O2 in stream water, forexample). Phases are comprised of species—forexample, the phase calcite is comprised of thespecies Ca+2 and CO3

–2, or of the species CaOand CO2 (calcite can be defined in either way).While species are generally substances thatcan or do exist in nature, components do notnecessarily exist in nature. Sometimes they aresimilar to species, but in other cases they maybe mathematical expressions that can be usefulin thermodynamic calculations, one examplebeing KNa–1, a mathematical operator used toindicate addition of K and subtraction of Na.

So, a soil with the mineral dolomite (CaMg[CO3]2), water, dissolved Ca+2, dissolved Mg+2,dissolved CO3

–2, CO2 gas and quartz could bedefined as having four phases (dolomite, water,quartz, CO2 gas) and 6 species or components(Ca+2, Mg+2, CO3

–2, CO2, H2O, and SiO2).Lastly, the phase rule (sometimes referred to

as Gibbs’ phase rule) relates components (C),phases (P) and degrees of freedom (F) accordingto this simple equation:

F=C−P+2 ð1:32Þ

Degrees of freedom represent tangible changesto a system, typically temperature and pressure.In a system with 2 degrees of freedom, tempera-ture and pressure can both change without pro-ducing a change in the state of the system. Theclassic example involves a simple system invol-ving only one component, H2O, which can existin 3 phases (solid, liquid and vapor forms ofH2O), as displayed in the H2O phase diagram(Fig. 1.13).

There is one point in the H2O phase diagramwhere all three phases coexist – the triple point.

Solid

(ice)

Water vapor

Liquid water

0 0.01 100 374

T °C

218

1

0.006

Triplepoint

P(a

tm)

Fig. 1.13 Phase diagram for H2O depicting onecomponent (H2O) and three phases (solid, liquid andgaseous water).


At the triple point, there is one component(H2O; C= 1) and three phases (solid, liquid andvapor; P= 3). The phase rule indicates that forthis system,

F=1 –3+2=0

In other words, there are no degrees of freedom.Any change in either T or Pwill produce a changein the state of the system. Increasing T at constantP will cause ice to melt and liquid to vaporize.Increasing P at constant T will cause vapor tocondense and ice to melt, producing a systemwith only one phase. Consider for a moment a sys-tem with only one phase, liquid water at 100 �Cand 200 atm of pressure. In this case, the phaserule indicates that (for a system where C= 1and P= 1) there will be two degrees of freedom,and this is borne out in the diagram above – ifeither T or P changes, or even if both T and Psimultaneously change, there will be no changeto the state of the system. It will remain liquidwater until the system either cools or decreasesin pressure to the point where P–T reaches thephase boundary (line) between liquid and solidor between liquid and vapor.

1.10.2 First law of thermodynamics

The term thermodynamics implies an approachbased on changes in heat, and the field didoriginate in studies of the transformation of heatenergy to mechanical energy during the 1800 s.Energy can change forms, e.g. potential energyto heat (or vice versa), as in the combustion oforganic matter, a process where storedchemical potential energy is converted to heatenergy. Thermodynamics deals with transfers ofenergy, and one of the fundamental principles isthe First Law of Thermodynamics, oftenknown as the Law of Conservation of Matterand Energy, which can be stated as follows:

ΔE=Q −W ð1:33Þ

Or, for small increments of change, this equationcan be stated as:

dE=dq – dW ð1:34Þ

E is the internal energy of a system, Q or dQ rep-resent heat flux, andW representswork done onthe system. The first law essentially states thatenergy is neither created nor destroyed – matterand energy may change forms, new phases maybe produced, gases may convert to solids, or solidsmay dissolve, but the net energy of a closed sys-tem never changes (or, for an open system, thenet energy of the open system and its surround-ings never changes). The kinetic energy of frictionalong a geological fault may produce heat, or thechemical potential energy of the bonds in hydro-carbons may produce heat during combustion,but the net change in energy is zero.Consider an experiment you could perform

while sitting in your overstuffed chair reading thistext, the example of heat (Q) produced by thework (W) of rubbing hands together. The systemcan be defined as your body and its immediatesurroundings. The energy you need to move yourshoulders and arms to produce friction comesfrom the stored energy in the chemical bonds inthe food you recently ate, and in a simplifiedapproximation, this chemical energy is trans-formed to mechanical energy. The frictionbetween your hands converts mechanical energyto heat energy, causing your hands to warm up(and raising the temperature of the system). Thisheat will soon be lost to the surroundings (anissue we will soon address with the 2nd Law),but if we consider the net energy of the system(body and immediate surroundings), it has notincreased or decreased, but rather merely haschanged forms. Of course, the heat will eventuallyescape from your surroundings and ultimatelyyou will need to eat more food to provide thechemical energy needed to reinitiate the cycle


(conversion of chemical energy in food to heatenergy is an irreversible reaction in our bodies).In terms of the equation of the 1st Law, the work(W) done will have a positive sign, as will the heatterm Q (heat is gained by the system, so the termQ will have a positive sign, because in geochem-istry, the flow of heat is considered positive forany system that gains heat). So, in a semi-quantitative sense, W and Q will both be positiveand the values will be equal, andΔE= Q –W = 0,reflecting the fact that there is no change in thenet energy of our body and its surroundings (atleast in the short term).Work can be defined as follows:

dw=P∗dV ð1:35Þ

where P is external pressure on a system and dVis the change in volume of the system. A typicalway to consider work relative to volume is toquantify the work done by expansion duringthe change of state from liquid water to watervapor. At constant P, a positive dV term (expan-sion increases V) will produce a positive dw, indi-cating that work has been done on thesurroundings by the system. If you substitutethe PdV term for W into the equation of the firstlaw presented above, you will arrive at this equa-tion for the first law, expressed in terms of changein internal energy:

dE=dq –PdV ð1:36Þ

1.10.3 Second law of thermodynamics

The Second Law of Thermodynamics dealswith entropy, a measure of the degree of disor-der within a system. Any system tends toward astate of increasing randomness unless energy isadded to the system to increase order. Increasingentropy is a spontaneous process and energymust be added to produce order. In geochemistry,

an example of entropy as a spontaneous process ischemical weathering of a granite, for example,where minerals with ordered crystal lattice struc-tures (a low entropy state) are decomposed intosoluble aqueous species such as Na+, K+, andSi(OH)4 that are then scattered across the globe.Only with addition of energy (e.g. the internalheat of the Earth) can some order be restored(i.e. entropy decreased). An everyday exampleof entropy is this: the natural state of a kitchenor living room will progress toward a state of dis-order (greater entropy) characterized by dirtydishes, potato chip bags, music scores and oldnewspapers scattered about unless we expendenergy to restore order.

Entropy (S) can be represented by the followingequation, where q is heat and T is temperatureand the process is reversible:

dS=dq=dT ð1:37Þ

Entropy can be quantified in terms of change inheat content per change in temperature, andunits of S are joules/deg (non-Si units are cal/deg; 1 cal = 4.18 J). Consider the entropy of liquidwater and water vapor at standard temperature(25 �C) and pressure (1 atmosphere) – standardT and P are considered standard state condi-tions and are commonly used to make calcula-tions immensely simpler. S for H2Ol is 69.9 J/mol/K, and for H2Og it is 188.7 J/mol/K(Appendix III). Intuitively, the value of S is greaterfor the gaseous, more disordered state of H2O thanfor the liquid state.

1.10.4 Enthalpy

Another important consideration in thermody-namics is enthalpy (H), the heat content of asystem. In some cases enthalpy is expressed inunits of calories (cal) or kilocalories (kcal), whichmakes sense because calories are a common


measure of heat in everyday life; however, the SIunits are joules (J) (or kilojoules, kJ).Enthalpy is typically expressed as follows:

H =E+PV ð1:38Þ

or

dH =dE+PdV ð1:39Þ

But often the most important considerationrelated to enthalpy is the change in H (ΔH=H2 – H1) during a reversible reaction, where thetwo values of H represent enthalpies associatedwith different states of matter, like liquid waterand water vapor, or with elements in differentbonding arrangements, e.g. Fe and O2 vs.Fe2O3. ΔH is an important parameter in geo-chemistry because it expresses heat absorbed orreleased during changes of state (e.g. evapora-tion) or during chemical reactions that produceminerals, ions or molecules. If heat is gainedduring a reaction (i.e.ΔH is positive), the reactionis endothermic; conversely, if heat is lost (ΔH isnegative), the reaction is exothermic.Combustion of organic matter is clearly exo-

thermic – we burn firewood and hydrocarbonsto produce heat. Boiling of water, the transforma-tion of andalusite to sillimanite during progrademetamorphism, and the maturation of petroleumin sedimentary basins are all endothermic pro-cesses – they absorb heat. The stored heat canthen later be released; for example, stored heatin petroleum is released during the exothermicreaction known as combustion. It is worth point-ing out the difference between two terms thatrepresent heat, Q and H. Q represents flow of heator heat transfer, for example from hot Hawaiianlava into cool ocean water; H represents heatstored within a system, such as the stored heatin petroleum, water vapor or sillimanite.Given that reactions either consume or pro-

duce heat, we can calculate the difference inenthalpy between the reactants and productsand determine the amount of heat produced or

consumed by the reaction. This is importantbecause it is one of the ways that thermo-dynamics can help to predict the behavior ofenvironmental systems such as soils and waters;typically, spontaneous processes produce (i.e.release) heat, i.e. spontaneous processes usuallyare exothermic (caveat: while this is generallytrue it is not always the case – for example, whensome salts dissolve in water the solution getscolder. The process absorbs heat from the wateryet it is a spontaneous process because theincrease in entropy is more important than thepositive ΔH).Enthalpies of reactions (ΔHo

R) are deter-mined by summing the enthalpies of formation(ΔHo

f) of all reactants (standard state conditions)and subtracting this term from the sum of ΔHo

f

values of all products (standard state conditions):

ΔHoR =Σnx ∗Ho

fx productsð Þ–Σnx ∗Ho

fx reactantsð Þ ð1:40Þ

Enthalpies of formation are available from varioussources and selected examples are presented inAppendix III. By convention, Ho

f = 0 for ele-ments in their pure state (e.g. Fe, Si, Na) andfor gases such as H2, N2 and O2. In theΔHo

R reac-tion above, the Hof for each reactant or product(represented by the variable x) is multiplied bythe number of moles (n) expressed in the reaction.We can examine the reaction of Fe and O2 to

form hematite (Fe2O3) by the chemical reaction2 Fe + 1.5 O2 = Fe2O3. Values of H

of (in kJ/mol):

Hof Feð Þ=0

Hof O2ð Þ=0

Hof Fe2O3ð Þ= –824:2

It is crucial to remember to multiply Hof values

by molar abundances presented in the chemicalreaction.

ΔHoR = 1∗ –824:3ð Þ – 2∗0+1:5∗0ð Þ= –824:3kJ=mol:


The negative ΔHoR for this reaction implies that

it is spontaneous at standard temperature andpressure, and that iron will oxidize to formiron oxide.Under certain conditions, calcite (CaCO3) and

quartz (SiO2) react to form wollastonite (CaSiO3)plus CO2 according to this reaction:

CaCO3 + SiO2 =CaSiO3 +CO2 ð1:41Þ

Is this a spontaneous process at the Earth’ssurface? (i.e. 1 atm and 25 �C). Values of Ho

f (inkJ/mol) are (Appendix III):

Hof CaCO3ð Þ= –1207:4Ho

f SiO2ð Þ= –910:7Ho

f CaSiO3ð Þ= –1630Ho

f CO2ð Þ= –393:5

The enthalpy of the reaction can then be deter-mined as follows:

ΔHoR = ½ð1mol ∗ –1630kJ=molÞ

+ ð1mol ∗ –393:5kJ=molÞ�– ½ð1mol ∗ –1207:4kJ=molÞ+ ð1mol ∗ –910:7kJ=molÞ�= +94:6kJ

This reaction requires addition of 94.6 kJ ofheat (per mol of each reactant given the stoichi-ometry of the reaction), indicating that it is notspontaneous, and evidence from geology con-firms this conclusion because calcite and quartzdo not react to form wollastonite and carbondioxide until systems reach medium-grademetamorphic temperatures and pressures(~500 �C, > 1 kb).

1.10.5 Heat capacity

Heat capacity is the amount of heat required toraise the temperature of a given amount (1 mol or1 g) of a substance by 1 �C. It is defined as the ratio

of heat added to resulting temperature change,where the greater amount of heat required toresult in temperature increase corresponds tohigher heat capacity.

C=dq=dT ð1:42Þ

C= heat capacity and dq and dT are changes inheat and temperature. Rearranging emphasizesthe point that, for a given amount of heat added(dq), the magnitude of temperature increase willbe lower if the heat capacity is high.

dT=dq=C ð1:43Þ

At constant volume, heat capacity can beexpressed as:

Cv = δq=δTð Þv ð1:44Þ

Given equation 1.36, and realizing that PdV= 0at constant volume:

Cv = δE=δTð Þv ð1:45Þ

(note: here, the symbol δ indicates that the deri-vation is performed with a restriction, in this caseconstant volume. The subscript v indicates thatvolume of the system is constant).

At constant pressure:

Cp = δq=δTð Þp ð1:46Þ

and equation 1.36 can be rearranged slightlyto give:

dq=dE+PdV ð1:47Þ

Now, substituting the right side of equation 1.47intothenumeratorof theright sideof equation1.46:

Cp = δE=δTð Þp +P δV=δTð Þp ð1:48Þ


And given equation 1.39 (effectively, δH= δE+ PδV),

Cp = δH=δTð Þp ð1:49Þ

Examples of heat capacities of substances (at 25 �C)include liquid water, which has a relatively highvalue (C= 4.19 J/g.K) relative to dry rock, whichhas values in the range of 0.7 to 0.9 J/g.K, or steel,which has an even lower value of 0.47 J/g.K. Thepaucity of water in deserts means that rock is thedominant control on temperature change and thisresults in drastic temperature swings, both diur-nally and seasonally; by comparison, in moisterregions such as tropical forests or temperate coastalregions, temperature extremes are minimized bythe high heat capacity of water.Heat capacity is an extensive property, i.e. it is

dependent on the amount of the substance in ques-tion – the greater the amount of substance, thegreater the amount of heat needed to be added toachieve the same change in temperature. That said,values of heat capacity are typically normalized to amassof1gram,meaningthateffectivelyheatcapac-ity is an intensive property, i.e. when normalized(divided by mass), it reflects a characteristic of agiven substance independent of amount (i.e. mass).

1.10.6 Gibbs free energy

The Gibbs Free Energy (G) of a system accountsfor changes in both enthalpy and entropy duringreactions.

ΔGoR =ΔHo

R – TΔSoR ð1:50Þ

Any reaction that produces a decrease in theGibbs free energy is spontaneous – any reactionfor which ΔGo

R is negative is spontaneous.Systems tend towards lower-energy states in theabsence of new addition of energy, so a decreasein G will produce a more stable system. In aschematic way (Fig. 1.14), stability of a physical

system can be used to illustrate this point. The ballat point A is unstable with respect to location, andto decrease this instability (high energy state), itwill roll down to point B, and if it can overcomethe slight energy barrier at point C, it will eventu-ally achieve its most stable configuration byrolling down to point D.Point B might be referred to as a metastable

condition, one that is not the most stableconfiguration (that would be D), but one thatmay play a strong role in system behavior.An example of metastability in the geochemical

realm is the occurrence of the mineral halloysite,a disordered form of kaolin that can form in trop-ical soils when highly unstable igneous mineralssuch as olivine and pyroxene rapidly dissolve(over hundreds to thousands of years). Rapiddissolution may lead to formation of metastablehalloysite rather than kaolinite due to kinetic fac-tors; once formed, halloysite may require hundredsof thousands to millions of years to finally trans-form into thermodynamically stable kaolinite.DeterminingGibbs free energies forgeochemical

systems allows prediction of their behavior, muchlike the simple analysis of the ball between pointsA and D. For example, if magnetite (Fe3O4), waterand oxygen gas exist in a soil at standard condi-tions, we might ask, is the assemblage stable or

A

BC

D

Fig. 1.14 Schematic diagram illustrating relativestability conditions, from unstable at points A and C,to metastable at point B and stable at point D. Notethat the metastable point B can be reached frompoints A or C, and that to shift from metastable(B) to stable (D), some energy must be introduced.If not, the metastable condition may persistfor a long time.


of out of equilibrium? Ismagnetitemore or less sta-ble in this soil than an iron hydroxide such as goe-thite? The possibility that the system will react toform goethite (FeOOH) to achieve a lower-energystate could be assessed according to this reaction:

2Fe3O4 +0:5O2 +3H2O=6FeOOH ð1:51Þ

ΔGoR for the reaction is computed according to

this equation:

ΔGoR =Σnx∗Go

fx productsð Þ – Σnx ∗Gofx reactantsð Þ

ð1:52Þ

Values for Gof (in kJ/mol) are (Appendix III):

Gof Fe3O4ð Þ= –1015:5Go

f O2ð Þ=0Go

f H2Olð Þ= –237:2Go

f FeOOHð Þ= –488:6

These values produce the following equation:

ΔGoR = 6mol∗ –488:6kJ=mol½ �

– 2mol∗ –1015:5kJ=molð Þ½+ 0:5mol∗0kJ=molð Þ+ 3mol∗ –237:2kJ=molð Þ�

= –189:0kJ

What this negative Gibbs free energy valuedemonstrates is that magnetite is unstable inthe presence of water and oxygen at the Earth’ssurface and will react to produce goethite, a com-mon soil mineral. What this reaction does not tellus is how fast this reaction will take place. Is therean energy barrier? Does the system need to over-come an activation energy like that at point C inthe ball example above? Does diffusion of O2 orH2O to magnetite surfaces limit reaction rate?In reality, magnetite can persist in soils in a met-astable state for at least thousands of years, andthat fact is not evident from thisΔGo

R calculation.Section 1.11 on kinetics addresses some questionsrelated to rates of processes.

1.10.7 Gibbs free energy and theequilibrium constant

The Gibbs Free Energy of a reaction can also beexpressed in terms of the equilibrium constant(Keq), which is much like the Ka discussed earlierin this chapter. Consider a general reaction atequilibrium where:

aA+ bB= cC+ dD ð1:53Þ

A and B, and C and D are reactants and products,respectively, and the lowercase a, b, c and d repre-sent molar fractions of reactant or product. In thisgeneral case:

Keq = C½ �c ∗ D½ �d= A½ �a ∗ B½ �b ð1:54Þ

[C]c represents the concentration of reactantC raised to the c power, and so on. Products arein the numerator and reactants in the denomina-tor. Or, for a more tangible example, we can writethe Keq for oxidation of ferrous iron (Fe+2). Thechemical reaction is:

4Fe+2 +O2 +10H2O=4Fe OHð Þ3 + 8H+

ð1:55Þ

and:

Keq := Fe OHð Þ3� �4∗ H+½ �8= Fe+2� �4 ∗ O2½ �8∗ H2O½ �10

ð1:56Þ

However, by convention, the concentration ofwater is given the value of 1, and pure solid phaseslike ferrihydrite (Fe(OH)3) are similarly assignedvalues of 1, so in this case the Keq simplifies to:

Keq : = H+½ �8= Fe+2� �4∗ O2½ �8 ð1:57Þ

which means that the main controls on the reac-tion are the concentrations of H+ and Fe+2 andthe availability and reactivity (fugacity) of O2.


A high value for the Keq indicates that the reac-tion will proceed in the direction of products. Inother words, for reactions that favor products,the numerator will be greater than the denomina-tor, producing a high Keq. In cases where reac-tants are more favored (i.e. reactants arepredicted to occur in higher concentrations thanproducts), Keq values are small. For example, thedissolution of gibbsite in the presence of acid,represented by the following reaction,

Al OHð Þ3 + 3H+ =Al+3 +3H2O ð1:58Þ

has a Keq = 108.11 at 25 �C and 1 bar, which isoften expressed in log form as log Keq = 8.11(Langmuir, 1997). Expressed in Keq form:

Keq = Al+3� �

= H+½ �3 = 108:11 ð1:59Þ

it is clear that the system strongly favors dissolu-tion of Al(OH)3 and formation of dissolved Al+3.(Of course, pH of the solution will strongly controlthe probability of this reaction). Conversely, thedissolution of fluorite (CaF2 = Ca+2 + 2F–) has alog Keq = –10.6, and this low value indicates thatfluorite is very insoluble in water. The product ofthe concentrations of [Ca+2] and [F–]2 is extremelylow (Keq = 10–10.6), indicating that the reactantside of the equation is favored.ΔGo

R for this reaction or any reaction can beexpressed as:

ΔGoR = –RTlnKeq ð1:60Þ

This is the Van’t Hoff equation, where R (thegas constant) is equal to 8.314 J/cal/mol K (or8.314 × 10–3 kJ/mol K) and T is temperature(in K), which produces units of j/mol (or kJ/mol) for ΔGo

R.In most cases, Gibbs free energy data are

determined for systems at 25 �C (298.15 K), sofor systems at 25 �C (reasonably representativeof surface environments), ΔGo

R can be furthersimplified and converted to log10 as:

ΔGoR = –5:708∗ log Keqð Þ ð1:61Þ

Typically enthalpy and entropy changes are notstrongly temperature-dependent, so given equa-tion 1.54, the Gibbs Free Energy, ΔG, should varyin a linear manner with temperature. Rearran-ging equation 1.50 slightly and presenting thevariables in a generic sense gives us:

ΔG=T =ΔH=T –ΔS ð1:62Þ

By then substituting the right side of equation1.60 for ΔG:

RTlnK=T = –ΔH=T +ΔS ð1:63Þ

Then dividing and rearranging:

lnK = –ΔH=RT +ΔS=R ð1:64Þ

Viewed in the following manner, equation 1.64 isan equation of a line: lnK= y, –ΔH/R is m (slope),1/T is x and ΔS/R is b (constant). Two character-istic types of plots (Fig. 1.15) result when differentvalues for T are plugged in to equation 1.64.

1/T

lnK

1/T

lnK

Endothermic Exo

therm

ic

Decreasing T

m = –ΔH/R

m = +ΔH/R

Decreasing T

Fig. 1.15 Arrhenius plots showing relationshipbetween equilibrium constant (K) and temperaturefor endothermic and exothermic reactions. Forendothermic reactions, increasing temperatureincreases K, i.e. adding heat increases reaction rate;for exothermic reactions, the opposite is true.


Taking the derivative of equation 1.64 withrespect to temperature gives a different formula-tion of the van’T Hoff equation:

d lnKð Þ=dT =ΔH=RT2 ð1:65Þ

Note that the term ΔS/R disappears because bothterms are constants (assuming S does not changewith temperature).Integrating equation 1.65 between tempera-

tures T1 and T2 produces equation 1.66, anotherexpression of the van’t Hoff equation, one whichmakes it possible, given a K value (e.g. K1) atstandard temperature (e.g. T1), to determine anunknown value of K (e.g. K2) at a different tem-perature (i.e. T2).

ln K2=K1ð Þ= –ΔH=R 1=T2 –1=T1ð Þ ð1:66Þ

For calculations using this equation, R is thegas constant (8.314 J/K.mol) and temperature isin Kelvin. It is straightforward and can beuseful (note: using this equation K is unitless, asit also is when expressed in equations 1.60through 1.64).

1.11 KINETICS AND REACTIONRATES: DISTANCE FROMEQUILIBRIUM, ACTIVATIONENERGY, METASTABILITY

Thermodynamic information is often insufficientto fully understand behaviors or states of ionsand compounds in nature, especially in low-temperature systems like soils, waters and theatmosphere. Soils, for example, almost alwayscontain minerals that are thermodynamicallyunstable at the Earth’s surface, such as the olivine,augite and plagioclase that form at T> 800 �C incooling basalt lava, as well as minerals that arethermodynamically stable at low temperatures(25 �C, for example), such as calcite, kaolinite

and hematite. While the latter three can cometo equilibrium with surface waters, olivine, augiteand plagioclase do not. Thermodynamics tells usthat. What thermodynamics does not quantify israte. Consider magnetite in tropical soils; thermo-dynamics (e.g.ΔG) indicates that it will transformto goethite, but it does not indicate how fast.

Questions about rates of processes areaddressed by the study of kinetics. Reactionsthat occur slowly, are not reversible or do not takeplace in a system at equilibrium for whatever rea-son (e.g. soils) are often best understood usingkinetic approaches. The same can be said forheterogeneous reactions, those that involvevarious states of matter such as solid mineralsand amorphous solids mixed with liquid and gasphases, conditions that tend to occur in soils,sediments, streams, aquifers and the atmosphere.

Rates of geochemical reactions depend onnumerous factors, including temperature, pres-sure, redox conditions, pH, mineral composition,abundance of organic matter, pore water composi-tion, diffusion rates, bond types, biotic factors,system composition, and more. In many cases,geochemical rates have been determined in labora-tory studies that require extrapolation to naturalenvironments and are often prone to large uncer-tainties (in some cases, an order of magnitude ormore). The multitude of potential biotic factorscan be harder to quantify than the main inorganiccontrols, yet biotic effects can be very strong. Forexample, plants can alter the pH of soil by exudingH+ from their roots to enhance dissolution of nutri-ent-bearing minerals and microbes may accelerateoxidation and reduction reactions. Thus, the com-plex array of variables influencing reaction ratescan complicate kinetic analysis.

Often geochemical processes are controlledby a rate-limiting step, a step in a processthat is much slower than others Consider thedissolution of a mineral grain, a process thatcould involve five steps: (1) diffusion ofreactant(s) toward the mineral surface (where acommon example of a reactant is H+); (2) sorption

Distance from Equilibrium, Activation Energy, Metastability 33

of reactants onto the mineral surface; (3) forma-tion of a bond between the reactant and the partof the mineral grain under attack (perhaps theO atom bonded to a Ca ion in a feldspar); (4)desorption of the newly formed complex betweenreactant and mineral component (e.g. OH–), andfinally (5) transport of the newly formed productaway from the mineral surface by diffusion. Ifany of one of the steps 1 through 5 is slowerthan others, it will be the rate limiting step. Inclay-rich sediments, diffusion can be a rate-limiting step, whereas in highly insoluble miner-als, step 3 (and/or 4) can be the rate-limitingstep. Kinetic limitations on crystal growth aresimilar, and can be envisioned by reversing steps1 through five.

1.11.1 Reaction rate, reaction order

Reaction rates are in part controlled by the orderof the reaction, where reaction order is defined asthe dependence of reaction rate on concentrationsor molar ratios of species involved in reactions.Reactions can be of zeroth, first, second, or thirdorder (0th, 1st and 2nd are by far most commonin environmental geochemistry). Rates of 0th-order reactions occur at rates independent of theconcentration of the reactant(s); rates of 1st-orderreactions are controlled by the concentration of areactant or product; rates of 2nd-order reactionsare controlled by concentrations of two reactantsor products, or a reactant (or product) squared.A 0th-order reaction, one where reactant

A undergoes transformation to product P, canbe represented by the simple chemical equationA ! P. The following equation describes changein concentration of A (represented as [A]) withtime (dt):

d A½ �=dt= – k ð1:67Þ

where k is a constant, probably determinedby laboratory experiments or by studies of

natural systems, and the negative sign indicatesdecreasing abundance of A with time. In inte-grated form, this equation is:

A½ �= A½ �o – kt ð1:68Þ

[A]o represents the concentration of A at time zero(where examples of to could be when sedimentswere deposited in a floodplain or when a gaseouscompound was emitted into the atmosphere),and the negative sign for kt correlates to decreasingA with time. So the concentration of A at any timeis merely determined by the amount of A at to [Ao],how much time has passed (t), and the rateconstant (k) that governs reaction of A into P.But the amount of A does not control the rate,and this distinction is important (when comparedto 1st-order reactions, e.g.). Furthermore, the con-centration of P does not affect reaction rate. Plottedgraphically (Fig. 1.16), it becomes clear that, for0th-order processes, [A] decreases in a linear fash-ion with time, and rate does not vary with time.Examples of 0th-order reactions include the dis-

solution of many types of minerals in aqueoussolutions, notably salts.A 1st-order reaction is one where the rate

depends on the concentration of a reactant(or product) raised to the first power – that is,

[A]

[A]Time

Rate

Fig. 1.16 Two graphical representations of zeroth(0th)-order chemical reactions, where [A] representsconcentration of a chemical species (e.g. ion,mineral or compound). In the example on the left,decrease in concentration of A is linear, and onthe right, the rate of change is constant.


1st-order reactions are proportional to the con-centration of the reactant. A good example of a1st-order reaction is radioactive decay, an exam-ple of which is provided below. The equation fora 1st-order reaction representing decreasingA with time is:

d A½ �=dt= – kA ð1:69Þ

Literally, the rate of decrease of A depends on therate constant (k) multiplied by the amount ofA present at time t. In its integrated form, thisequation is:

A½ �= A½ �o∗e – kt ð1:70Þ

Graphically, 1st-order reactions depict non-linearchanges in concentration with time as well asrates that vary as a function of the concentrationof reactants (Fig. 1.17) – note that the rate(expressed as, e.g., mol/year) varies but the rateconstant (expressed as a percent or proportion,does not).In the case of radioactive decay, the amount of

daughter isotope (e.g. 222Rn produced by decay of226Ra) produced per unit time decreases as theamount of parent isotope (e.g. 226Ra ) decreases– as reactant is used up during the reaction, therate of formation of products decreases (which isnot to say that the amount of product decreases– it continues to increase, but at a progressivelyslower rate). In Fig. 1.17, [A] decreases exponen-tially (and the product increases parabolically, fol-lowing the dashed line) while the rate of change iscontrolled by the amount of A (reactant) – lessA equals lower rate. In the graph on the left,the solid line would represent amount of parentisotope (and the dashed line amount of daughterisotope) in a radioactive decay reaction.Rates of 2nd-order reactions are proportional

to the concentration of a reactant squared(dependent on Awhere 2A! B), or in some casesto the product of the molar concentrations of tworeactants (where A + B ! C).

d A½ �=dt= – kA2 or d A½ �=dt= – kAB ð1:71Þ

and integrated, the first equation can expressed as:

1= A½ �= k∗ t+C ð1:72Þ

And given that [A] = [Ao] when t= 0,

1= A½ � –1= Ao½ �= k∗ t ð1:73Þ

When depicted graphically (Fig. 1.18), 2nd-orderreactions exhibit rapid initial changes in concen-trations and progressively slower rates with time,and rate increases exponentially with higher con-centrations of reactant(s).

An example of a 2nd order reaction is that ofnitrogen dioxide to nitrogen monoxide plusoxygen, one that plays an important role inatmospheric chemistry:

2NO2 =2NO+O2 ð1:74Þ

We can summarize reaction rates with thisequation:

Time [A]

[A]

Rate

Fig. 1.17 Two graphical representations of first(1st)-order chemical reactions, where [A] representsconcentration of a chemical species. On the left,change in concentration of A is logarithmic (solidline) or exponential (dashed line); on the right, forboth cases (increasing or decreasing concentration),rate of change decreases in a linear fashion withtime. Arrows in diagrams indicate direction offorward reaction (emphasizing that, on the right,the solid line represents decreasing [A] with time).

Distance from Equilibrium, Activation Energy, Metastability 35

2A+B=A2B ð1:75Þ

If the reaction rate depends on the concentrationof none of the components, then it is a 0th-orderreaction. If rate depends on the concentration ofB, it is 1st order, and if it depends on the concen-tration of A, it is 2nd order. If it depends on theconcentration of A, B and the product A2B, it is3rd order (and 1st order with respect to B andA2B, and 2nd order with respect to A), but thisis a rare occurrence in geochemical systems.

1.11.2 Temperature and the Arrheniusequation

Temperature also plays an important role inreaction rate – higher temperatures foster higherreaction rates. Generally, reaction rates doublefor every increase of 10 �C. The greater energyimparted by higher temperature tends toenhance interactions between reactants, thusenhancing the probability of productive interac-tions. Higher temperature can also overcomeactivation energies that can sometimes serveas barriers, like the one depicted schematicallyin Fig. 1.14. The Arrhenius equation relates tem-perature to the rate constant:

k=A∗e –Ea=RT ð1:76Þ

where Ea is activation energy, R is the gasconstant, T is temperature (K), and A is thetemperature-independent term known as thepre-exponential factor or the A factor – it servesto convert the product term into values appropri-ate for k of different reaction orders. In a qualita-tive sense, and because e is raised to the negativeEa/RT, increasing temperature will increase k,speeding up the reaction. Graphs of the effect oftemperature on reaction rate are typically plottedas inverse of T (1/T, where T is in Kelvin) versuslog of the rate constant k, a type of graph knownas an Arrhenius plot. Figure 1.19 depicts thedissolution rate of powdered rhyolite (composi-tionally similar to granite; Table 1.1) as a functionof temperature:The rates shown in Fig. 1.19 increase more

than ten-fold with a temperature increase ofapproximately 35 �C (note log scale of y-axis).The preceding introduction to equilibrium

thermodynamics and kinetic constraints ismeant to present some of the concepts,approaches and limitations contained withinthese two important fields. In the following chap-ters some of these approaches will be applied topredicting and understanding behaviors of envi-ronmental systems. For more detailed treatmentsof this topic, the reader is referred to excellent

Time

[A]

[A]R

ate

Fig. 1.18 Two graphical representations of second(2nd)-order chemical reactions, where [A] representsconcentration of a chemical species. In the exampleon the left, [A] decreases exponentially, and on theright, we see that rate of change decreasesexponentially with time.

–14.5

–15.0

–15.5

–16.0

–16.5

–17.0

50 °C

30 °C

15 °C

1000 / T (K)

Log

rate

(m

ol c

m–2

s–1

)

3.0 3.2 3.4 3.6

Fig. 1.19 Effect of temperature on chemicalweathering rate of rhyolite as a function oftemperature. (Yokoyama and Banfield 2002.Reproduced with permission of Elsevier.)


and much more thorough treatments of thesetopics in geochemistry texts by James Drever(The Geochemistry of Natural Waters), GunterFaure (Principles of Geochemistry) and DonaldLangmuir (Aqueous Environmental Geochemistry.The Kinetics of Geochemical Processes issue ofReviews in Mineralogy edited by Lasaga andKirkpatrick (1981) also presents various pers-pectives on this topic.

REVIEW QUESTIONS

1. Magnetite, hematite and goethite all occur intropical soils. Rank them in order of thermody-namic stability in oxidized soil at 25 �C.Whichis more likely to be released into soil water, atrace element (e.g. arsenic) substituted intothe structure of hematite or goethite? Explain.

2. What is (are) the oxidation state(s) ofmanganese in each of the following minerals?MnOOH MnO2 Mn3O4 MnO MnCO3

3. Recalculate the chemical compositions of theigneous rocks given below in terms of theweight percent concentrations of the elementsand rank them in terms of their abundance.Also calculate concentrations of each elementin terms of mg/kg and mol/kg.

#1 #2

SiO2 45.0 68.0Al2O3 3.5 15.0FeO 8.0 3.2Fe2O3 0.0 1.3MgO 39.0 1.7CaO 3.2 3.4Na2O 0.32 3.1K2O 0.04 3.6Cr2O3 0.42 1.1 × 10–3

NiO 0.25 9.2 × 10–4

4. Why are elements with even atomic numbersmore abundant than their neighbors with oddatomic numbers? Explain with appropriatenuclear reactions.

5. Why is Fe more abundant than its neighborswith similar atomic number? Describe withappropriate nuclear reactions.

6. Why is Pb more abundant than its neighborswith similar atomic number? Describe withappropriate nuclear reactions.

7. List the units of concentration that are com-monly used in the following cases:A. Acids in water.B. Salts in water.C. Metals in soil.D. Metals in water.E. SiO2 in rock.

8. A. Write the chemical reaction for thedissolution of barite.

B. Calculate the change in enthalpy associ-ated with the dissolution of BaSO4

(barite).C. Based on your result, predict how the sol-

ubility of barite varies with temperature.D. Calculate the Gibbs free energy for the

dissolution of barite into ions.E. Calculate the solubility of barite at 25 �C

and 1 atm.F. Given that Ra co-precipitates with Ba,

predict the solubility of Ra in ground-waters (a) ~ lacking Ba (e.g. 1 ppb) and(b) rich in Ba (e.g. 100 ppb).

9. Which compound is more soluble in water,MgO or CaO? Explain.

10. Examine the kinetics of dissolution of a saltlike NaCl by experimentation using anapproach like that described by Vel-bel (2004).

REFERENCES

Drever, J.I., 1997. The Geochemistry of Natural Waters:Surface and Groundwater Environments, 3rd edn.Simon and Schuster, Upper Saddle River, New Jersey,USA.

Faure, G., 1986. Principles of Isotope Geology. JohnWiley & Sons, Canada.

References 37

Faure, G., 1998. Principles and Applications ofGeochemistry (2nd edn). Prentice-Hall, Upper SaddleRiver, New Jersey, USA.

Gromet, L.P., Dymek, R.F., Haskin, L.A., and Korotev,R.L., 1984. The “North American shale composite”:Its compilation, major and trace element character-istics. Geochimica et Comochimica Acta 48:2469–2482.

Langmuir, D.L., 1997. Aqueous EnvironmentalGeochemistry. Prentice Hall, Upper Saddle River,New Jersey, USA.

Lasaga, A.C., Kirkpatrick, R.J. (eds.), 1981. Kinetics ofGeochemical Processes: Reviews in Mineralogy(Vol. 8). Mineralogical Society of America, Washing-ton, DC, USA, 398 pp.

Lu, H., Dai, D., Yang, P., Li, L., 2006. Atomic orbitals inmolecules: general electronegativity and improve-ment of Mulliken population analysis. PhysicalChemistry Chemical Physics, 8: 340-346. DOI:10.1039/B511516G

Railsback, L.B., 2003. An Earth scientist’s periodictable of the elements and their ions. Geology 31:737–740. ∗∗ Versions in English and other lan-guages are available at: http://www.gly.uga.edu/railsback/PT.html

Rudnick, R.L. Gao, S., 2003. The Composition of theContinental Crust, in: The Crust, Rudnick, R.L.(Ed.), in: Holland, H.D., Turekian, K.K. (eds.), Treatiseon Geochemistry, Vol. 3. Elsevier-Pergamon, Oxford,pp. 1–64.

Taylor, S.R., McLennan, S.M. 1985. The ContinentalCrust: Its Composition and Evolution. BlackwellScientific Publishers, Oxford, England, 312 pp.

Turekian, K.K., Wedepohl, K.H., 1961. Distributionof the elements in some major units of the Earth’scrust. Geological SocIety of America Bulletin 72:175–192.

Velbel, M.A. 2004. Laboratory and homework exercisesin thegeochemical kinetics ofmineral-water reaction:rate law, Arrhenius activation energy, and the rate-determining step in the dissolution of halite. Journalof Geoscience Education 52: 52–59.

Vinogradov, A.P., 1962. Average contents of chemicalelements in the principal types of igneous rocks of theEarth's crust. Geochemistry 7: 641–664.

Yokoyama, T., Banfield, J.F., 2002. Direct determina-tions of the rates of rhyolite dissolution and clayformation over 52,000 years and comparison withlaboratory measurements. Geochim. Cosmochim.Acta 66: 2665–2681.


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Background and Basic Chemical Principles: Elements, Ions ...€¦ · in environmental geochemistry require under-standing of the relationships among aqueous solutions, geological