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Selected Topics

Geochemistry

Abbreviated Course Notes

Main tasks of geochemistry:


Part 1: Cosmochemistry

Elemental abundance in Universe determined from

  1. analyzing spectra of outer portions of stars and from nebulae
  2. then averaging many stars of different kinds

Estimates constantly change.

Specifics almost certainly not known (exact %, for example) but some general facts (or "rules") known.

We can explain relative abundances of elements by seeing how elements form in stars.

A typical star might go through several stages - for example: nebula, proto-star, main sequence star, red giant, supernova/neutron star, black hole

Main sequence stage of star's life: Fusion of hydrogen into helium
Several possible routes -

in one route, C acts as a catalyst; intermediate steps result in the production of N and O, explaining their abundance
Some elements (such as Li, Be, Bo, Sc) used up as part of H fusion process

Red giant stage: In core, He converted to C and O, explaining their abundance
C + O yields Ne, Na, P, Mg, Si, S, Cl and Ar

Mg, Si, S then form Fe, Ni

Supernova stage: Implosion to form neutron star forms some exotic heavy elements.
In fact, can correlate half lives of some short lived heavy elements (like Cf254) with half life of luminosity of supernova.
Explosion results in dispersion of elements into space.

Neutron star: Addition of neutrons to Fe forms heavier elements which are less stable than Fe.
For these heavier elements, relative abundance depends on relative stability.

Elements with even numbers of protons and neutrons are more stable because they have greater nuclear binding energy.

For second, third, etc. generation stars, different processes occur because they have different starting materials as well as being at different stages in their lives.

Composition of our Sun results from its present state of evolution
Most striking feature is extreme abundance of H and He.
About 70 elements found in Sun so far.


Part 1a: Solar System Composition

Brief look at composition of other bodies in solar system in order to compare with Earth:

Meteorites - chunks of rock from space that land on Earth

Common minerals in meteorites:

Kinds of meteorites:

Iron meteorites:
Predominantly Ni-Fe alloys
Minor amounts of other minerals such as troilite (FeS)
Types classified according to % Ni:

Stone meteorites:
Chiefly silicates, mostly ferromagnesian
Up to 1/4 metallic Ni-Fe
Types:

Chondrites: Contain chondrules (BB-sized round bodies)
Composed chiefly of silicates such as olivine, pyroxene, and plagioclase or glass
Important type is carbonaceous chondrite.

Achondrites: no chondrules

Stony-iron meteorites: Equal amounts of silicates and Ni-Fe alloys
Many are crystallized silicates which have been brecciated, then invaded by metallic and sulfide melts
Classified according to kind of silicate:

Some questions on meteorite origins:

Moon:

Other planets, etc.:


Part 2: Composition and Early History of Earth

Most information about Earth's interior comes from geophysical studies:

Outer core

Mantle

Crust:

4 approaches to obtain overall composition of crust:

  1. averaging of available rock analyses
  2. as #1, but weighted in proportion to rocks' occurrences
  3. abundance based on crustal models
  4. indirect assessment - estimating mass balance of mafic and felsic rocks to give, through weathering, etc., the observed average composition of sediments (or selected sediment)

Approach #2 used by Clarke and Washington, 1889
Calculations based on more than 5000 "superior" rock analyses
Problem of how to weight different kinds of rocks to strike a reasonable average

Ronov-Yaroshevsky, 1969, used approach #3.
Based on many thousands of specimens from continents, ocean floors, continental shelves and slopes

Goldschmidt used approach 4
Based on glacial clays in Norway
Gives lower values for Na, Ca (easily removed in weathering)

Material Earth formed from most easily studied by considering the Fe-Mg-Si-O-S equilibrium system, when O>>S and O+S<Fe+Mg+Si (you will get to this kind of equilibrium system later)
Results in 3 essentially immiscible
phases (iron present in all):

Arrange these in order of density and this looks like Earth.
Other elements concentrated in these phases according to their chemical properties (not density)

Chemical properties of elements determined primarily by their electronic configurations; that's why elements in the same group on the periodic table have very similar properties.
Examples:

Goldschmidt first recognized primary geochemical differentiation of the elements in 1923.
Coined terms siderophile, chalcophile, lithophile, atmophile


Part 3: Currently accepted theory for origin and early history of Solar System and Earth:

Step 1: Rotating cloud of gas and dust (nebula) over volume larger than orbit of Pluto

Questions:

Step 2: Condensation of nebula

Questions:

Step 3: Material collects in central portion of nebula to form Sun.
In remaining material, small groupings of particles form randomly
Because of combined gravitational pull of the group, they attract other particles.
Many bodies grow until they are several kilometers in diameter
Larger bodies sometimes join together by collision and sometimes become fragmented.
Bodies which continue to grow become planets and satellites; those that fragment become interstellar dust, asteroids, comets, etc.

Question: Did Earth form from cold fragments or hot fragments?

Question: Was early Earth homogeneous or heterogeneous to start with?
(Reference: Grossman, 1972, Condensation in the primitive Solar nebula,
Geochim. Cosmochim. Acta, 36, 597-619)

Step 4: planets heat up

Questions:

Two sources of heat:

  1. Gravitational potential energy from collapse
  2. Radioactive decay (most important)

How fast and how far temperature rose is debatable
Depends on:

Also not known whether present temperature rising, falling, or remaining constant
Depends on:

None of which known accurately

Heating up could produce:

Step 5. Differentiation to form core and mantle? and crust formation.
"iron catastrophe"

Questions:

Overall chemistry of crust -

This is reminiscent of layering in differentiated igneous rock bodies and of differentiation observed in labs.
Suggests similar process for formation of crust
Further evidence from Moon - anorthosite layer, flooded by basalts derived from underlying ultramafic layer

When Earth heated, heavier Mg, Fe silicate would move to or remain in mantle; scum of material rich in Al, Si, K, Ca, etc. would rise to surface
Process might take place piecemeal over Earth, because of slightly different temperature, so entire outer part of Earth not necessarily molten at once

None of original crust preserved
Time between accretion and formation of oldest existing rocks would have been tumultuous.
Moon's surface, as well as Mercury, Venus and Mars, indicates intensive meteorite bombardment producing cratered surface; almost certainly Earth's was too
Some of the chemical heterogeneity of Earth's crust and upper mantle may be heritage from the variety of meteor masses that were incorporated into surface material.
Might explain some odd concentrations of tin, iron, and copper ore.
Could it have produced present (and past?) lopsided distributions of continents and ocean basins?

Question: How did continents originate and grow?

Questions:

It is currently commonly thought that continents:

Mafic lavas came to surface from deep in upper mantle, formed by partial melting of garnet peridotite
Solidifying, lavas became
oceanic crust, carried away from spreading centers
At other places, oceanic crust and sediments which had accumulated on it were carried back down and became hot enough to generate felsic and andesitic magmas which produced volcanoes and batholiths on the edges of granitic islands, thus enlarging continents
Process much smaller in scale than at present
Slow and inefficient process but result was silica, alumina, and alkali metals enriched in continents and olivine and pyroxene enriched in upper mantle

Question: Is differentiation to produce continents continuing or are continents simply rearranging themselves?
Other ways of asking same question:

Evidence for one side:

Evidence for other side:

Questions:
What kind of change in composition might occur in one way process?

Step 6. Origin of atmosphere and ocean
Cold primitive Earth had virtually no atmosphere because gases tied up in compounds or in frozen particles.
As Earth heated up (due to gravitational collapse and radioactivity) gases would be released and accumulate at surface
(This would presumably include water vapor which would condense to form oceans, to be looked at later.)
Gravity would be sufficient to retain all gases except H and He

Example of evidence for degassing: Argon comes from decay of Potassium. Not enough K in surface rocks to supply all Ar in atmosphere. Some Ar must have come from interior.

Questions:

Clues:
Evidence from elemental abundance studies suggest that isotopes of atomic number greater than about 40 are present in about equal amounts.
So when elements in solar system formed, Ar
36 = Ar38 = Ar40
But Ar
40 also forms from radioactive decay of K so ratios Ar40/Ar36 or Ar40/Ar38 constantly increasing
In Earth's interior, ratio is about 5000/1 at present (found in diamonds which are heated to carbonization)

Look at 2 possible extremes:

Ratio in atmosphere at present is about 300/1 (that is, small); indicates most of atmosphere formed early
Estimates that about 80 % formed in first 500 my; remaining 20 % from volcanic eruptions since then.

Average product of current degassing (from analysis of gases from volcanoes):

Bears little resemblance to composition of present atmosphere. Why?

We have implied that early atmosphere may have been different than present one.
Look at CO
2 and oxygen for evidence

Question: Was there less CO2 in past?
Continued evidence of life for more than 3 billion years establishes a fairly definite minimum for the amount of CO
2 in the atmosphere for at least 3 billion years

Question: Was there more CO2 in past?
Extremely high amounts would produce highly acidic surface waters and greatly speed chemical weathering
(will look at results now, chemical reasons why later)

Question: Is there evidence of silicates broken down into free silica?
Banded iron formation - high amounts of silica (chert) in alternating layers with various iron minerals such as magnetite and hematite; restricted to period 2.5 to 1.8 by in age

Question: Evidence of high ion concentration in oceans?
Significant evaporites older than latest preCambrian not found

Question: Evidence of high acid waters preventing carbonate deposition?
Carbonate rocks uncommon in most of preCambrian; become abundant near close of era (especially dolomite) and are continuing to form in large amounts today

Question: Was there less oxygen in past?

Conclusion:
Early preCambrian atmosphere had more carbon dioxide, less oxygen (about 0.1 % of present value)

Great increase in abundance and complexity of organisms in late preCambrian changed carbon dioxide rich atmosphere to oxygen rich atmosphere

Life probably began in oceans.
Water would shield from ultraviolet radiation (ozone does now but little oxygen in primitive atmosphere)
Organisms produce oxygen; some is used to form oxides, some released to atmosphere
PreCambrian banded-iron formations probably formed when iron weathering from rocks in an almost oxygen free environment was washed into plant-containing oxygenated water, oxidized, and precipitated
Later oxygen in atmosphere provided shield, life forms expanded dramatically and moved from water environment.
Little change in atmospheric composition since then.

Origin of oceans:
Water condensed from atmosphere
pH dependent on composition of atmosphere because of dissolved gases included

present pH 8.1- 8.4 in surface waters, slightly lower at depth

Early ocean more acidic (because more carbon dioxide in atmosphere, and thus more dissolved)
Ocean should have obtained dissolved material fairly rapidly in history (due to extensive early weathering)

Question:
Any large changes since?

Rock record shows no evidence of change since late preCambrian

For a few elements (notably chlorine, bromine, sulfur, and boron), concentrations in seawater much higher than in surface rocks
Common
products of volcanoes and probably added directly to seawater by volcanic activity rather than by erosion
Elements involved with
biological activity (C, O, N, P) vary horizontally and vertically because of concentration of organisms in surface layers in warmed waters and because of decaying organisms returning material to seawater


Part 4: Radiometric Dating

Geochronology - concerned with determining age and history of geologic materials by studying their isotopes

Radioactivity
Discovered in 1896
Natural change from one element to another by emission of particles from nucleus or addition of particles to nucleus

Particles include:

Decay occurs at constant rate and is not affected by temperature, pressure, chemical combination or any other known thing

Radioactive isotopes - an element capable of spontaneously changing into another element by the emission or addition of particles to its nucleus
Stable isotopes - an isotope which is not radioactive

Radiogenic isotopes - an isotope produced by radioactive decay
Non-radiogenic isotopes - an isotope not produced by radioactive decay

Half-life - time for half of element to decay

Parent - the radioactive element which decays
Daughter - an element formed from another by radioactive decay

T (half life) = ln 2/ = 0.6931/

The equation which represents radioactive decay is (derived in most geophysics texts for those who are interested and know a little calculus):

solved for t (age of rock):

Assumptions made in radiometric dating:


Part 4a: RbSr dating

Rb87 -> Sr87 (could also write 87Rb, etc.)
Rb commonly substitutes for K in minerals; so method used on K-bearing minerals or rocks which contain them

Decay equation reads:

(Subscript m stands for measured, or in other words, now; o stands for original)

It is easier to measure ratios of atoms rather than absolute numbers so expression usually written:

Could solve for t (age of mineral):

Now measure Sr87/Sr86 and Rb87/Sr86 ratios and for reaction ( = 1.39 x 10-11/yr)
Then
estimate (Sr87/Sr86)o (Can measure this ratio in coexisting undisturbed minerals which contain no Rb)

Note: Sr86m = Sr86o since Sr86 is stable and non-radiogenic
Sr
86, Sr84, and Sr88 are all stable and non-radiogenic.
Any could be used; Sr
86 most abundant and therefore most often used.

Easier mathematics and more accurate way of determining (Sr87/Sr86)original:
Equation for straight line is y = ax + b, (where a is slope, b is intercept on y axis)
Equation is in that form (actually y = b + ax) when t is constant (for several minerals in a rock or several rocks of the same age)

If we plot (Sr87/Sr86)m vs (Rb87/Sr86)m, the values should be different for different rocks and minerals because they would have different initial amounts of Rb.
The slope of the line obtained by connecting these points is -1 and the intercept is (Sr
87/Sr86)o
Thus we can
obtain both the age of the suite and the initial strontium ratio.
These plotted lines are called
isochrons.

Isochrons can also be used to determine age of metamorphism.
If whole rock hasn't lost Rb or Sr, but minerals have passed them around during metamorphism, two ages will be obtained - one from dating whole rock and one (metamorphism age) from dating individual minerals in rock.

Another Sr isotope use:
First must know Sr
87/Sr86 in material that made up primitive Earth.
Usually assume it was same as non-Rb
87 bearing meteorites or about 0.699

During differentiation of crust, behavior of Rb and Sr would be different (different charge, different size).
Rb concentrated in crust, Sr evenly distributed between crust and mantle.
Production of Sr
87 should thus be faster in crust than in mantle and Sr87/Sr86 ratios should be higher for crustal material.
Difference in Sr87/Sr86 ratios, then, is a means for distinguishing igneous rocks that have formed by partial melting of crustal rocks from those that have their origin in differentiation or partial melting of mantle material

Present Sr87/Sr86 ratio for mantle rock estimated from analyses of recent basalts and gabbros from oceanic environments (direct origin from mantle assumed and no contamination by continental material)
Value is about 0.704
Extrapolation between 0.699 and 0.704 gives reasonable estimate for ratio in mantle at any time in past.
Look at Sr87/Sr86 ratios for rocks when they formed to determine origin.
(ratio above or consistent with expected mantle ratio?)
(Remember can get Sr
87/Sr86 ratios from isochrons.)


Part 4b: Uranium, Thorium - Lead dating:

U238 -> Pb206

U235 -> Pb207

Th232 -> Pb208

Commonly use ratio with stable Pb204

One equation might be written:

or:

Must determine ratio (Pb206/Pb204)o and .
Can find
original ratio from associated lead minerals (such as galena) or can use mineral for study that wouldn't have had any original lead (zircon, uraninite, sphene, apatite, monazite, etc.)

By using U238, U235, and Th232, theoretically you get three age determinations and they should agree (concordant ages).
If disagreement, ages are said to be
discordant.
This is probably
due to gain or loss of material.

Lead-lead method

If equation for U235 is divided by equation for U238, we get another equation:

Use of this equation called lead-lead method.
Handy because U
235/U238 ratio known, as are decay constants.
Can't solve remaining equation directly for t but ages corresponding to different isotope ratios have been plotted and can be obtained from published graphs or tables

Use of Pb-Pb method is good check on U235, U238, and Th methods because if lead lost, the ratio of isotopes of remaining lead should not be changed and valid age should still be given.

Can also directly use ratio:

These two quantities increase with time at different rates and if plotted against each other, a curved line is formed (called a concordia curve because all points on the curve have concordant U238/Pb206 and U235/Pb207 ages).

If a rock sample has lost no Pb, calculated ages from U238 and U235 would be concordant and a point representing the ratio of the above quantities would lie on the concordia curve.

If Pb has been lost, the ages will be discordant and the point representing the ratio will lie below the curve.

Since lead loss would presumably be different for different areas in the sample, several different analyses from different locations in the sample should give several different ratios and thus several different points below the concordia curve.
It can be determined mathematically that these several points will lie on a straight line (called a
discordia).

If the discordia line is extended to intersect the concordia curve, upper intersection gives age of rock.
Lower intersection supposedly gives time lead lost but almost never accurate since lead almost never lost all at once but gradually over long time.

Technically could use U238 -> He4, U235 -> He4, or Th232 -> He4
But, helium may be lost since a gas.

Assume that any He present when rock was molten escaped
Therefore, any He present now formed from U or Th after solidification.
He ages thus give solidification ages
(Example: how long it takes for granite batholith to solidify).


Part 4c: Other Pb uses

1. Can measure average amounts of U238 and Pb206, or U235 and Pb207 in rocks at the Earth's surface (usually use recent marine sediments).
Assume no radiogenic lead to start with, can calculate
age of Earth's outer portion.

2. Begin with primeval lead (lead present when Earth formed): Pb204, Pb206, Pb207, Pb208 in certain ratios for Earth as whole (usually assume this to be same as ratios in meteorites without U, Th).
With time, radiogenic lead increases, thus higher Pb
206/Pb204, etc., ratios with time.
Can get
age of Earth (4550-4750 my).

3. (variation on 2)
After a time, ore might form (example: galena).
This ore would "sample" the lead at time of formation, which would consist of the primeval lead plus all radioactive lead formed before the time of ore formation (total lead called the common lead).
Thus,
age of ore can be determined by comparing its lead ratios to the ratios which would have existed at various times.

4. Stable nuclei atomic weight about 40 and above are present in about same abundance.
Assume when elements formed, same rule applied to unstable elements.
Now U
238 is 140 times as abundant as U235.
If both once equally abundant, would take 6 billion years to reach present proportion.
Age of Universe? of our part of Universe? of our Solar System nebula?


Part 4d: Fission-track dating:

U238 spontaneously breaks down by fission (splits into two large parts).
This is a rare occurrence.
These fission particles pass through the surrounding material with very high energy and leave tube-shaped
damage tracks.

These tracks can be counted (etch mineral with HFl, look at under microscope) and thus the number of spontaneous fissions may be counted.
This
gives amount of daughter product in sample.

Can determine (generally from measurement of amount of radiation being emitted) current U238 content in sample.
Essentially
have number of daughter atoms and number of remaining parent atoms and can thus determine age.

Useful because can be used on wide variety of substances of wide range of ages.

Disadvantage which turns out to be an advantage:
Fission tracks are "healed" by prolonged heating (millions of years).
Temperature at which healing occurs is different for each mineral.
Each different mineral thus can yield a different age (apparent disadvantage) because each mineral has its clock "restarted" by healing at different temperatures and thus different times.
But
temperature history of sample can be determined by comparing different minerals in sample.


Part 4e: Potassium-Argon dating:

K40 undergoes 2 principal kinds of decay, to Ca40 and to Ar40.
Decay to Ca
40 not useful, because Ca40 most common isotope of Ca and small amount produced radiogenically would be undetectable.
Therefore, use K-Ar.

Since 2 separate decay types are possible, decay equation somewhat more complicated.

Let be total decay constant, Ar be decay constant for K-Ar reaction, and Ca be decay constant for K-Ca reaction.
Then decay equation can be written:

Ar40original = 0 for all but very exotic minerals (original Ar a gas, wouldn't survive formation except under very unusual circumstances, such as enormously high pressures).
Therefore, substituting 0 for original Ar and also substituting decay constants:

     t = 1.88 x 109 ln (1+ 9.07 Ar40/K40)

If metamorphism occurs, Ar40 already formed will probably be lost and clock reset.
K-Ar methods can therefore be
used to date metamorphic events.

Disadvantage to method:

Advantages to method:


Part 4f: Samarium-Neodymium dating:

Techniques same as for Rb-Sr or K-Ar.

Has advantage that both elements are members of rare-earth group and have virtually identical chemical properties.
Both similarly affected by weathering and metamorphic processes.
Sm/Nd ratios would remain unchanged, giving reliable date for original crystallization.


Part 4g: Carbon dating:

Carbon 14 dating (also called radiocarbon dating)

C14 formed in upper atmosphere by reaction of N2 with neutrons produced by cosmic rays.
Reaction is:
0N1 + 7N14 -> 6C14 + 1H1
then C
14 decays -> 7N14 + -10
Thus,
total amount of C14 in atmosphere constant.

Carbon in organism has same C14/C12 ratio as air or water does as long as organism alive.
When organism dies, C
14 not replenished, disappears, and C14/C12 ratio decreases to zero.
C14/C12 ratio thus gives age since death.

Limited to very young samples (less than 70,000 years) because of short half-life (5730 years).

Instead of measuring C14/C12 ratio in material directly, normally we compare C14 in sample to C14 in air by comparing radioactivity of the 2 samples (number of decays per minute per gram of carbon).
A is activity of C
14 in material to be dated and Ao is activity of air.
(Age of sample) t = 19,035 log Ao/A.

Is % C14 really constant?
Known that C14 content of atmosphere increased 10 % in period 6000 to 2000 years ago.
Found by studying tree rings.
Cause not known.
Now changing because of:


Part 5: Stable Isotopes

Commonly used stable isotopes in geology:

Atoms of 2 different isotopes of the same element in the same molecule or crystal structure will vibrate at different frequencies.
The
difference in frequency depends on the relative difference in mass between the isotopes.
It would be a significant difference for isotopes of hydrogen (where H
2 is twice the mass of H1) and insignificant for isotopes of elements such as tungsten (W186 is 1.03 times mass of W180).

The lighter isotope vibrates at a higher frequency than the heavier one and is therefore less strongly bonded.

It can, therefore, be more easily removed by physical, chemical, or biological processes.
Degree to which separation occurs is called
fractionation.
Theoretical calculations for fractionation are very difficult or impossible.
Therefore, data is usually merely observed experimentally and then one uses tables, graphs, etc. to study natural occurrences.

The extent of isotopic separation (or fractionation) can be expressed by a ratio called the fractionation factor, .
= (Ra/Rb) - 1
where R
a is ratio of isotopes in phase a, Rb is ratio in phase b.

Example: for oxygen at 25 degrees C,

No fractionation number is very far from zero.

Differences between fractionation numbers are very small, so we usually use comparisons between isotopic separation in a particular sample and the corresponding isotopic separation in some standard.
Standards used:

These comparisons are given in terms of delta values, , in permil units (parts per thousand).
A positive means the sample contains a higher % of the heavier isotope than a standard and a negative means the sample contains a lower % of the heavier isotope than the standard.

Examples:

1. physical process
Evaporation of water leads to concentration of lighter isotopes of H and O in the vapor phase and heavier isotopes of H and O in the fluid phase.
O in fresh water has values down to -60 permil because light isotope concentrated in vapor escaping from the sea (and then precipitating to form fresh water).

2. chemical process
If one mixes water made of only O
16 with CO2 made of only O18, the isotopes will exchange to yield water and CO2 with both kinds of isotopes but the water will wind up with a slightly different ratio of O16/O18 than the CO2, because of different bond strengths in the two compounds.

3. biological process
Biological processes are complex and not well understood.

Temperature effects:
At high temperatures, the difference between vibrational frequencies of different isotopes is much smaller than at low temperatures.
Thus less separation of isotopes occurs at high temperatures.

This fact can sometimes be used to determine the temperature that prevailed when a certain process was occurring.
Example:
Temperature at time of formation of CaCO
3 precipitated in ocean can be determined by measuring oxygen isotope composition of the carbonate.
Ideally, temperature differences of less than 1 degree C can be determined.

Shows decrease in average temperature of seawater of 5 to 10 degrees C in past 150 my.

Use of stable isotopes to determine source of fluids: (Big ore deposit debate).
Comparison of H
2/H1 ratios and O18/O16 ratios for hydrothermal fluids to same ratios for rainwater in that area and to magmatic water.
Ratios of hydrothermal fluid almost same as rainwater but not quite.
Indicates most fluid from rainwater in vicinity but some possibly from magmatic source.


Part 5a: Meteorite Impacts and Extinctions

Geochemical evidence for impact:
Rocks deposited in the Cretaceous/Tertiary boundary clay layer contain substantial amounts of elements such as iridium which are common in extraterrestrial rocks but rare in Earth rocks.

Geochemical evidence against impact:
These anomalies could also be produced by:


Part 6: Low Temperature Geochemistry or Water Geochemistry

Will now look at some chemical reactions of rocks at the surface of the Earth:

Important reminder: "Chemists" usually deal with very simple solutions where they can control the conditions.
Geologists must deal with several
complications:

Mechanical weathering:

Chemical weathering:

At this point, you should review your introductory geology coursework on weathering of rocks and minerals.


Part 6a: Chemical Equilibrium

For the reversible reaction aA + bB + ... <-> yY + zZ + ...

the equilibrium constant K is defined as

          [Y]y [Z]z ...
     K = ---------------
          [A]a [B]b ...

where values in brackets are concentrations and the coefficients in the equation determine the powers to which the concentrations are to be raised.

Example:

Values of K are given for 25 degrees C and 1 atmosphere pressure, unless otherwise specified.

Concentrations of gases are expressed as partial pressure in atmospheres
(In a mixture of gases, each gas behaves approximately as though it were alone and exerts a pressure independently of the the others. This is its partial pressure.)

Concentrations of solutes in water solutions are given either as moles of solute per liter of solution (molarity) or as moles of solute per kilogram of water (molality)
In dilute solutions (types usually dealt with in geology), the difference between molarity and molality is negligible.

Concentrations of pure solids and pure liquids and of water are included in the equilibrium constant and hence do not appear in the formula.

Once equilibrium is established, a change in concentration of one constituent will lead to changes in the others
(le Chatelier's Principle)

Example: What would be the effect of more CO2 in the atmosphere? (as in the early history of Earth)
From 2nd equation, increasing [CO
2] results in increased [H2CO3].
From 1st equation, increased [H
2CO3] means more solution of CaCO3.
Thus, explains why decrease in CO
2 in late preCambrian resulted in more carbonate deposition.

Note that in discussion of equilibrium, nothing has been said about rates of reaction.
Has significance in geology.

Some rules:

(Important)
Many natural systems not in equilibrium but may be considered "stable" because reactions are so slow


Part 6b: Solubility Products

Solubility product - equilibrium constant for solution of a substance in water

Example: CaSO4 <-> Ca+2 + SO4-2

K = [Ca+2] [SO4-2] = 3.4 x 10-5

Example of how used:
Determine solubility of CaSO
4
Solubility is maximum amount that can be dissolved.
Solubility (in moles per liter) = concentration of Ca
+2 in saturated solution [Ca+2]; also = [SO4-2]
(Solubility)
2 = [Ca+2] [SO4-2] = K = 3.4 x 10-5
Solubility = 5.8 x 10
-3 M

Solubilities are given for ideal solutions at 25 degrees C.

Effect of raising temperature is usually increased solubility
Some exceptions if solubility reaction gives off heat.

For real solutions:

Uses of solubilities:
From table, determine solubility of ion.
Compare to measured concentration (or activity of non-dilute solution).
This will tell degree of undersaturation and often indicate whether a small change will lead to precipitation.

Example of use:
order of precipitation of evaporites when water evaporates is least undersaturated first to most undersaturated last

Common ion effect -
the decrease in solubility of a salt due to presence of one of its own ions already in the solvent

Example:
(remember, solubility of CaSO
4 in water is 5.8 x 10-3 M)
Calculate solubility of CaSO
4 in 0.1 M CaCl2
Solubility of Ca
+2 = x (from CaSO4) + 0.1M (since 0.1M already in solution)
Obviously, solubility of SO
4-2 is also x.
K (from table) = 3.4 x 10
-5 = (x + 0.1) (x)
Use quadratic formula and x = 3.4 x 10
-4 M
Thus solubility of CaSO
4 less in CaCl2 solution than in water.

Most important solvents at Earth's surface are rainwater and other natural waters.
Rainwater and surface waters are not neutral but actually
weak acids or bases.
(reminder: pH = - log [H
+]
In nature, observed pH's lie mostly in the range of 4 to 9; a typical value for surface waters might be 5.7
Since natural waters are weak acids and bases, we must consider solubility of rocks in weak acids and bases rather than in pure water.


Part 7: Carbonates

An example of a substance dissolving in a weak acid, calcium carbonate dissolving in surface waters, is represented by the following compound reaction (Equation is not balanced.):

      H2O + CO2 <-> H2CO3 <-> H+ + HCO3-
 
                     +
 
                    CaCO3 <--> Ca+2 + 2HCO3-

This reaction describes the following:

We will use this reaction to look at the effect of real life complications on what at first glance might seem to be a relatively simple chemical reaction:

1. effect of pH
Dissolving of CaCO
3 uses up H+ (to form HCO3-)
Addition of H
+ (lowering the pH) should increase the solubility
Addition of OH
- (raising the pH) should favor precipitation

2. forest fires, air pollution, volcanic eruptions (all sources of CO2)
More CO
2 in atmosphere, more H2CO3
More CaCO
3 dissolves.

3a. decay of organic matter in air or in aerated water
Produces CO
2
More CO
2, more CaCO3 dissolves.

3b. decay of organic matter when air is restricted or cut off
Produces CO
2 and/or H2S and/or ammonia (process varies)
CO
2 and H2S would make CaCO3 more soluble
NH
4OH would raise pH and make precipitation of CaCO3 more likely.
Hard to predict what will happen.

4. temperature changes

  1. the solubility of most salts increases with increasing temperature
  2. the opposite is true of CaCO3; its solubility decreases with increasing temperature.
  3. the solubility of CO2, like any other gas, decreases with increasing temperature.

Effect b is greater than effect c.
Thus increasing temperature leads to precipitation.

As an illustration of the effect of temperature, CaCO3 dissolves at great depth in the ocean, where the water is perennially cold, but precipitates near the surface, especially in the tropics, where the water is warm.

5. effect of pressure
Increase of pressure increases solubility by increasing solubility of CO
2 in water.
In deep parts of ocean, pressure alone would increase the solubility of CaCO
3 to about twice its surface value.
In near surface environments, changing atmospheric pressure can theoretically change amount of CO
2 that dissolves and thus effect solubility of CaCO3.

6. organic activity

6a. Many organisms use calcium carbonate in the construction of their shells. How they accomplish this is not certain.
Biological precipitation of CaCO
3 is extremely important, however.

6b. Green plants cause precipitation of CaCO3 by removing CO2 from water in the process of photosynthesis.
Example: Abundant green algae in the warm waters of the Bahama Banks aid in the precipitation of the limy mud and sand with which the banks are covered.

More exotic complications:

7. CaCO3 is polymorphous.
The two commonly occurring natural forms are calcite and aragonite; at least one other form (vaterite) can be prepared artificially and is known as a rare mineral in nature.

Solubilities of the various forms are different.
K = 4.5 x 10
-9 for calcite
K = 6.0 x 10
-9 for aragonite

Suppose we establish equilibrium between solid aragonite and its saturated solution.
Then the product of the concentration of the two ions is: [Ca
+2] [CO3-2] = 6.0 x 10-9
This number is larger than the equilibrium product for calcite; that is - there are too many ions present for the solution to be in equilibrium with solid calcite.
The excess Ca
+2 and CO3-2 should therefore combine to form calcite.

This process does happen but not rapidly enough to be observed under natural conditions.
Main reason is that in many cases nuclei are not present for calcite to precipitate around.
Therefore if aragonite is present, it will remain apparently stable for a long time but eventually turn into calcite.

Important question:
If calcite precipitates at lower ion concentrations than are necessary for the precipitation of aragonite, how does aragonite get produced in the first place?
In fact, experiments show that calcite and aragonite often precipitate together.
No satisfactory answer to this question has been found.
It is an observed fact that many polymorphic substances show this same tendency to precipitate first in metastable forms, which change only slowly to the stable varieties.
Living creatures that use CaCO
3 in their shells may precipitate either polymorph, some species favoring one and some the other. Many pelecypods precipitate both in alternate layers.

8. effect of grain size
Very tiny grains show a greater solubility than do large crystals.
(Ions escape most easily from corners and edges; more of these per volume on small grains.)
We should then have two solubility products for each reaction, one for small grains and one for large grains.
(Technically, we should have an infinity of solubility products.)

A solution in equilibrium with small grains will be oversaturated with respect to large grains and therefore the small grains should eventually dissolve and the large ones grow at their expense.
This is common and produces such effects as growth of large grains in limestone and conversion of opal to chalcedony and quartz after precipitation.

9. effects of other electrolytes (electrolyte - something that produces ions when dissolved)

9a. Common ion effect (covered previously) - decrease in solubility of a salt due to the presence of one of its own ions.

9b. Electrolytes which do not supply a common ion generally have the opposite effect; they make the solubility greater.

Consider a Ca+2 ion in water.
Water is a polar molecule; one end positively charged, the other negative.
All the negative ends of the H
2O molecules surrounding the Ca+2 ion should be close to the ion, the positive ends pointing outward.
The CO
3-2 ions should be similarly surrounded.
Thus the H
2O molecules serve as a shield between the positive ions and the negative ions, making it harder for them to get together and precipitate.
(This is why electrolytes tend to dissolve better in water than in non-polar compounds.)
Now if another electrolyte (say NaCl) is added, the Na
+ and Cl- ions as well as the water molecules will be attracted to the Ca+2 and CO3-2 ions.
This will mean each Ca
+2, for instance, will be at the middle of a cluster of water molecules and Cl- ions.
Thus the shielding is increased and the Ca
+2 and CO3-2 ions will have even more trouble getting together.

In general, the greater the concentration of the second electrolyte, the more ions of it are present for shielding, and the greater the solubility of the CaCO3.
Divalent ions shield better than monovalent ions and this effect means that CaCO
3 is more soluble in a solution of 0.1 molar MgSO4 than in a 0.1 M solution of NaCl.

Of course, the presence of electrolytes should have little effect on the solubility of things that don't ionize much when dissolved, such as H2CO3.

10. possible associations of ions to form complexes such as UO2(CO3)2-2
This removes CO
3-2 ions from solution and makes CaCO3 more soluble.

Thus there are so many things that can effect the solubility of CaCO3 in H2CO3 that in many cases it is impossible to predict whether CaCO3 will dissolve or precipitate or only qualitative reasoning can occur.
Example: We might be able to predict that precipitation would occur as temperature was raised but not be able to predict at exactly what temperature the precipitation would occur.

The fact that CaCO3 is observed to be dissolving in some parts of the ocean and precipitating in other parts and that only slight shifts in the ocean environment can change precipitation to dissolving implies that seawater is essentially saturated in CaCO3.

Other simple carbonates enter into equilibria like the CaCO3 equilibrium according to reactions that can be symbolized:
XCO
3 + H2CO3 <--> X+2 + 2HCO3- with X standing for Fe, Sr, Mn, etc.

Replacement of one carbonate by another is often observed, both in sedimentary rocks and veins.
For example, the replacement of calcite by siderite should follow the equation:
CaCO
3 + Fe+2 <--> Ca+2 + FeCO3

We can obtain the equilibrium constant for this reaction as follows:

            [Ca+2]      [Ca+2][CO3-2]      4.5 x 10-9
       K = -------- =  -------------  = ------------ = 225
            [Fe+2]      [Fe+2][CO3-2]      2.0 x 10-11

This means that a solution of 1 M [Ca+2] would be at equilibrium with 1/225 M [Fe+2].
If the ratio of Ca
+2 to Fe+2 were decreased (more Fe+2 added or Ca+2 removed), in order to reestablish equilibrium, some of the iron would replace the Ca+2 in CaCO3 to produce more siderite.

The Dolomite Problem:
Dolomite rock is one of the commonest sedimentary materials.
There are thick and extensive beds in strata of all ages.
But - Efforts to prepare dolomite in the laboratory under normal sedimentary conditions have failed.
No dolomite is observed forming in nature today in ordinary sedimentary environments.
There is no geologic evidence to indicate that formation took place in the past under unusual conditions of temperature and pressure.
So how does dolomite form?

Clue 1 - look at structure of dolomite
Anions are CO
3-2 group.
Cations are regularly alternating Ca
+2 and Mg+2.
The regular alternation is important.
This is a special, highly-ordered crystalline structure; perhaps it takes a long time to grow.

Clue 2 - free energy levels indicate the following reactions which produce dolomite can occur spontaneously:
(will explain later how free energy works)

Clue 3 - geologic evidence
Poor preservation of fossils in dolomite
Coarseness of grains
Commonly observed cavities and pore spaces

All clues suggest that dolomites in older strata did not form as primary deposits but are instead altered CaCO3 deposits.

Currently accepted theory:
Dolomite is formed by a reaction between Mg
+2 ions and a CaCO3 sediment.
The Mg
+2 may come from seawater in contact with the limy sediment or buried with it, from ions taken up in the original CaCO3 structure (particularly in shells), or from later solutions moving through the sediment.


Part 8: Colloids

Colloidal dispersion - particles in size range between that of true solution (10-7 mm) and suspensions (greater than 10-3 mm)

How to tell colloidal dispersion from suspension:

How to tell colloidal dispersion from true solution:

Electric charges characteristic of colloids.
Due to adsorption of ions on particles
What determines amount of charge and sign of charge not understood.

Colloids are thermodynamically unstable.
According to free energy calculations (will look at later), the particles should spontaneously grow larger.
Electric charge is what keeps colloid particles dispersed.
Reducing or removing charge causes flocculation.
Thus most colloidal dispersions destroyed by adding electrolytes.

Apparent confusion:
Must have electrolyte to form colloid (to get ions which are adsorbed) but more electrolyte causes precipitation.
Key apparently is amount and kind of ions present, but process not completely understood.

Protective colloids - the presence of a second colloid sometimes makes a colloid more stable
Examples:

The role of electric charge in colloidal flocculation leads to three general rules useful in geology:

Some ions adsorbed on colloids are held tightly; some weakly.
Depends on type of force, type of colloid and type of ion.
Usually attachment is weak enough so some ions are easily replaced by others.
Important fact is that
some surfaces or some kinds of colloidal particles hold adsorbed ions more strongly.

In geology, ion exchange is very important.
Examples:

Especially important for clays.
Clays with adsorbed Na
+ are sticky and impervious to water; clays with Ca+2 are granular, easily worked, readily permeable.
That's why gypsum often added to yards to convert Na
+ clays to Ca+2 type.

The effect of heat on colloids not understood.
Important because of hydrothermal studies.

Two forms of colloidal dispersions:
1.
sol

2. gel

Which form depends on:

Two types of colloidal substances:
1.
hydrophilic

2. hydrophobic

Some compounds difficult to classify.
Example: silica

Some common natural colloids:

1. silica

Colloidal characteristics:

silica in the laboratory:

silica in nature:

2. clay minerals

Colloidal characteristics:

Clays are very important in the study of differences in weathering environments (warm, humid, arid, polar, underwater, etc.)
Decay is faster and penetrates deeper in warm humid climates but the products of weathering in all environments are the same except for differences in clay minerals.

Why do we have different clay minerals in different environments?

Formation of clay minerals during weathering is essentially a hydrolysis reaction.
Example:

     4KAlSi3O8 + 22H2O --> 4K+ + 4OH- + Al4Si4O10(OH)8 + 8H4SiO4

Evidence from laboratory work:

Mechanism for producing clay minerals not understood.
How do we go from chain structures (amphiboles, etc.) or 3 dimensional structures (feldspar, etc.) to sheets?
Two
possibilities:

  1. modification of structure perhaps by formation of colloidal particles which then recombine
    Illite currently being studied for clues.
  2. breakdown into Al-O and Si-O which then recombine

Evidence from nature:

Laterites - soils with little or no clay but with AlO and FeO (and SiO2 which is removed in solution)
Why do we get laterites in tropics?
Apparently where weathering is intense, clays break down further.
This is an important process in forming 75 % or world's aluminum ore, 80 % of Ni, 50 % of Fe, etc.


Part 9: Hydration

Other examples of hydration reactions common in nature:
CaSO
4 . 2H2O (gypsum) <--> CaSO4 (anhydrite) + 2H2O
Fe
2O3 (hematite) + H2O <--> 2FeOOH (limonite)

Both of these reactions are simple but not well understood.
Common occurrence of both hydrated and anhydrous forms suggests that the difference in stability is not large and the reactions are extremely slow.


Part 10: Oxidation

Oxidation reactions in nature are simple but not well understood.

At ordinary temperatures, oxidation is slow.
Water speeds up oxidation (probably by dissolving to produce ions which then react more quickly).

Oxidation reactions probably take place in stages.
To understand how complex the situation may be, use pyrite as an
example:
In an arid region:

Other reactions are possible.
Also, can have significant
time intervals between steps.


Part 11: Eh-pH Diagrams

In chemistry, oxidation is defined as a loss of electrons.
Reduction is defined as a gain of electrons.
A substance which is oxidized is said to be a
reducing agent (because it reduces something else) and vice versa.

Examples of oxidation:

Currents flow when there is a difference in electrical potential energy (voltage).
This voltage can be measured with a galvanometer.
If we take 2 metals, attach a wire between one end of each and put the other two ends in a solution which contains ions, a current will flow along the wire.
Electrons lost by one metal are being gained by the other.
An
oxidation-reduction reaction is thus occurring.

The oxidation potential (or technically, the oxidation-reduction potential, symbol Eo) of a substance is that voltage which would occur between an electrode of that substance and an electrode of hydrogen (formed by letting H2 at 1 atmosphere pressure bubble over a platinum rod) at standard conditions.
For the oxidation potential of a natural solution the symbol used is not E
o but Eh (called redox potential).
A positive Eh means hydrogen would be oxidized by the substances in the solution.

Example of use:
Consider a natural water (symbolized by x) with an Eh of 0.5 volt.
Would dissolved iron be present as Fe
+2 or Fe+3?
Equation for oxidation of natural water:

     x --> x+ + e-           Eh = 0.5 volt

Equation for oxidation of iron:

     Fe+2 --> Fe+3 + e-      Eo = 0.77 volt

Interpretation:
Fe is more likely to oxidize hydrogen than the natural solution is and thus Fe is more likely to oxidize the natural solution and be itself reduced.
We would then expect to find Fe
+2.

The usual natural values of Eh and pH are often plotted on graphs with Eh as the vertical axis and pH as the horizontal axis.
These are called
Eh-pH diagrams.
The upper and lower limits of water stability are often indicated as well as the pH-Eh relationships at these limits.

These diagrams can be interpreted to


Part 12: Thermodynamics

Internal energy (E) - total energy of all kinds contained within a system (sum of kinetic and potential energies of its constituent atoms)
Cannot easily be determined in absolute values
Change in internal energy is factor which is most significant.

1st law of thermodynamics - energy can neither be created nor destroyed (Law of Conservation of Energy)

If system undergoes a change of state, and E1 is internal energy before change and E2 is internal energy after change, then E = E2 - E1.
The change of energy is
determined by

By convention, + used for heat in, - used for work done by system.

So E = Q - W, or for infinitesimal change: dE = dQ - dW

Since W = F x distance, pressure = F/A, and V = A x distance, W = PV
or, at constant pressure, dW = PdV
thus dE = dQ - PdV and dE = dQ at constant volume

Enthalpy (H) defined as E + PV (Enthalpy sometimes called heat content.)

Since H = E + PV, dH = dE + PdV + VdP
or, at constant pressure, dH = dQ - PdV + PdV + VdP = dQ + VdP
or since pressure constant, dH = dQ
Also note that if pressure and volume both constant, dH = dE

The change in enthalpy (or in heat at constant pressure) is given special names in certain circumstances:

The amount of heat absorbed or given off (dH) during a chemical reaction can be measured.
(For standardization, unless told otherwise, assume all reactions occur at 25 degrees C and 1 atmosphere pressure, and quantities of reactants are given in moles.)

Heat unit used is usually kcal and is often written as part of the equation.

Example:
H
2 + 1/2 O2 -> H2O + 68.3 kcal
and, since energy lost, dH = -68.3 kcal for this reaction.

(Reminder - the enthalpy change involved in forming a compound from its elements is called the heat of formation.)
Thus the heat of formation of water is -68.3 kcal/mole.

By adding and subtracting heats of formation, the enthalpy change for any reaction can be computed.
For
example - the burning of hydrogen sulfide may be written:
2H
2S + 3O2 -> 2H2O + 2SO2
The heats of formation are:

dH = sum of heats of formation for products minus sum of heats of formation of reactants
= 2 (-68.3) + 2 (-70.9) - 2 (-4.9)
= -268.6 kcal

Entropy (S) - the degree of disorder in a system
In general, spontaneous processes in nature tend toward the formation of less ordered structures, or entropy increases.

2nd law of thermodynamics:
For a system in equilibrium, dS = dQ/T or dQ = TdS
or since dE = dQ - PdV, dE = TdS - PdV
or since dH = dQ + VdP, dH = TdS + VdP
For non-equilibrium processes, dS > dQ/T

Helmholz Free Energy (A) defined as E - TS

Gibbs Free Energy (G) defined as H - TS

Look at Gibbs Free Energy: G = H - TS
Since H = E + PV, G = E - TS + PV
at constant pressure and temperature, dG = dE - TdS + PdV
and, if in equilibrium, dE = TdS - PdV (2nd law of Thermodynamics)
so dG = TdS - PdV - TdS + PdV
or dG = 0

Therefore the criterion for a system to be in equilibrium at constant pressure is that dG = 0.
Another way of saying this is that the Gibbs free energy of the reactants must be equal to that of the products.
If dG for reaction not equal to 0, then system not in equilibrium.
The
farther the value is from 0, the farther the reaction is from equilibrium (at standard concentrations).

If dG is a negative number, then reaction will proceed spontaneously in direction written (although rate may be so slow than no reaction is apparent.)
If dG is a positive number, then reaction will proceed in reverse direction (or energy must be supplied from external source to make reaction go in direction written.)

Refer back to dolomite problem.
Remember we said Gibbs Free Energy calculations showed that dolomite could form from several reactions involving calcite and Mg ions.

Similarly, we can determine that for a system in equilibrium at constant volume, dA = 0
Since most chemical reactions of geologic interest take place at constant pressure rather than constant volume, Gibbs Free Energy more important to us than Helmholz Free Energy.

When substances have standard concentrations, free energies of the products and the reactants can be combined to obtain the free energy change for the reaction (like combining enthalpies of formation)
Values for free energies can be found in tables. Free energies, like enthalpy, are expressed in kcal.

Example of use: Take H2 at 1 atm., HI at 1 atm., and solid Iodine.
1/2 H
2 + 1/2 I2 <-> HI
Is this mixture in equilibrium?
dG = 0.3 kcal
Thus approximately at equilibrium, but not quite. A little bit of HI would tend to break down.

Another example: S at 1 atm., O2 at 1 atm., SO2 at 1 atm.
S + O
2 <-> SO2
dG = -71.8 kcal
Reaction far from equilibrium.
Mixture will react to produce much more SO
2 before equilibrium attained.

Normally tables give free energies of formation for substances as follows:

How often are standard partial pressures and concentrations found in nature? never

At other concentrations, various formulas have been developed by which free energies of reactions can be calculated.
Won't derive formulas, but merely give examples:

                           [Y]y [Z]z ...
     dG = dG0 + 1.364 log -----------------
                           [A]a [B]b ...

dG0 = free energy of reaction for standard "activities" (will define activity later)

Example:
PbS + 2H
+ <-> H2S + Pb+2
at H
+, H2S, and Pb+2 all 1 molar, dG0 = + 11.1 kcal

                                     (1)(1)
at H+ = 10-3 M, dG = 11.1 + 1.364 log ------- = +19.3 kcal
                                     (10-3)2

Tables give free energies at 25 degrees C plus often give list of coefficients A, B, and C
These coefficients used in an equation to calculate dG at
other temperatures.
dG = A + BT log T + CT

Example: Cu2O (s) <-> 2 Cu (S) + 1/2 O2
Table gives A = 40.5, B = .004, C = - 0.03
At 400 degrees C, dG = 40.5 + (.004) (673) (2.9) - (.03) (673) = 28.1 kcal

Additonal information (to show preceding has actually been an over-simplification):

  1. When 2 or more components are in a system, total Gibbs Free Energy is not a simple sum of components' Gibbs Free Energies because the process of mixing changes the Gibbs Free Energy of each component.
    Instead we have an "effective Gibbs Free Energy" for the whole system called the chemical potential.
    Rather complicated equations exist for calculating chemical potential for a multi-component system.
  2. For an ideal gas (no force acting between atoms, etc.), chemical (including thermodynamic) equations are relatively simple.
    For real gases, equations are horrendous.
    Fugacity is fudge factor needed to make the properties of a real gas satisfy the equations of an ideal gas.
    Can be found experimentally for each gas.
    Tables are published.
    Whenever pressure (P) occurs in an equation, we use fugacity instead to get correct answer for real gases.
  3. A similar fudge factor, called activity or effective concentration, is needed for a real solution vs an ideal solution.
    Whenever concentration is called for, we use activity instead to get correct answer for real solutions.
    Activities not accurately known in most cases.