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Equilibrium Chemistry
Equilibrium may be defined as the state of a chemical or physical system where no
further measurable change occurs. It is important to note that, while it may appear that
the reaction has stopped, the forward and reverse reactions are simply proceeding at the
same rate. Equilibrium is a dynamic, not a static, condition.
A B  C  D
Some reactions reach equilibrium after forming on a small amount of product, while
others proceed until only small amounts of reactant remain. Many reactions end up
somewhere in between with significant amounts of both reactant and product present at
equilibrium. The composition of the reactant-product mixture at equilibrium is called the
equilibrium position.
Some reactions reach equilibrium almost immediately, i.e. in fractions of a second. Other
reaction proceeds more slowly, taking days, decades, millennia to reach equilibrium. In
the case of these ‘slow’ reactions we take a kinetic approach to quantifying chemical fate.
In this lecture we will focus on the ‘fast’ reactions, where chemical equilibrium
calculations can help us to understand chemical fate.
A reversible chemical reaction in a closed system,
aA  bB  cC  dD
will proceed to equilibrium, with the final composition of the reactant-product mixture
defined by the equilibrium position. The character of that equilibrium can be quantified
by the equilibrium constant, the product of the molar concentrations on the right hand
side of the equation divided by the molar concentrations on the left hand side of the
equation,
C D
d
A B
b
c
a
 K
In environmental engineering, it is important to be able to predict and control the
equilibrium position defined by the equilibrium constant. The larger part of this lecture
will provide examples of such applications. Our ability to do so is based on the
equilibrium constant and an understanding of Le Chatelier’s Principle. Stated in general
terms, Le Chatelier’s Principle states that if we impose a change on a system at
equilibrium, that system will respond in such a way as to undo the change imposed upon
it. For example, in the equation,
A B  C  D
if we were to remove some “C” from the system, the reaction would move to the right,
reducing the levels of “A” and “B” and increasing the levels of “C” and “D” until the
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balance described by the equilibrium constant is restored. Similarly, if we were to add
some “D” to the system, the reaction would move to the left, increasing levels of “A” and
“B” and decreasing levels of “C” and “D” until the balance described by the equilibrium
constant is restored.
Equilibria can be manipulated by chemical means (changing concentrations of products
or reactants) and also by physical means (temperature, pressure, volume). We employ
these manipulations to bring the principles of equilibrium chemistry to bear on
environmental issues. Here, we will consider five applications of equilibrium chemistry
important to the environmental engineer.
1. Volatilization: synonymous with evaporation, volatilization is the transformation of a
compound from its liquid state to its gaseous state. The reverse reaction is termed
condensation. The process of volatilization in a closed container will proceed until there
are as many molecules returning to the liquid as there are escaping, i.e. equilibrium. At
this point the vapor is said to be saturated, and the pressure of that vapor is called the
saturated vapor pressure.
For example, consider a spill of gasoline, a mixture of over 500 hydrocarbons, in a closed
classroom. If the spill occurred on Friday, how much liquid gasoline would remain on
Monday for spills of a teaspoon, a cup and a thousand gallons?
Saturated vapor pressure is defined as the partial pressure of a substance that exists in
equilibrium with the liquid phase of the substance at a given temperature.
Consider water in equilibrium between the liquid and gas phases:
H 2 O( l )  H 2 O( g )
The equilibrium constant (K), called the saturated vapor partial pressure is given by:
K 
H 2 Obg
g
H 2 Obg
l
 PH 2O
Since the concentration of a pure liquid is defined as 1:
K  H 2 Obg
 PH 2O
g
the equilibrium constant is simply equal to the concentration in the vapor phase, i.e. the
saturated vapor pressure. We are familiar with this concept through relative humidity,
i.e. the ratio of the partial pressure of water in the air to the saturated vapor pressure,
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RH 
PH2O
PH2O , sat
100
expressed as a percent.
The saturated vapor pressure for all compounds increases with temperature. Thus there is
more water in the air (and it is more uncomfortable) for a relative humidity of 90% in
summer than in winter.
Dew point looks at this from a different direction, specifying the temperature at which the
water content of the air would equal its saturation vapor pressure and condensation
begins. Thus if a warm, moist air mass moves over a cold surface or if a warm air mass
cools at night, the temperature of the air will drop to the dew point and water will
condense.
Examples:



fog
condensation on a glass of ice water
ice on car windows
The saturation vapor pressure not only varies with temperature, but also among
chemicals. The saturation vapor pressure or tendency to volatilize, decreases as the size
of the molecule increases.
Organic Chemical
acetone
toluene
trichloroethylene
tetrachloroethylene
benzo(a)pyrene
PCB congener #77
MW
58
92
131
166
252
292
Use
nail polish remover
gasoline component
industrial metal cleaning solvent
dry cleaning fluid
coal tar, diesel fumes
banned industrial chemical
2. Air-Water Partitioning: Henry's Law describes a chemical's equilibrium between the
air and water phases and the equilibrium constant (K) is termed the Henry’s Law
constant. For oxygen, the Henry’s Law Constant is given by:
O2( g )
 O2(l )
O2(l ) 
O2( l ) 
KH  
 
PO2
O2( g ) 
O2(l )  PO2  K H
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Example – calculate, at 25 C, the concentration of oxygen in water at equilibrium with
the air. The Henry’s Law constant for oxygen at 25 C is 1.29x10-3 moles/L·atm.
O2(l )  PO2  K H

since oxygen accounts for 21% of air, its partial pressure is 0.21 atm.
O2( aq )  PO2  K H  0.21 atm 1.29 x103
2.7 x104
moles
moles
 2.7 x104
L  atm
L
gO2
gO2
moles
mg
 32
 0.00864
 8.64
L
mole
L
L
The concentration of a gas in water varies with the partial pressure of the gas and with
temperature as it influences KH.
Examples,



oxygen in rivers … critical conditions
oxygen in biological treatment … pure oxygen
air-stripping … bomb train
3. Acid-Base: the equilibrium constant (K) here is referred to as the dissociation
constant,
HA  H 2 O  H 3O  A 
HA  H   A 
Ka 
H A
HA
The most fundamental of these is the dissociation of water, where the equilibrium
constant is termed Kw:
H 2 O  H   OH 
Kw 
Since the [H2O] in [H2O] is ~1,
H  OH 
H 2O
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K w  H  OH   1014
and thus the concentration of one ion can be calculated if the concentration of the other is
known. Remember the definition of pH:
pH   log H 
There are hundreds of applications of the dissociation of acids and bases in
environmental engineering. Among these are cases relating to:
pH of an HCl solution
HCl  H   Cl 
[ H  ][Cl  ]
K
 103
[ HCl ]
Determine the pH of a solution prepared by adding 10-2 moles of HCl to 1L of water.



based on the magnitude of K, we can conclude that [H+] >>> [HCl]
since we added 10-2 moles of HCl, we have 10-2 moles H+/L
since pH = -log[H+], the pH = 2
lemon
juice
battery
acid
0
1
2
3
rain
4
5
baking
soda
solution
pure
water
6
7
8
household
ammonia household
lye
9 10 11 12 13 14
Ammonia nitrogen (NH3 is the toxic, pH strippable form)
NH 4  NH 3  H 
K a  10
9 .3

NH 3 H 
NH 4
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What percentage of the NH4+/NH3 system is present as NH3 at pH 7?
% NH 3 
NH 3

4
NH  NH 3
 100
Need to express NH4+ in terms of NH3,
10
9 .3

NH 4 
NH 3 107
NH 4
NH 3 107
109.3
NH 4  200  NH 3
and substitute this to the "%" equation,
% NH 3 
NH 3
 100  0.5%
200  NH 3  NH 3
and for pH 6 (0.052%), 8 (4.8%), 10 (83.3%), 11 (98%).
Chlorine (HOCl is 80-200 times better at disinfection than OCL-)
Cl 2  H 2 O  HOCl  H   Cl 
HOCl  OCl   H 
K a  10
7 . 6

OCl  H 
HOCl
What percentage of the HOCl/OCl- system is present as HOCl at pH 8?
% HOCl 
Need to express OCl- in terms of HOCl,
HOCl
HOCl  OCL
 100
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107.6 
OCl


OCl  108
HOCl
HOCl 107.6
108
OCl   2.5 HOCl
and substitute this to the "%" equation,
% HOCl 
HOCl
 100  28.6%
2.5 HOCl  HOCl
and for pH 6 (97.4%), 7 (79.2%), 9 (3.7%), 10 (0.4%).
4. Precipitation - Dissolution: here, the equilibrium constant (K) is referred to as the
solubility product and refers to the tendency to partition between the solid and dissolved
phases.
For the case of iron addition for phosphorus removal in wastewater,
FeCl 3  Fe3  3Cl 
K sp 
FePO4 ( s)  Fe3  PO34
Fe 3 PO 34
FePO 4 ( s)
the concentration of a solid is, by definition equal to 1 and thus,
K sp  Fe 3 PO 34  1017.9
For the case of calcium carbonate precipitation in drinking water,
CaCO3 ( s)  Ca 2  CO23
K sp 
Ca 2  CO 23 
CaCO 3 ( s)
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K sp  Ca 2  CO 23   108.5
Typically, the value of Ksp increases as temperature increases. A notable exception to
this is the coefficient for CaCO3 which decreases with increasing temperature. This has
important impacts on water supply and treatment.
[T] Scale formation in piping
5. Sorption and Ion Exchange: these processes deal with the interaction between
chemicals and solid surfaces and are important in both natural (air, soil, surface- and
groundwater) and engineered (water and wastewater treatment) systems.
Adsorption is defined as the physicochemical process in which a substance accumulates
at a solid-liquid interface. Adsorption is a physical phenomenon and occurs because the
molecular attraction (van der Waals’ forces) of a solute/sorbent pair are greater than that
of a solute/solvent pair.
Solutes can also become associated with the solid phase by partitioning into the organic
matter which constitutes or coats particles. The combined processes of adsorption and
partitioning are termed sorption and most often involves organic chemicals.
Here, the chemical equilibrium is that between the organic chemical in the solid phase
and that associated with the solid surface (particulate matter):
K 
Csolid
Caqueous
 sorption coefficient
Discuss examples in water treatment and contaminated sediments.
The ion exchange process consists of a chemical reaction between ions in the liquid phase
and ions in the solid phase. Certain ions in solution are preferentially sorbed by the solid
phase on the ion exchanger. Because electroneutrality must be maintained, the solid
releases replacement ions back into solution.
The equilibrium may be described by:
aA  bBads  aAads  bB
and the equilibrium constant, termed here selectivity coefficient, as:
K 
 B
b
A  Bads
b
Aads
a
a
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The selectivity of an ion for a given resin depends on the ion’s charge and size.
In water softening, a synthetic resin, charged with sodium ions, is brought in contact with
a water supply containing calcium:
Ca  Naads  Caads  Na
the equilibrium is given by:
K 
Caads  Na
Ca  Naads
Because calcium is larger (atomic number = 20) and more highly charged (Ca2+) than
sodium (atomic number = 11, Na+), the selectivity coefficient is large and the equilibrium
tends to the right, i.e. exchange calcium for sodium.
Thus when water containing calcium is brought in contact with the resin, calcium is
exchanged (associates with the solid phase) for sodium (is released to the liquid phase).
After some period of time, all of the exchange sites on the resin become saturated with
calcium and the system no longer removes hardness (calcium ions). At this point, the
system must be recharged.
Consider, by inspection of the equilibrium equation, how the resin might be recharged.
What does the ‘recharge solution’ look like and what happens to it.
Reviewing 




Volatilization - the equilibrium between pure compounds and the air
Air/Water - Henry's Law
Acid/Base - ionization of acids and bases
Precipitation/Dissolution - formation and disappearance of solids
Adsorption - exchange of chemicals with solid surfaces