AQUEOUS GEOCHEMISTRY:

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AQUEOUS GEOCHEMISTRY Lecture 4
(11/14/2009) http//ees2.geo.rpi.edu/abrajanoCo
urses/Aqueous2000/AqGeochem.html

• Last Time
• Chemistry of Natural Waters
• Portraying Water Composition
• Water Flow
• Thermodynamics Introduction

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Background Thermodynamics (11/14/2009)

• Tools of thermodynamics.
• Equilibrium and Keq
• Activity and g
• IAP and saturation index SI
• Keq and temperature
• Mineral solubility and gi
• Mineral solubility and complexation

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WHAT CAN THERMODYNAMICS TELL US?

• Whether a mineral/compound should dissolve in or
  precipitate from a solution.
• What types of other reactions control water
  chemistry (e.g., acid-base, redox) might occur.

What cant thermodynamics tell us?
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THE MEANING OF EQUILIBRIUM

• A system with none of its properties changing
  with time, no matter how long it is observed.
• A system that will return to the same state after
  being disturbed.
• Thermodynamically speaking, a system is at
  equilibrium when ?rG 0

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Thermodynamic Background

• Thermodynamic concepts apply perfectly to
  idealized systems. Given infinite time, all
  systems will tend towards equilibrium.
• In real systems, we make assumptions to apply
  thermodynamic concepts (what approximations are
  valid and when?)
• Thermodynamics determine the abundance,
  distribution, and fate of aqueous chemical
  species at equilibrium.
• Conversely, the presence and abundance of species
  in systems can tell us if the system is out of
  equilibrium.

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Whats the big deal??
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Terms

• System A system is the part of the universe that
  is of interest for a particular problem, e.g., a
  puddle of water, a lake, an aquifer, etc.
• Isolated, Closed and Open systems No exchange
  of energy and mass, exchange energy but not mass,
  exchange energy and mass with surroundings
  (outside of system), respectively.
• Extensive and Intensive Variables Function of
  and not a function of system magnitude (or size),
  respectively.
• M, V versus P, T. M/V (ratio of extensive) r
  (intensive).
• Phase portion of the system with distinct and
  homogeneous physical and chemical properties.
• Components chemical entities required to
  describe completely the composition of all phases
  present in a system.

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Gibbs Phase Rule

• In a given system, how many phases can be at
  equilibrium at any one time?
• The question was addressed by J. Willard Gibbs
  using the "Phase Rule"
• P F C 2 (can be derived - see background
  notes)
• P is the number of phases
• C is the number of components, and
• F is the variance (i.e., the number of physical
  or chemical variables that need to be fixed for
  the system to be specified).

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Gibbs Phase Rule
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Equilibrium and the Laws

• Thermodynamics is primarily concerned with energy
  changes, and in geochemical thermodynamics, these
  energy changes are associated with atomic
  rearrangement or change in atomic configuration.
  These chemical changes are conventionally
  represented by chemical reactions.

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Equilibrium and the Laws

• aA bB ... mM nN
• a, b, m, n are the amounts of A, B, M, N
  participating in the rxn. A, B, M, N are the
  components of our system.
• Rule The left hand side are reactants, the right
  hand side are products, so that what we call
  energy change, say Gibbs free energy (Go), is
• ?rG ?Goproducts - ?Goreactants mGoM
  nGoN .. aGoA bGoB
• If ??rG 0, the reaction is at equilibrium if
  ??rG lt 0, the reaction will proceed to the right
  if ?? rG gt 0, the reaction will proceed to the
  left.

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Equilibrium and the Laws

• A system is at equilibrium if no gradients of
  energy exist, i.e. there is no driving force to
  make it change.

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Two Types of EQUILIBRIUM

• Stable equilibrium - System is at its lowest
  possible energy level.
• Metastable equilibrium - System satisfies
  criteria for equilibrium, but not E minimum.
• Metastable Low T, Kinetic

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Other Types of EQUILIBRIUM

• Partial Equilibrium
• Local Equilibrium

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Equilibrium and the Laws
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Equilibrium and the Laws
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Equilibrium and the Laws
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Equilibrium and the Laws
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Equilibrium and the Laws
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Enthalpy (DH)

• 1st Law DU Dq - Dw 2nd Law DS Dqrev/T
  DW PDV
• DU TDS - PDV
• DU PDV TDS
• D (U PV) TDS DH ? DEnthalpy (heat
  change at constant P)
• H U PV
• General form DH DU PDV VDP TDS VDP
• ? T (?H/?S)P, V(?H/?P)S, (?T/?P)S (?V/?S)P
• DH expresses the heat (q) that is exchanged
  during transformations at constant P. Practical
  application it can be measured.

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Gibbs Free Energy (G)

• From above, DU - TDS PDV 0
• _at_ constant P T D (U - TS PV) 0 DG
• G H - TS (H U PV)
• _at_ constant T DG DH - TDS
• General form
• DG DU - TDS - SDT PDV VDP -SDT VDP
• Partial Derivatives!!
• At equilibrium DG 0 or SDT VDP
• or
  S/V DP/DT

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Gibbs Free Energy (G)

• For a reaction to occur _at_ constant T
• DG DH - TDS lt 0
• ?Hrx ?Srx ?Grx Example
• if ??Hrx? gt T??Srx? rx doesnt occur
• - if ??Hrx? lt T??Srx? Melting
• - rx doesnt occur
• - - Burning
• - - if T??Srx? gt ??Hrx? rx doesnt occur
• - if T??Srx? lt ??Hrx? Freezing
  (crystallization)
• We can predict whether a reaction will occur by
  calculating ?Grx.

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Calculating ?Grx
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UNITS

• Thermodynamic Variable English SI
• G, H, Q (energy) calories joules
• S cal/? joules/?
• V cal/bars m3
• P bars pascals
• R (gas cnst) 1.9872 cal/? 8.314 joules/?
• T in ?K
• 1 cal 4.184 joules
• cal/bar cm3/41.84

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Equilibrium Constant

• Chemical reactions proceed until chemical
  equilibrium is achieved, i.e., ?Gr? 0, when the
  rates of forward and backward reactions are
  equal A B ? C D
• vf kf(A)(B), vb kb(C)(D)
• At equilibrium, vf vb, therefore
• where K is the equilibrium constant. In the case
  of a general reaction aA bB ? cC dD, the
  equilibrium constant expresses the constant
  relationship between activities of reactants and
  products

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• aA bB ? cC dD
• Assume
• activity concentration

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EXAMPLE Fluorite Dissolution

• CaF2 (s) Ca2 (aq) 2 F- (aq)
•  
• aCa2,aq a2F-,aq
• Keq ------------------------
• aCaF2s
• If CaF2 is pure (activity mole fraction1)
•  
• Keq (Ca2)aq(F-)2 solubility product Ksp

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Solubility

• The amount of solid that can be dissolved in
  water without precipitating it (i.e., before
  saturation).
•  
• For CaF2 _at_ 25 oC in pure water, we can dissolve
  0.017 g of CaF2 /1 kg of water.
• In molality, the solubility is 0.017/78.08/kg
  (molecular weight of CaF2)
• 0.00022 moles/kg

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Solubility

• CaF2 solubility 0.00022 moles/kg
• What is the Equilibrium Constant?
• Keq Ksp (Ca2)aq(F-) aq 2
  (0.00022)(0.00044)2
• 4.26 x 10-11
•  Keq is often written in logarithmic form
• log Keq -10.4

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Solubility and Equilibrium Constant
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Common Ion Effect

• CaF2 (s) Ca2 (aq) 2 F- (aq)
• aCa2,aq a2F-,aq
• Keq ------------------------
• aCaF2,s
• Approaching this from undersaturation Keq gt
  aCa2,aq a2F-,aq
•  
• Addition of a "common ion salt" (e.g., CaSO4 or
  BaF2) leads to an approach to saturation even
  when CaSO4 and CaF2 were added to the water in
  amounts lower than their solubility.
• (e.g., Floridan Carbonate Aquifer)

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Common Ion Effect

• CaF2 (s) Ca2 (aq) 2 F- (aq)
• aCa2,aq a2F-,aq
• Keq ------------------------
• aCaF2,s
• Approaching this from undersaturation Keq gt
  aCa2,aq a2F-,aq
•  
• Addition of a "common ion salt" (e.g., CaSO4 or
  BaF2) leads to an approach to saturation even
  when CaSO4 and CaF2 were added to the water in
  amounts lower than their solubility.
• (e.g., Floridan Carbonate Aquifer)

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ALTERNATE RETRIEVAL OFSOLUBILITY PRODUCT

• The equilibrium constants for reactions of the
  type
• CaSO4(s) ? Ca2 SO42-
• is called a solubility product (KSP). The KSP can
  be calculated according to
• ?? rG ?f GCa2 ?f GSO42- - ?f GCaSO4(s)
• and

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SAMPLE CALCULATION OF KSP FROM DGs

• From the Appendix of Kehew (2001) we obtain
• ?f GCa2 -553.6 kJ mol-1
• ?f GSO42- -744.0 kJ mol-1
• ?f GCaSO4(s) -1321.8 kJ mol-1
• so
• ??rG -553.6 (-744.0) - (-1321.8) 24.2 kJ
  mol-1

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Solubility and Equilibrium Constant
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A SECOND EXAMPLE CALCULATION OF KSP

• Calculate the solubility product of mackinawite
  at 25C
• FeS(s) ? Fe2 S2-
• ??rG ?f GFe2 ?f GS2- - ?f GFeS(s)
• ??rG -90.0 (85.8) - (-93.0) 88.8 kJ mol-1

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A THIRD EXAMPLE CALCULATION OF KSP

• Calculate the solubility product of dolomite at
  25C
• CaMg(CO3)2(s) ? Mg2 Ca2 2CO32-
• ??rG ?f GMg2 ?f GCa2 2?f GCO32- - ?f
  GCaMg(CO3)2(s)
• ??rG -455.5 (-553.6) 2(-527.0) - (-2161.3)
  
• 98.2 kJ mol-1

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Anhydrite
Mackinawite
Dolomite
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THE ION ACTIVITY PRODUCT (IAP)

• Consider once again the reaction
• CaSO4(s) ? Ca2 SO42-
• The equilibrium constant is expressed in terms of
  the activities of the reactants and products at
  equilibrium
• However, a real solution may or may not be in
  equilibrium. The ion activity product (IAP ) or
  reaction quotient (Q ) has the same form as the
  equilibrium constant, but involves the actual
  activities

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THE SATURATION INDEX (SI)

• The saturation index (SI) is defined according
  to
• If IAP KSP, then SI 0, and the water is
  saturated with respect to the mineral.
• If IAP lt KSP, then SI lt 0, and the water is
  undersaturated with respect to the mineral.
• If IAP gt KSP, then SI gt 0, and the water is
  supersaturated with respect to the mineral.

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APPLICATION

• Suppose a groundwater is analyzed to contain
  5x10-2 mol L-1 Ca2 and 7x10-3 mol L-1 SO42-. Is
  this water saturated with respect to anhydrite?
  (assume activityconcentration)
• KSP 10-4.24 mol2 L-2
• IAP (5x10-2)(7x10-3) 3.5x10-4 10-3.45 mol2
  L-2
• In this case, SI gt 0, i.e., IAP gt KSP, so the
  solution is supersaturated and anhydrite should
  precipitate.

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A SECOND APPLICATION

• A water contains the following 3.13x10-4 mol L-1
  Mg2
• 8.48x10-4 mol L-1 Ca2 and 2x10-6 mol L-1 CO32-.
  Determine whether this water is saturated with
  respect to dolomite (assume activityconcentration
  ).
• KSP 10-17.20 mol2 L-2
• IAP (8.48x10-4)(3.13x10-4)(2x10-6)2
• 1.06x10-18 10-17.97 mol2 L-2
• Because SI lt 0, the solution is undersaturated
  with respect to dolomite the mineral should
  dissolve.

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Keq of H2O DISSOCIATION OF WATER AND NEUTRAL pH

• In this case, we have the reaction
• H2O(l) ? H OH-
• for which
• but because usually a H2O ? 1,
• _at_25 oC
• ??rG ?f GH ?f GOH- - ?f GH2O
• ??rG 0 (-157.3) - (-237.1) 79.80 kJ mol-1

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Keq of H2O DISSOCIATION OF WATER AND NEUTRAL pH

• If this reaction were occurring in pure water
  with no other solutes, then it would have to be
  true that at equilibrium
• Thus

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Keq of H2O DISSOCIATION OF WATER AND NEUTRAL pH

• When the solution is said to
  be neutral. Hence, neutral pH at 25C is pH 7.
• Note that, a pH of 7 is only neutral at 25C and
  1 bar, because Kw is a function of both pressure
  and temperature.
• For example, at 0C neutral pH is 7.47 and at
  50C neutral pH is 6.63. By 300C, neutral pH is
  5.7!

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DISSOCIATION CONSTANT OF WATER AT VARIOUS
TEMPERATURES
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THE DISSOCIATION OF WATER AND NEUTRAL pH

• Is a solution with and
• in equilibrium at 25C?
• To answer this we need to calculate the IAP
• Because IAP gt Kw, the solution is not in
  equilibrium. Some OH- and H need to be consumed
  to bring the reaction back to equilibrium.