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Module 5
Thermodynamics
1. Introduction
2. Internal Energy
3. First Law of Thermodynamics
3.1 Enthalpy
4. Second Law of Thermodynamics
4.1 Entropy
5. Gibbs Free Energy
6. Equilibrium Constant (K)
6.1 Temperature and Pressure Dependence of K's
7. Fugacity and Activity (concentration) Terminology
READINGS: pp 49-71, 86-89, and 134-138 in “Introduction to
Geochemistry Principles and Applications” by Kula Misra
(2012).
1
What is Thermodynamics?
Branch of science concerned with the transformation of energy
In this course we are interested specifically in energy
transformations associated with geochemical reactions
Chemical reactions occur because systems can lower their
energy by such changes
Chemical systems like all other physical systems always move
toward the lowest energy state
2
Some Definitions
System: any particular part of the physical universe
that we wish to study.
Surroundings: all parts of the universe that are not
part of the system.
Boundary: the surface that separates a system from
its surroundings.
3
Phase: A homogeneous, physically distinct part of system.
e.g., A mineral phase (e.g. olivine); or a gas and liquid phase
(different state)
Component: The chemical constituents needed to form the
phases in a system.
e.g., A system of H2O (liquid + vapour + solid) has 3 phases
but only 2 components (O, H)
Equilibrium: A condition or state where the system (at
equilibrium) does not change with time.
e.g. Consider the reversible reaction:
A + B <------> C + D
At equilibrium, both the forward and reverse reactions occur but
the effects cancel out.
4
Equilibrium is a "relative concept".
Is a diamond sitting on the desk at equilibrium?
It may appear to be stable but diamonds forms at high P, T
and, therefore, is only METASTABLE (appears stable) when
sitting on the desk at STP.
Predicting whether a phase spontaneously reacts is governed
by the ENERGY of the SYSTEM (study of thermodynamics),
But the actual RATE of REACTION is the study of KINETICS
(not part of this course).
The a diamond on the desk is thermodynamically unstable at
STP but the slow kinetics associated with a transformation
reaction makes it metastable (i.e. appear to be stable).
5
For many thermodynamic calculations, we define
STANDARD STATE CONDITIONS as being at T = 298 K
and P = 1 atmosphere (referred to as STP). The concept
of standard state is applied differently depending on the
type of geochemical reaction being considered. For
example:
In the case of solids (e.g., one mineral reacting to form
another), STP applies to pure materials
In the case of gases, STP applies to the pure gas behaving
ideally (units are partial P (atm))
In the case of dissolved ions, STP applies to ideal, 1 molal
H2O solutions (units are molality) (moles of ion/kg of
water)
6
Module 5
Thermodynamics
1. Introduction
2. Internal Energy
3. First Law of Thermodynamics
3.1 Enthalpy
4. Second Law of Thermodynamics
4.1 Entropy
5. Gibbs Free Energy
6. Equilibrium Constant (K)
6.1 Temperature and Pressure Dependence of K's
7. Fugacity and Activity (concentration) Terminology
7
Internal Energy
Thermodynamics deals with the physical and chemical changes of matter
due to HEAT FLOW and WORK. These are governed by the
INTERNAL ENERGY of a system.
The total energy that a system contains is known as the internal energy of the
system.
Faure’s textbook uses the letter “E” to represent internal energy. (Other
books often use the letter “U”.)
E accounts for all types of energy in the system (heat, electrical, chemical,
nuclear etc.)
It is impossible (or at least very impractical) to measure the actual value of E
for any given system.
Fortunately, thermodynamics is concerned with changes in E that
accompany physical and chemical reactions.
8
• Fortunately, thermodynamics is concerned with changes in
E that accompany physical and chemical reactions,
• and it is considerably easier to measure the change in
energy (∆E) for a reversible change from state 1 to state 2,
and to express the change as:
∆ E = E2 - E1
• Remember, reactions occur because systems can lower
their energy (i.e. ∆E < 0).
– Meaning E2 < E1
9
In the situations we will be concerned with in this course, the
internal energy of a system changes if:
Heat passes into or out of the system
Work is done on or by the system
Heat, or thermal energy, results from random motions of
molecules or atoms in a substance and is closely related to
kinetic energy. Heat is the form of energy that flows in
response to a difference in temperature.
We will designate heat by the symbol “q”
10
Work is most commonly defined in its mechanical form.
If a force F acts on an object, work is done in order to move
the object through a distance, say from position x1 to
position x2.
For example, work is required to push a boulder up hill against
the force of gravity.
Work is the integral of the force over the specified distance.
x2
w = ∫ Fdx
x1
11
In chemical thermodynamics work is more commonly expressed
in terms of the variables pressure and volume than in terms of
the variables force and distance. Work, rewritten in terms of
pressure and volume, is as follows.
x1
V1
x0
V0
w = ∫ Fdx = ∫ PdV
This expression shows that work is done by a system on its
surroundings in order to increase the system’s volume from V0
to V1 against the influence of pressure P being exerted by the
surroundings.
12
Module 5
Thermodynamics
1. Introduction
2. Internal Energy
3. First Law of Thermodynamics
3.1 Enthalpy
4. Second Law of Thermodynamics
4.1 Entropy
5. Gibbs Free Energy
6. Equilibrium Constant (K)
6.1 Temperature and Pressure Dependence of K's
7. Fugacity and Activity (concentration) Terminology
13
First Law of Thermodynamics
The 1st law of thermodynamics is also known as the law of
conservation of energy. It can be stated in words as:
“Energy can be neither created nor destroyed, but only
transferred from one form to another.”
Let “q” = amount of heat energy that is exchanged between the
system and its surroundings
Let “w” = the amount of work that is exchanged between the
system on its surroundings.
Any changes to internal energy E will be due to only q and w.
14
The net change in internal energy between two states, 1 and 2,
of the system is given by:
( E2 − E1 ) = ∆E = q − w
This is a mathematical expression of the 1st Law of
Thermodynamics
Internal energy of a system will change by the difference
between the heat energy (q) added to the system from its
surroundings, and the work (w) done by the system on its
surroundings.
15
Insert the explicit expression for work w in the expression of
the 1st law to get
V2
∆E = q − w = q − ∫ PdV
V1
If the system’s volume V is held constant, the
integral on the right side of this equation is zero,
and so
∆E = qv
“qv” = amount of heat energy that is exchanged between the
system and its surroundings at constant volume
16
∆E = qv
In a situation where a system undergoes a transformation at
constant volume, the change in internal energy can be
determined by measuring the heat (q) absorbed by the system.
It is easy to measure the change in internal energy if we can
measure the amount of heat absorbed or released by a
geochemical reaction occurring at constant volume.
17
Definition of Enthalpy (H) in terms of E
Geochemical transformations usually occur under constant pressure
conditions rather than constant volume conditions. It would be useful to
have some characteristic of a system that would equal the heat absorbed
by the system at constant pressure rather than constant volume. This
characteristic is known as “enthalpy”. The change in internal energy
between two states, 1 and 2, at constant pressure can be expressed as:
V2
∆E = E2 − E1 = qP − P ∫ dV = qP − PV2 + PV1
V1
E2 − E1 = qP − PV2 + PV
( E2 + PV2 ) = qP + ( E1 + PV1 )
this simplifies to: H2 = qp + H1
(because the letter “H” designates enthalpy, and the definition is related
to internal energy (E) and a work term (PV): H = (E + PV )
18
For a system reacting from state 1 to state 2 –
H2 = qP + H1
∆Η = H2 – H1 = q p
or ∆Η = q p
“qp” = amount of heat
energy that is exchanged
between the system and
its surroundings at
constant pressure
An expression very similar to that defined previously:
∆E = qv
In a situation where a system undergoes a chemical reaction at
constant pressure, the change in enthalpy (∆H) can be
determined by measuring the heat absorbed or released by the
system.
19
The heat absorbed or released by a chemical reaction
occurring at constant pressure qP is known as the “heat of
reaction” or “enthalpy of reaction” (∆HR ).
For the special case of a chemical reaction in which a
compound is formed from its constituent elements at STP,
the heat of reaction (∆H o ) is given the name enthalpy of
R
o
formation (H f ) to emphasize the special nature of the
chemical reaction. For any reaction:
∆Ho = Σ (n i* H o ) products - Σ (n i * H o ) reactants
R
f
f
∆H°R = change in enthalpy of a reaction at STP
where ° implies at STP
20
Known values of Hof for different compounds can then be used to
determine the heat absorbed or released for any chemical
reactions that involve these compounds.
Suppose that two compounds, A and B, react to form a product
A2B3 according to the following reaction:
2 A + 3B → A2 B3
If we knew the value of “Hof” for each of the compounds A and
B, and for the resulting compound A2B3, we could calculate
the enthalpy of reaction rather than measuring it.
∆H Ro = H of ( A2 B3 ) − [2 H of ( A) + 3H of ( B)]
∆HoR positive = endothermic reactions (i.e. absorbs heat from
surroundings).
∆HoR negative = exothermic reactions (i.e. releases heat to 21
surroundings).
Example: olivine + water goes to serpentine + brucite
2Mg2SiO4 (s) + 3H2O <---> Mg3Si2O5(OH)4 (s) + Mg(OH)2 (s)
Given the enthalpy of formation values as follows:
Olivine = -519.6 kcal/mole
Water = -68.315 kcal/mole
Serpentine = -1043.12 kcal/mole
Brucite = -221.14 kcal/mole
Heat of reaction = [products] - [reactants]
= Ʃ (ni * Hof ) products - Ʃ(ni * Hof ) reactants
=[-1043.12 + (-221.14)] –[2(-519.6) +3(-68.315)]
= -20.12 kcal/mole
This reaction is exothermic so heat is released when the reaction
occurs in the forward direction
22
• A negative value (exothermic reaction) means that the products of
reaction have less enthalpy than the sum of the reactants and that heat
is released to the surroundings when the reaction occurs in the forward
direction.
• Exothermic reactions don’t require the addition of heat energy from
their surroundings, and the system has less enthalpy after an
exothermic reaction than before.
• In way of contrast, endothermic reactions do require addition of heat
from their surroundings, and the system has more enthalpy after the
reaction than before.
• Recall that thermodynamics requires that chemical systems move
(spontaneously) toward the lowest energy state.
• An exothermic reaction, by definition, has the products of reaction
with less heat energy so one might be tempted to equate, albeit
incorrectly, exothermic reactions to spontaneous reactions.
23
Example:
NaCl
(s)
<-----> Na+ (aq) + Cl- (aq)
Hof for NaCl = -98.268 kcal/mole
Hof for Na+ = -57.39 kcal/mole
Hof for Cl- = -39.9 kcal/mole
The enthalpy change (∆HoR) when 1 mole of halite dissolves into its
constituent ions :
∆HoR = (-57.39 + (– 39.9)) - (-98.268) = +0.978 kcal/mole
(endothermic reaction when running as a forward reaction)
The enthalpy change (∆HoR) when 1 mole of halite precipitates from
solution = (-98.268) – (-57.39 – 39.9) = -0.978 kcal (exothermic
reaction when running as a reverse reaction)
24
Everyday experience suggests that if we pour salt into a glass of
drinking water, the salt will dissolve spontaneously.
We now know that this dissolution is endothermic, meaning that
heat energy must be supplied by the surroundings in order for
this dissolution to occur.
Why should an endothermic reaction that requires energy occur
spontaneously?
There must be more to the thermodynamic issue than simply a
reduction in internal energy or enthalpy – also need to
consider ENTROPY.
25
• This example demonstrates that a ∆H cannot (by itself) predict
whether a reaction is spontaneous, the ∆H does permit one to
predict whether the reaction is favoured to proceed (or not) at a
higher or lower temperature.
• An endothermic reaction absorbs heat from the surroundings, so in
a scenario where the temperature of a system increases, one would
expect such a reaction to be MORE favourable (i.e. the reaction
wants heat, and supplying more heat by raising the T is helpful).
• In contrast, exothermic reactions release heat and supplying more
heat by raising T is not helpful (i.e. these are going to be LESS
favourable at higher T).
• Importantly, for a non-spontaneous reaction that happens also to be
endothermic, such a reaction will indeed be more "favourable" as T
increases (even a little bit), but this does not mean that the reaction
becomes "spontaneous" as T increases, because the reaction may
need a very, very high T before it become spontaneous.
26
Question (i) Consider the reaction where liquid water evaporates to water
vapour:
H2O(l) -------> H2O(g)
For this simple reaction, ∆ HoR = + 10.5 kcal.
Is this reaction exothermic or endothermic?
Is the sign of ∆ HoR consistent with your everyday experience?
Explain.
27
Module 5
Thermodynamics
1. Introduction
2. Internal Energy
3. First Law of Thermodynamics
3.1 Enthalpy
4. Second Law of Thermodynamics
4.1 Entropy
5. Gibbs Free Energy
6. Equilibrium Constant (K)
6.1 Temperature and Pressure Dependence of K's
7. Fugacity and Activity (concentration) Terminology
28
Second Law of Thermodynamics (introduces Entropy)
There are several different ways of expressing the second law
of thermodynamics. A useful way of expressing it is:
Energy spontaneously dissipates from areas of higher
concentration to areas of lower concentration unless
prevented from doing so. (e.g. hot objects spontaneously
cool, never get spontaneously hotter)
Examples:
1) A hot object placed in cooler surroundings gets cooler
spontaneously, but never spontaneously gets hotter.
2) Air spontaneously rushes out of a punctured tire but
never spontaneously rushes into a punctured tire.
29
Entropy (designated by the symbol “S”) is the quantitative
measure of how much energy is dispersed, per unit
temperature, in any process. This energy is not available to
do work.
In a reversible process
dS = dq / T
A reversible process is one that takes place slowly enough that a
small change in temperature or pressure could cause the
process to reverse.
Note the units of S are:
calories / (mole × K)
30
Like enthalpy (H), standard values of entropy (S°f) have been
determined for a wide range of compounds. These entropy
values are determined from a knowledge of heat capacity.
∆S0R = change in entropy of a reaction at STP
Like enthalpy (H), this can be expressed in terms of molar
entropy values for the compounds involved.
∆S R = ∑ ni Si (products) − ∑ ni Si (reactants)
0
0
0
31
Example:
Go back to the example of halite dissolving in water
NaCl
(s)
<-----> Na+ (aq) + Cl- (aq)
∆S0R = {1 * S0 Na+ + 1 * S0 Cl-} – 1 * S0 NaCl
= 14.1 + 13.5 – 17.2 = +10.40 cal mole-1 deg-1
Entropy increases when NaCl dissolves into its ions.
We determined earlier that the dissolution of NaCl in water results
in an increase in enthalpy.
How do these changes in entropy and enthalpy combine to make
this a spontaneous reaction? (need to consider Gibbs Free
Energy which combines H and S)
32
Module 5
Thermodynamics
1. Introduction
2. Internal Energy
3. First Law of Thermodynamics
3.1 Enthalpy
4. Second Law of Thermodynamics
4.1 Entropy
5. Gibbs Free Energy
6. Equilibrium Constant (K)
6.1 Temperature and Pressure Dependence of K's
7. Fugacity and Activity (concentration) Terminology
33
Gibbs Free Energy (G)
Reactions occur spontaneously so as to decrease some
combination of the system’s enthalpy and entropy.
This combination is known as the “Gibbs free energy” of the
system, designated by the letter “G”.
G = H – TS
(Note that we are simply defining this as G. It can be derived.)
Spontaneous reactions occur so as to decrease the Gibbs free
energy in a system.
The change in Gibbs free energy for a chemical reaction
occurring at STP is given by:
∆G0R = ∆H0R – T∆S0R
Reactions that have negative values of ∆G0R proceed
spontaneously in the forward direction.
34
Like enthalpy (H) and entropy (S), standard values of Gibbs
free energy of formation (Gof) have been determined for a
wide range of compounds.
The standard Gof value of a compound is the change in Gibbs
free energy associated with the reaction where the compound
is formed from its constituent elements at STP.
These Gof values are listed in appendices in many
geochemistry texts and can be used to calculate the change in
Gibbs free energy of a reaction (∆GoR) in an analogous
fashion as used previously to calculate ∆HoR and ∆SoR :
∆GoR = Ʃ(ni * Gof ) products - Ʃ(ni * Gof ) reactants
35
• If ∆GoR < 0 Reaction is SPONTANEOUS (i.e. favours
forward reaction) at STP.
• If ∆GoR = 0 Reaction is at equilibrium with neither the
forward nor reverse reaction favoured.
• If ∆GoR > 0 Reaction is NOT SPONTANEOUS (i.e.
favours backward reaction) and energy will have to be
supplied to the system to make the forward reaction occur
at STP.
36
Example 1:
NaCl (s)  Na+ (aq) + Cl- (aq)
∆G0R={G0f(Na+) + G0f(Cl- )} – G0f(NaCl)
∆G0R={(-62.593) + (-31.3)} – (-91.815) = -2.08 kcal/mole
∆GR < 0 so, forward reaction proceeds spontaneously
This is consistent with our everyday expectation that halite
dissolves in water at STP.
The magnitude of ∆G0R indicates how far the system is from
equilibrium. The larger the value, the more free energy will be
released as the reaction has to goes to equilibrium
37
Question 2: Consider the thermodynamic properties as
calculated for the reaction where water evaporates at
STP:
Example:
H2O(liq) <--------> H2O(gas)
∆H0R = + 10.5 kcal
∆S0R = +28.4 cal/deg
∆G0R = + 2.05 kcal
Do the results make sense based on our everyday experience?
38
• In the "real world", many geological processes and
geochemical reactions occur at temperature and pressure
conditions other than STP.
• The change in Gibbs free energy for a reaction at non-STP
is written simply as ∆GR.
• Although ∆GoR is a constant at STP for any reaction, ∆GR
changes as a function of T and P.
• Regardless of T and P, a reaction occurs spontaneously in
the direction that will reduce the value of ∆GR because
∆GR is the driving energy for the chemical reaction.
• For any T and P conditions, a reaction at equilibrium has
∆GR = 0 because when a reaction is at equilibrium there is
no energy driving it one way or the other. In order to
calculate ∆GR at any T and P, we need first to understand
the relationship between Gibbs free energy and equilibrium
39
constants (K).
Module 5
Thermodynamics
1. Introduction
2. Internal Energy
3. First Law of Thermodynamics
3.1 Enthalpy
4. Second Law of Thermodynamics
4.1 Entropy
5. Gibbs Free Energy
6. Equilibrium Constant (K)
6.1 Temperature and Pressure Dependence of K's
7. Fugacity and Activity (concentration) Terminology
40
Free Energy Change (G) and Equilibrium Constants (K)
Consider the reaction:
aA + bB <=> cC + dD
Recall that Le Chatelier's Principle states that when a reaction at
equilibrium is disturbed, it re-establishes equilibrium by
counteracting the disturbance
Before the reaction has re-established equilibrium, the Law of Mass
Action governs the reaction quotient (Q) according to
[products] / [reactants]:
Q = [C]c [D]d / [A]a [B]b
41
• Q changes as this reaction proceeds towards equilibrium
until, at equilibrium, Q becomes a constant value called the
equilibrium constant (K) - a numerical expression of
the Law of Mass Action.
• At equilibrium, K = (C)c (D)d / (A)a (B)b
For ideal gases, ( ) = partial pressure in atm
• This expression of K was developed for ideal gases.
• An ideal gas is defined as one in which all collisions
between atoms or molecules are perfectly elastic and in
which there are no intermolecular attractive forces.
(Visualize an ideal gas as a collection of perfectly hard
spheres which collide but which otherwise do not interact
42
with each other.)
Change in Gibbs Free Energy is related to Q:
∆GR = ∆GR + RT ln (Q )
0
where R = ideal gas constant
At equilibrium Q = K and ∆GR = 0
Substituting these values in the last expression gives a
relationship between the change in Gibbs Free Energy at STP
and the equilibrium constant at STP:
∆GR = − RT ln (K )
0
43
∆GR = − RT ln (K )
0
If R = 1.987 cal/(K·mole) and T = 298 K, then
∆GoR (cal) = - 592.126 ln (K)
If you convert cal to Kcal, then
∆GoR (kcal) = - 0.592126 ln (K)
If ln(K) = 2.303·log(K), then
∆GoR (kcal) = - 1.364 log (K)
44
Some properties of K's
When K is a "large" value (i.e. K >>1), the forward reaction (or
products) are favoured.
When K is a "small" value (i.e. K<<1), the reverse reaction (or
reactants) are favoured.
Example: For a reversible aqueous reaction between two species:
A (aq) <------> B (aq)
If K = 1000
If K = 0.001
If K = 1
[B] / [A] = 1000 therefore [B] = 1000×[A]
[B] / [A] = 0.001 therefore [B] = 0.001 [A]
[B] / [A] = 1 therefore [B] = [A]
45
In Summary:
∆GR of a reaction tells us whether a forward or backward
reaction is favoured, but
knowing the value of K of a reaction permits one to
QUANTIFY the activities (aka thermodynamic concentrations)
of the reactants and products at equilibrium where [activities]
are expressed in units of (m) molality (mol/kg).
46
Some algebraic approaches to manipulation of reactions and their effect
on K's:
(1)
reverse
A + B <-------> C + D
C + D <-------> A + B
K1
(reaction #1)
1 / K1
(2)
2n rxn
Add
Simply
A + B <-------> C + D
K1
(reaction #1)
E <-----------> B + C
K2
(reaction #2)
A + B + E <---------> B + 2C + D
A + E <--------> 2C + D
K1 * K2
(3)
2n rxn
Subtract
Simplify:
A + B <-------> C + D
A <-----------> C + E
B <--------------> D - E
B + E <--------------> D
K1
K3
K1 / K3
(reaction #1)
(reaction #3)
47
Temperature dependence of K’s:
Not all reactions occur at STP
One can design experiments to measure the effect that
changing P and T has on K of any particular reaction, but
there are a near-infinite number of reactions and P and T
conditions such that this approach is impractical.
Instead, we have developed thermodynamic models to
estimate the value of K (or ∆GR) at elevated T and P.
One such model involves the “van’t Hoff Equation”
1
∆H R  1
log( K ) − log( K ) = −
− 0
1

2.3025 R  T T 
0
1
0
48
Module 5
Thermodynamics
1. Introduction
2. Internal Energy
3. First Law of Thermodynamics
3.1 Enthalpy
4. Second Law of Thermodynamics
4.1 Entropy
5. Gibbs Free Energy
6. Equilibrium Constant (K)
6.1 Temperature and Pressure Dependence of K's
7. Fugacity and Activity (concentration) Terminology
49
∆GR = − RT ln (K )
0
Problem - This expression was developed for ideal gases.
Recal: An ideal gas is defined as one in which all collisions
between atoms or molecules are perfectly elastic and in
which there are no intermolecular attractive forces.
However: It can be applied to non-ideal gases as well as ions
and molecules in aqueous solutions, BUT one must modify
the equilibrium constant K is these cases would be ratios of
fugacity values (for non-ideal gases) or activity values (for
ionic solutions).
50
• The expressions given above for K and G can also be applied
to non-ideal gases and non-ideal ions and molecules in aqueous
solutions.
• The difference is that the ratio of concentration values that
define K for ideal gases are replaced by ratios of fugacity values
(for non-ideal gases) or activity values (for ionic solutions).
• The fugacity and activity terms for non-ideal gases and
aqueous species are, in effect, modified "thermodynamic
concentration" terms that take into account non-ideal behaviour
of gases and ions in "real world" systems and reactions.
• We will better define these "thermodynamic concentration"
terms later but, for now, they are commonly represented in text
books using a [square bracket] notation or the letters "f" and
"a" instead of the usual (round brackets).
51
• For our hypothetical reaction now involving non-ideal gases or
aqueous ions:
aA + bB <-------> cC + dD
• The form of the K expression is very similar to that used before
for ideal gases, only now:
K = [activity or fugacity of products] / [activity or fugacity of reactants]
K = [C]c [D]d / [A]a [B]b
where [ ] = fugacity is expressed as partial P in atm; or
[ ] = activity expressed in moles {usually molality [m] or
molarity [M]}
m = moles/Kg
M = moles/L
52
Fugacity and Activity (concentration) Terminology:
Fugacity is a "fictitious" pressure, which represents the pressure
needed at a given temperature to make the properties of a real
gas satisfy the equations of an ideal gas.
Suppose we have a mixture of M different gases, each with
partial pressure Pi (i = 1, 2, …M).
Each of these gases has a fugacity fi = γiPi. The quantity γi is
known as the fugacity coefficient.
For an ideal gas the fugacity coefficient = 1.
In applying thermodynamic formulae to real gases we should
substitute fugacity values fi wherever partial pressure Pi
occurs.
53
Activity:
Activity is an analogue of fugacity for ionic solutions.
The concentration of ions in an aqueous solution is often not a
good measure of their ability to enter into ionic reactions
with other ions in the solution.
Rather than substituting concentration values for partial
pressures, we need to substitute activity values.
The activity of an ionic component is given as the product of
its concentration and an activity coefficient.
[activity] = γ (molality)
54
Example: If one looks up the concentration of calcium ion (Ca2+)
dissolved in seawater, it will be reported as having a total measured
concentration of 413 ppm (ppm=mg/kg) (which is equivalent to 0.01
m or 0.01 mol Ca2+ / kg H2O).
BUT, some of the calcium in seawater occurs in the form of
other aqueous (dissolved) Ca complexes such as: CaHCO3- ,
CaCO30, CaCl- , CaCl20, CaNO3+, CaOH+, CaPO4-, CaSO40
among many others.
For Ca2+ ion in seawater: (where 413 mg/kg = 0.01 molal)
[Ca2+]= γ (Ca2+)
= 0.28 (0.01 m)
= 0.0028 m
= 0.0028 moles of Ca2+ / kg of water
(a reduction of 72%)
55
Whereas the total concentration of Ca2+ = 0.01 m, the
activity of Ca2+ is 0.0028 m which represents the
thermodynamic ("real") concentration of Ca2+ available to
participate in various chemical reactions in seawater, such
as:
Ca2+ (aq) + CO32- (aq) <----------> CaCO3 (s)
i.e. the precipitation of limestone
The value for the activity coefficient (γ) comes from
tables in geochemical texts, or it can be calculated using
various theoretical models (see below)
In very dilute aqueous systems, there is no difference
between "ideal" and "real" behaviour of ions or gases
because activity coefficients γ = 1.
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Conventions - Activities are expressed in moles (units of m = mole/kg
solvent or M = mole/L)
- In geochemistry, activity units of molality (mol / kg of H2O)
are much better than molarity (mol / litre) because volume
changes with temperature, whereas a kg of H2O is a
defined mass.
- Activities of pure solids and water are defined as = 1
- Fugacities of gases are expressed as partial pressures (units
of atm)
Reactions are assumed to take place at STP (25°C, 1 atm)
unless otherwise indicated
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How to calculate activity coefficients (γ) for solutions:
First need to calculate Ionic Strength (I)
I = 1/2 Σ mi zi2
(m = molality of ion; Z = charge of ion)
Example: Calculate I for a 0.1 m solution of CaCl2
CaCl2 (s)
I
= 1/2
Ca2+ (aq) + 2Cl-
(aq)
Σ [(0.1 x 22) + 2(0.1 x (-1)2)]
Ca
Cl
= 0.3
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Debeye Huckel theory
Activity Coefficient (γ )
For I < 5 x 10-3 (e.g. rivers and lakes)
- log γ = A z2 I1/2
For I < 0.1 (e.g. groundwater)
- log γ = [ A z2 I1/2 ] / [1 + a B I1/2]
For I ~ 0.5 (e.g. seawater)
- log γ = A z2 {( I1/2 / (1 + I1/2)) - 0.2I}
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Table 10.3: Values for the "A and B" constants used in the Debeye Huckel
and Davies equations for calculating activity coefficients (γ) for various
ions: (from Faure, 1998)
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Table 10.4: Values for the "a" constant used in the Debeye Huckel
equation for calculating activity coefficients (γ) for various ions: (from
Faure, 1998)
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Example: Effect of activity coefficients on solubility of minerals
Solubility of calcite in seawater: γ Ca 2+ = 0.28; γCO3 = 0.20
CaCO3 (s) <-------> Ca +2 (aq) + CO3-2 (aq)
Ksp = 10-8.35
Recall [a] = γ (m)
Ksp = 10-8.35 = γ* [Ca+2] * γ*[CO32-]
For the dilute case (where γ = 1): Let x = (Ca+2) = (CO32-)
=1*x * 1 *x
Ksp = 10-8.35
= x2
x = 6.68 * 10-5 m (or 0.0067 g calcite / kg H2O)
(given a molar mass of calcite ~ 100)
For the seawater case:
Ksp = 10-8.35 = 0.28 x * 0.2 x
10-8.35 = 0.056 x2
x = 2.82 * 10-4 m (or 0.028 g calcite / kg H2O)
% solubility change = 322 % increase in solubility of calcite in
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seawater compared to dilute water
% solubility change = 322 % increase in solubility of calcite in
seawater compared to dilute water
Ca2+ ions form many Ca complexes such as CaHCO3- , CaCl- ,
CaNO3+, CaOH+ etc. so the total solubility of calcite increases.
CaCO3 (s) <-------> Ca+2 (aq) + CO3-2 (aq)
Le Chatelier’s Principle requires reaction to shift to the right as the
calcium ion produced is used to form the many Ca complexes in
seawater.
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