Lecture 7 Chemical Equilibrium

Define equilibrium constant – KDefine the free energies: Gf

, Gr and Gr

Calculate K from Gr

Define Q – How is it different from K

Goal – Learn how to do equilibrium calculations!

Sounds hard but the tools to use are actually very straightforward

Why do we need to study chemical equilibria?

Two questions are asked

1. Is a geochemical system at chemical equilibrium?

2. If not, what reaction (s) are most likely to occur?

Example 1: Solubility - Diatoms (with opal shells) exist in surface seawater

SiO2 • 2H2O (opal) H4SiO4(diatom shell material) (silicic acid)

SiO2 (Opal) SiO2 (quartz)

opal is undersaturated yet diatoms grow

quartz is more stablewhy doesn’t it form?

Example 2: redox and complexation – Iron speciation and plankton growth

Oxidation ReactionFe2+ = Fe3+ + e-

Hydrolysis reactionsFe3+ + H2O FeOH2+ + H+

Organic complexation (e.g. with siderophores) reactionsFe3+ + H3CL FeCL + 3H+

FeSO4 is added to SWbut need to know Fe3+

one of several reactions

can have complexationwith inorganic andorganic ligands

Example 4: Carbonate System – CaCO3 in marine sediments

CaCO3 Ca2+ + CO32-

OrCaCO3 + CO2 + H2O Ca2+ + 2 HCO3

-

important for lysocline

solubility reaction

written in terms of the main species in SW

Chemical reactions can be characterized by an equilibrium constant, K.

This constant expresses the ratio of the product of the concentrations of the reaction products (right side) to the product of the concentration of the reactants (left side).

** Always check that a reaction is balanced for elements and charge**

Example: The solubility of gypsum is written as:

CaSO4.2H2O ===== Ca2+ + SO42- + 2 H2O

Gypsum forms when SWis evaporated 5x. There aregeological formations withgypsum.Used as dry wall in houses!

The solubility of gypsum is written as:

CaSO4.2H2O = Ca2+ + SO4

2- + 2H2O

The equilibrium constant is written as:

K = aCa2+ . aSO42- . aH2O2 / aCaSO4

.2H2O

The magnitude of the equilibrium constant is:

K = 10-4.58 pK = 4.58 (pK = -logK = -log (10-4.58) = 4.58)K = 2.63 x 10-5 (10-4.58 = 10+0.42 . 10-5.00 = 2.63 x 10-5)

written in terms of a = activity = effective conc.

note:2H2O = a2

H2O

log 1.5 = 0.2log 2 = 0.3log 3 = 0.5log 5 = 0.7log 8 = 0.9

What does this K mean?

Ks = aCa2+ . a SO42- = (Ca2+) (SO4

2-) For Ideal Solutions (means activity = concentration) = [Ca2+ ][SO4

2-]

If [Ca2+] = [SO42-] = 10-3 M

If [Ca2+] = [SO42-] = 10-2.29 M

If [Ca2+] = [SO42-] = 10-2 M

activity of solvent = 1 thus aH2O = 1activity of pure solids = 1 thus aCaCO3.2H2O = 1

K = 10-4.58

solution is:undersaturatedat saturation (equilibrium)supersaturated

How do we calculate equilibrium constants?

Answer: From Gibbs Free Energy of reaction Gr

We need to define three types of free energy.

1. Gf ---- Standard Free Energy of FormationDefinition: the energy content of one mole of a substance at standard temperature and pressure (STP = 1atm, 25C). See Appendix I from Drever (1988)

It is the energy to form one mole of substance from stable elements

under standard state conditions.

By definition Gf = 0 for all elements in their stable form at STP. For

example, the most stable form of elemental Ca and O at STP are elemental Ca and O2 gas respectively.

Example: the free energy of formation of CaO is given from the following reaction where the elements Ca and O (in their most stable

form) react to form CaO:

Gr2Ca + O2 = 2CaO

Gf = 0 0 2(-604.2) kJ/mol

the free energy of formation of 2CaO can be obtained from this reaction. Thus:

Gr = 2(-604.2) = -1208.4 kJ/molGf CaO = -1208.4 2 = -604.2 kJ/mol

2. Gr ---- Standard Free Energy of ReactionDefinition: Free energy released or absorbed by a chemical reaction with all

substances in their standard states. This means P = 1 atm, T = 25C, activity = 1 (pure solids = 1, solvent = 1, ideal gas = 1, ions = 1M)

Gr = (Gf)products - (Gf)reactants

For a generalized reaction where B and C are reactant compounds with stoichiometric coefficients b and c and D and E are product compounds with coefficients d and e: bB + cC = dD + eE

Then:Gr = {d Gf D + e GfE } - { b Gf B + c GfC }

Example: for CaSO4.2H2O ===== Ca2+ + SO42- + 2 H2O

Gr = GfCa2+ + GfSO42- + 2 GfH2O - GfCaSO4.2H2O

Gr = (-553.58) + (-744.53) + 2 (-237.13) – (-1797.28) (from Drever, 1988)

Gr = (-1773.37) + (1797.28)

Gr = + 24.91 kJ/mol

The equilibrium constant (K) is calculated from Gr as follows:

Gr = - RT ln K = -2.3 RT logK (note: ln K = 2.3log K)where R = gas constant = 8.314 J deg-1 mol-1 or 1.987 cal deg-1 mol-1

T = absolute temperature = C + 273 = 298° for 25° C

at 25C, K can be simply calculated from: Gr = -5.708 log K (for G in kJ/mol)

= -1.364 log K (for G in kcal/mol)

Example: For the solubility of gypsum +24.91 = -5.708 log K K = 10-4.364

The equilibrium constant K gives the ratio of the products to the reactants at equilibrium, with all species raised to the power that corresponds to their stoichiometry.

For the generalized reaction, K would be written as:(where a = activity)

K = aDd aEe / aBb aCc

Thus Gr and K tell you the equilibrium ratio of products to reactants

but not the concentration of any one species!

3. Gr ---- Free Energy of Reaction There is a whole class of problems where we know the concentrations for a specific set of conditions and we want to ask the question a slightly different way.

Is a specific reaction at equilibrium, and if not, which way will it spontaneously want to proceed for these conditions.

The way we do this is to calculate the reaction quotient using the observed concentrations (converted to activities).

We call this ratio Q to distinguish it from K, but they are in the same form. Comparison of the magnitude of Q with K tells us the direction a reaction should proceed.

The free energy change for a given set of conditions is written asGr = Gr + 2.3 RT log (aDd aEe / aBb aCc)in SW

Gr = Gr + 2.3 RT log Qwhere Q = aDd . aEe / aBb . aCc = Ion Activity Product (IAP)

in seawater

We can replace Gr with its log K expression. Then:

Gr = -2.3 RTlogK + 2.3 RT log Qor

Gr = 2.3 RT log Q/K = 2.3 RT log where = Q / K

if K = Q = 1 Gr = 0 reaction is at equilibriumsystem's free energy is at a minimum

Q > K >1 Gr > 0 reaction goes to leftsystem has too many products and shifts to the left.

Q < K <1 Gr < 0 reaction goes to rightsystem has too many reactants and shifts to the right

G = Total Energy versus extent of reaction

Gminimum at equilibrium

G = Change in Energy versus extent of reaction

G = 0 at equilibrium

only B + C at 0%only D + E at 100%