Chapter 2: Equilibrium Thermodynamics and
Equilibrium Thermodynamics: predicts the
concentrations (or more precisely, activities) of various
species and phases if a reaction reaches equilibrium.
Kinetics tells us how fast, or if, the reaction will reach
System: that portion of the universe we wish to study. It
could be as simple as a beaker containing solution or as
complex as the universe. A system can be open
(exchanging matter and energy with its surroundings),
closed (not exchanging matter with its surroundings), or
isolated (exchanges neither matter nor energy with its
Intensive properties: independent of the magnitude of the
system (i.e. temperature and pressure.)
Extensive properties: dependent on the magnitude of the
system (i.e. volume and mass.)
Phase: defined as a uniform, homogeneous physically
distinct and mechanically separable portion of a system.
Components: chemical constituents (species) needed to
completely describe the chemical composition of every phase
in a system.
For example: the solid, liquid and vapor phases of H2O would
be formed by combining the two components (H and O) in the
proportions 2H + O → H2O.
The First Law of Thermodynamics: (a.k.a. Conservation of
Energy) Energy can neither be created nor destroyed, it can
only be changed from one form to another.
The change in internal energy of a system is equal to the
heat added to the system minus the work done by the
ΔE = q – w
Where ΔE is the change in internal energy of the system, q is
the heat added to or removed from the system, and w is the
work done on or by the system.
By convention, heat added to a system is positive and work
done by a system is positive. Thus, the internal energy of a
system will increase if heat is added and will decrease if work
is done by the system.
Internal energy: the energy associated with the random,
disordered motion of molecules.
Heat: energy in transit from a high temperature object to a
lower temperature object.
Work: forms of energy
transfer which can be
accounted for in terms of
changes in the macroscopic
physical variables of the
ΔE = q – w
What is the system?
Where does the heat come from and what is the work being
Is the heat being added to the system, or removed from the
Is the work being done to or by the system?
Remember: ΔE = q – w
Now to the math…
If the work done by or on a system causes a change in
volume at constant pressure, then the equation for ΔE
ΔE = q – PΔV.
Enthalpy(H) is equal to the heat flow when processes occur
at constant pressure and the only work done is pressure-
volume work. Note: dH = dq at constant pressure. (see page
28 of text)
Exothermic reactions: reactions that release heat energy
(enthalpy is negative for the reaction).
Reactants → products + heat
CH4 + 2O2 → CO2 + 2H2O + heat
Endothermic reactions: reactions that use heat energy
(enthalpy is positive for the reaction).
Reactants + heat → products
Ba(OH)2·8H2O(s ) + 2NH4SCN(s ) → Ba(SCN)2(s ) + 10H2O(l ) + 2NH3(g )
Heat of formation: enthalpy change that occurs when a
compound is formed from its elements at a specific
(standard) temperature and pressure. The heat of formation
for the most stable form of an element is arbitrarily set equal
Second law of thermodynamics: for any spontaneous
process, the process always proceeds in the direction of
Another way to look at entropy is that during any spontaneous
process, there is a decrease in the amount of usable energy.
Consider the combustion of coal, an ordered complex organic
molecule is through the process of combustion, broken down
into ash, CO2, H2O, SOx & NOx. Resulting in a dramatic
decrease in the amount of usable energy.
Mathematically, entropy (S) is described in the following
DS = q/T
Where DS is the change in entropy and T is the
temperature in Kelvin.
When the entropy equation is combined with the enthalpy
equation and only pressure-volume work is considered,
dH = TdS + VdP
Systems can exist in several states: unstable, metastable
Figure 2-1 in text
A system at equilibrium is in a state of minimum energy. This
energy can be measured either as:
Helmholtz free energy when the reaction occurs at constant
temperature and volume.
Gibbs free energy when the reaction occurs at constant
temperature and pressure.
Let’s first concentrate on Gibbs free energy.
For a system at constant T & P. Gibbs free energy can be
G = H – TS
Where H is enthalpy(kJ mol-1), S is entropy(J mol-1K-1) and T
is temperature (K)
For changes that occur at constant T & P the expression for
Gibbs free energy becomes:
DG = DH – TDS
If DG is (-), the process occurs spontaneously.
If DG = 0, the process is at equilibrium
If DG is (+), the reaction does not occur spontaneously
To determine DGRº (for the entire reaction) you must subtract
the SDG of the reactants from the SDG of the products.
DGReaction = SDGProducts – SDGReactants
See appendix II in the back of the book (page 474).
So… DGR = DHR - TDSR
mi = (DG / Dni)
Where mi is the chemical potential of a certain component in a
system and Dni is the change in moles of that component. For a
system at equilibrium, mi is the same in all phases.
The chemical potential, µ, of a component in a solution can be
thought of in many ways:
1. A measure of the "escaping tendency" for a component in a
2. A measure of the reactivity of a component in a solution;
3. The chemical potential of a component in a solution is
defined as the rate at which the internal energy of the solution
increases as the number of moles of the component in question
increases, for a given entropy and volume of the solution.
Activity and Fugacity (for a gas): the apparent (or effective)
concentration of a species as opposed the the actual
concentration. They are a measure of the departure of a
system from ideal behavior and need to be taken into
account even when dealing with relatively dilute solutions.
Activity (or fugacity) is related to concentration through the
activity coefficient (gi).
gi = ai / mi
Where ai is the activity and mi is the actual concentration.
Incorporating the chemical potential into the activity yields
the following equation:
mi = mo + RT ln(ai)
Where mo is the chemical potential of component i in its
Now, a word or two about partial pressure…
John Dalton studied the effect of gases in a mixture. He
observed that the Total Pressure of a gas mixture was the sum
of the Partial Pressure of each gas.
P total = P1 + P2 + P3 + .......Pn
The Partial Pressure is defined as the pressure of a single gas
in the mixture as if that gas alone occupied the container. In
other words, Dalton maintained that since there was an
enormous amount of space between the gas molecules within
the mixture that the gas molecules did not have any influence
on the motion of other gas molecules, therefore the pressure of
a gas sample would be the same whether it was the only gas in
the container or if it were among other gases.
So, if you ask "what is the fugacity, and how do I think about
it", just think of it as a partial pressure: it is a strong function
of the mole fraction of the component in the gas phase, and
of the total pressure of the gas phase, just like a partial
pressure; more precisely, remembering that chemical
potential is a quantitative measure of the reactivity of a
component in a phase, we can think of fugacity as a
measure of how much the chemical potential of the
component in the gas deviates from the chemical potential
of some reference, namely, the standard state, due to
changes in P and/or the mole fraction of the component i.
Chemical equilibrium applies to reactions that can occur in
both directions. In a reaction such as:
CH4(g) + H2O(g) <--> CO(g) + 3H2(g)
Many chemical reactions are reversible. The products formed
react to give back the original reactants, even as the
reactants are forming more products. After some time, both
the forward and reverse reactions will be going on at the
same rate. When this occurs, the reaction is said to have
To determine the amount of each compound that will be
present at equilibrium you must know the equilibrium constant.
To determine the equilibrium constant you must consider the
aA + bB <--> cC + dD
The upper case letters are the molar concentrations of the
reactants and products. The lower case letters are the
coefficients that balance the equation. Use the following
equation to determine the equilibrium constant (Kc also written
For example, determining the equilibrium constant of the
following equation can be accomplished by using the Kc
Using the following equation, calculate the equilibrium
N2(g) + 3H2(g) <--> 2NH3(g)
A one-liter vessel contains 1.60 moles NH3, .800 moles N2,
and 1.20 moles of H2. What is the equilibrium constant?
Incorporating Gibbs free energy with the equilibrium
constant yields the following equation:
logKeq = -DGR / 5.708
Where 5.708 is a combination of the gas constant (R) and
standard temperature of 25°C.
Calculate the solubility product for gypsum at 25°C
Note: [activity] and (concentration).
Henry’s Law: The relationship between the equilibrium
constant and the solubility product is used to describe the
activity of a dilute component as a function of concentration.
ai = hiXi
ai is the activity of species i, hi is the Henry’s law proportionality
constant and Xi is the concentration of species i.
Henry’s law relates the fugacity of a gas to its activity in
solution. When a gas behaves ideally (1 atm and near surface
temperatures) fugacity of a gas equals its partial pressure.
Therefore, Henry’s law can be written:
ci = KHPi
Where ci is the concentration of the gaseous species i in
solution, KH is Henry’s law constant in mol L-1 atm-1 (see table
2-1 pg 33), and Pi is the partial pressure of gaseous species i.
Example 2-2 (page 33)
Calculate the solubility of oxygen in water at 20°C.
At sea level (a total atmospheric pressure of 1 bar; where one
atm. = 1.0135 bar) the partial pressure of oxygen is 0.21bar. At
20°C, the Henry’s law constant for oxygen is 1.38 X 10-3
O2(aq) = (1.38 X 10-3 mol/L·bar)(0.21 bar) = 2.9 X 10-4 mol/L
(2.9 X 10-4 mol/L)(32.0 gO2/mol) = 9.28 X 10-3 g/L or 9.28 mg/L
Free energies at temperatures other than 25°C
If the deviations in temperature from 25°C are small (~15°C
or less) we can make the assumption that DHR and DSR are
This brings us to the van’t Hoff equation:
lnKt = lnKr + (DHR/R)(1/Tr – 1/Tt)
Where Kt is the equilibrium constant at temperature t, Kr is
the equilibrium constant at 25°C, Tt is the temperature t and
Tr is 298.15K. Of course, DHR is the enthalpy for the reaction,
and R = 8.314 X 10-3 kJ/mol·K.
Example 2-3 page 34.
Calculate the solubility product of gypsum at 40°C using the
van’t Hoff equation.
For temperature changes greater than 15°C, computation
becomes very complex and requires the use of a spread sheet
or other computer program.
To determine the correct enthalpy under these circumstances,
heat capacity must be introduced into the calculation.
Recall that heat capacity is the amount of heat energy required
to raise the temperature of 1 gram of a substance 1°C.
Cp = a + bT – c/T2
a,b & c are experimentally determined constants.
(You may see cp which refers to heat capacity at a constant
pressure, or cv which refers to heat capacity at a constant
cp = a + bT – c/T2
With the heat capacity and several of the previous formulas,
we can use the following two equations to determine the
Gibbs free energy as a function of temperature (GT).
HT = H298 + a(T-298) + b/2(T2 – 2982) + c(1/T – 1/298)
ST =[a·ln(T/298) + b(T – 298) + c/2(1/T2 – 1/2982)] + S298
Le Chatelier's principle
If a change is imposed on a system at equilibrium, the
position of the equilibrium will shift in a direction that tends
to reduce the change.
CaSO4·2H2O Ca2+ + SO42- + 2H2O + heat
If there is an increase in Ca2+, what will happen?
If the system’s temperature is decreased, what will happen?
CaCO3 + SiO2 CaSiO3 + CO2(g)
If the pressure on the system is increased, what will happen?
Activity coefficient (g): measure of how a specific real system
deviates from some reference system that is taken to be ideal.
In an ideal solution, activity would equal concentration. The
departure from ideal behavior is caused mainly by:
•Electrostatic interactions between charged ions.
•The formation of hydration shells around ions.
There are a variety of models that are used to calculate activity
coefficients. The next few slides will address these models
without getting into the mathematical pathways.
One factor that is important to understand is ionic strength (I).
I = ½ Smizi2
Where mi = the moles/liter of ion i and zi is the charge of ion i.
loggi = (-Azi2( I )1/2)/(1 + Bai( I )1/2)
Where A and B are constants depending only on T and P
and ai is the hydrated radius of a particular ion.
loggi = (-Azi2( I )1/2)/(1 + Bai( I )1/2) + bI
Where b is an experimental constant. This equation includes
the observation that high-ionic strength experimental
systems the activity coefficients begin to increase with
increasing ionic strength.
Use Tables 2-2 and 2-3 for the necessary experimental
Given the following river water chemistry, calculate the activity
coefficient for Ca2+ at 25°C, using both the Debye-Hückel
ands the Truesdell-Jones equations.
Appropriate Ranges of Ionic Strengths
for Activity Coefficient Models
Model Ionic Strength
(mol/L) We should be aware of
Debye-Hückel 0 to 0.1 the different types of
Davies 0 to 0.6
models and their
Truesdell-Jones 0 to 2 limitations.
Specific ion interaction 0 to 4
Pitzer 0 to 6
Calculation of activity coefficients for uncharged species
g = 100.1I
Where I is ionic strength. This translates to:
loggi = KiI
Where Ki is a constant ranging in value from 0.02 to 0.23 at
Environment Ionic Strength Activity
Fresh water 0.002 1.00
Brackish 0.02 1.00
Sea water 0.7 1.17
Activity of water
When water is used as a solvent, pure liquid water at infinite
dilution is used as the standard state; -- the activity of H2O = 1
at infinite dilution. The activity of water is related to the mole
fraction of pure water, XH2O as follows:
mH2O = moH2O + RTlnXH2O
In most cases, we are dealing with dilute solutions and we can
set the activity of H2O = 1. In more concentrated solutions,
such as seawater, the activity will be slightly less than 1.
An aqueous complex is a dissolved species formed from
two or more simpler species, each of which can exist in
With the equation A+ + B- AB(aq)
The equilibrium constant, in this case, is considered the
stability constant (Kstab) because it is a measure of the
stability of the aqueous complex.
Kstab = [AB(aq)] / [A+][B-]
[AB(aq)] = Kstab·[A+][B-] = Kstab·Ksp
By setting the activity of the solid to 1, Ksp = [A+][B-]
A Measure of Disequilibrium
Consider the dissolution of gypsum:
CaSO4·H2O Ca2+ + SO42- + 2H2O
Ksp = [Ca2+][SO42-]
This simplifies matters to the point that we can consider the
activities of the products when determining Ksp. We would
call this the activity product [AP] or ion activity product if only
charged species are involved [IAP].
If the system is in equilibrium, the [AP] = Ksp.
If the [AP] is less that the Ksp, the solution is undersaturated
If the [AP] is greater than the Ksp, the solution is
supersaturated with respect to gypsum.
Equilibrium thermodynamics predicts the final state of a system.
Kinetics tells us if the system will actually achieve this state
within a reasonable time.
It’s all relative…
Homogeneous reactions: involve one phase (gas, liquid or
Heterogeneous reactions: involve two or more phases.
Reactions often occur in a series of steps. The slowest step
determines the rate at which the reaction will proceed.
The dissolution of gypsum
occurs in two main steps.
1st: energy input is
required to free the ions
from the crystal structure.
2nd: the ions must be
diffused. If they are not,
then the microenvironment
surrounding the gypsum CaSO4·H2O Ca2+ + SO42- + 2H2O
crystal becomes saturated
and the reaction stops.
Which ever these steps is the slowest, will determine the rate at
which dissolution proceeds. A common environment for gypsum
dissolution is in the presence of ground water. In this case,
which step would determine the rate of dissolution?
Order of reactions
Zeroth order: The reaction rate is independent of the
concentration of the reactant (A)
t1/2 = 0.5A0/k
First order: The reaction rate is dependent on the
concentration of the reactant A B.
t1/2 = 0.693/k
Second order: The reaction rate is dependent on the
concentration of the reactant 2A B.
t1/2 = - 1/kA0
Note that the units for k depend upon the order of the reaction.
See example 2-7
Comparison of 0th, 1st and 2nd orders
For a zero-order reaction, the rate of reaction is a constant. When
the limiting reactant is completely consumed, the reaction abruptly
Differential Rate Law: v=k
The rate constant, k, has units of mole L-1 sec-1.
For a first-order reaction, the rate of reaction is directly
proportional to the concentration of one of the reactants.
Differential Rate Law: v = k [A]
The rate constant, k, has units of sec-1.
For a second-order reaction, the rate of reaction is directly
proportional to the square of the concentration of one of the
Differential Rate Law: v = k [A]2
The rate constant, k, has units of L mole-1 sec-1.
Figure 2-4 Illustrates the variability of dissolution rates with
change in pH.
The Arrhenius Equation: relates the rate at which a
reaction occurs to the temperature.
k = A exp(-Ea/RT)
log k = log A – (Ea/2.303RT)
Where A is a experimentally derived factor, Ea is the
activation energy for the reaction, R is the ideal gas
constant and T is temperature in Kelvin.
Activation energy: the energy that must be overcome in
order for a chemical reaction to occur.
*Eby notes that a rough rule of thumb for activation energy
is that it doubles with every 10ºC increase in temperature.
Homogeneous nucleation occurs when a nucleus forms
spontaneously in an oversaturated solution.
Heterogeneous nucleation occurs when a nucleus forms in
contact with a, usually solid, surface.
The free energy of formation of a nucleus consists of the
energy gained from the formation of bonds and the energy
required to create the surface.
DGnuc = DGbulk + DGsurf
For an oversaturated solution, DGbulk is always negative.
DGbulk can also be written:
DGbulk = (4pr3/3V)kBT ln(a/ao)
Where (4pr3/3V) is the volume of a spherical nucleus, V is the
molecular volume, kB is the Boltzman constant (1.3805 X 10-23
J/K), T is the temperature in Kelvin, a is actual activity and ao is
the activity for a saturated solution.
DGsurf can also be written:
DGsurf = 4pr2g
Where g is the interfacial energy.
This yields the final equation of:
DGnuc = - (4pr3/3V)kBT ln(a/ao) + 4pr2g
The rate at which nuclei form can be determined from the
standard Arrhenius rate equation.
Rate of nucleation = A exp(-DG*/kBT)
Where A is a factor related to the efficiency of collisions of
ions or molecules, DG* is the maximum energy barrier, kB is
the Boltzmann constant and T is the temperature in Kelvin.