Download LECTURE 5 - CHEMICAL EQUILIBRIUM

Environmental Geochemistry, GLY 4241/5243, © David Warburton, 2016
LECTURE 5 - CHEMICAL EQUILIBRIUM
Note: Slide numbers refer to the PowerPoint presentation which accompanies
the lecture.
Chemical Equilibrium, slide 1 here
Whenever two substances are brought together the potential for a reaction
between them exists; whether the reaction takes place depends on several
factors.
Chemical Equilibrium, slide 2 here
We may define a chemical reaction as one or more substances, known as
reactants, combining chemically to form one or more different substances,
known as products.
Chemical Equilibrium, slide 3 here
Examples:
5-1
5-2
5-3
5-4
Given that a reaction might occur, a chemist is interested in two questions:
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Chemical Equilibrium, slide 4 here
A.
How far can the reaction proceed?
For example, the combination of a strong acid and a strong base (equation
5-1) will produce an almost complete reaction so that the amount of products
will greatly outweigh the amount of reactants. On the other hand, a reaction like
5-5
produces almost no Ag+ or Cl- ions; that is, AgCl is insoluble.
Most reactions lie somewhere between the extreme cases. The property
in question here is thermodynamic, and given the proper data we can calculate
the exact conditions that will exist when equilibrium is attained.
Chemical Equilibrium, slide 5 here
B.
How fast does the reaction proceed?
Some reactions are possible but may occur very slowly.
5-6
A mixture of hydrogen and oxygen gases in the proper proportions will react
explosively to form water vapor if a source of energy, such as a spark, initiates
the reaction. If no such energy source exists, the reactants may coexist for years
with very little product formation. The speed of a reaction is not a
thermodynamic property but instead is a kinetic property.
Chemical Equilibrium, slide 6 here
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Environmental Geochemistry, GLY 4241/5243, © David Warburton, 2016
LAW OF MASS ACTION
In the nineteenth century, two Norwegian chemists, Cato Maximilian
Guldberg & Peter Waage (1864), determined the key to the numerical handling
of chemical equilibrium. They discovered that when the driving forces of the
forward and back reactions become equal, equilibrium has been achieved.
Chemical Equilibrium, slide 7 here
Forward reaction:
5-7
Chemical Equilibrium, slide 8 here
Back reaction:
5-8
Driving force of the forward reaction = kf[NaCl]
Driving force of the back reaction = kb[Na+][Cl-]
Chemical Equilibrium, slide 9 here
At equilibrium:
kf[NaCl] = kb[Na+][Cl-]
so the equation may be rewritten:
K is called the equilibrium constant for the equation as written. This type of
formulation is known as the "Law of Mass Action." (The term concentration
should really be substituted for mass).
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Chemical Equilibrium, slide 10 here
The Law of Mass Action may be described as, "The rate of a reaction is
directly proportional to the concentration of each reacting substance."
(Krauskopf, 1979, p.2).
Chemical Equilibrium, slide 11 here
Unfortunately the situation described by the Law of Mass Action is more
complicated than the law suggests. If the reaction is written differently, such as
by doubling all quantities, we need to rewrite the equilibrium constant equation
using concentrations raised to the second power. Another problem with such a
formulation is that kf and kb are not physically measurable quantities.
Nevertheless, Guldberg and Waage gave us a tremendous step in the right
direction with their formulation.
Le Chatelier's Rule
Chemical Equilibrium, slide 12 here
Assume we are dealing with a very simple reaction such as is given in
equation 5-1. The equilibrium constant is given by equation 5-10:
5-10
If the system is at equilibrium, and some component B is added, so that [B]
increases, what happens to the system? Since K is a constant, the numerator
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must be altered so that K remains constant. Because [B] has increased, the
reaction must proceed to the right, thus reducing both [A] and [B], and
increasing [C] and [D]. When the right side is equal to the original value, the
system is again at equilibrium. If more C had been added to the original system,
exactly the reverse would have occurred - the reaction would have gone to the
left, reducing [C] and [D] and increasing [A] and [B].
The disturbance need not be the addition of a reactant or product species.
It could equally well be a change in temperature, or sometimes, in pressure.
Equilibrium constants are temperature, and often, pressure dependent. If a
reaction is exothermic (formation of products gives off heat) then an increase in
temperature will favor the reactants. The back reaction speeds up using the heat
generated by the forward reaction (the back reaction is endothermic).
Chemical Equilibrium, slide 13 here
This is the essence of Le Chatelier's Rule: When a system at equilibrium
is disturbed, the system will respond in a way that lessens the disturbance. This
is one of the most important rules in chemistry and one that we need to keep in
mind.
Chemical Equilibrium, slide 14 here
Le Chatelier first said it this way, “Any system in stable chemical equilibrium,
subjected to the influence of an external cause which tends to change either its
temperature or its condensation (pressure, concentration, number of molecules
in unit volume), either as a whole or in some of its parts, can only undergo such
internal modifications as would, if produced alone, bring about a change of
temperature or of condensation of opposite sign to that resulting from the
external cause.” (Le Chatelier, 1884)
Chemical Equilibrium, slide 15 here
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Four years later, he restated the principle more succinctly as, “Every
change of one of the factors of an equilibrium occasions a rearrangement of the
system in such a direction that the factor in question experiences a change in a
sense opposite to the original change.” (Le Chatelier, 1888)
The second point that this simple system shows goes back to the second
question raised at the beginning of the lecture. If the forward reaction is very
slow, when will equilibrium be reached?
Chemical Equilibrium, slide 16 here
5-11
At room temperature kf is very small; kb is far smaller and the system is not at
equilibrium. Rather, it is said to be metastable. A metastable system is one that
changes so slowly that it appears stable. Metastable systems are not at
equilibrium but may persist for very long times.
The reason that metastable systems exist is the presence of a significant
kinetic barrier. Kinetic barriers may be explained by examining the reaction on
an atomic scale. To form water, breaking two H-H bonds and one O-O bond is
necessary. The four resulting O-H bonds are stronger than the three broken
bonds. That is they release more energy when they form than it took to break
the other three bonds. However, this energy is not available at the beginning of
the reaction. If a spark is present at the beginning of the reaction, some
hydrogen and oxygen molecules are converted to free radical hydrogen and
oxygen. These free radicals are extremely reactive and initiate chain reactions:
5-12
5-13
Once initiated these reactions are very rapid and go to completion.
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Chemical Equilibrium, slide 17 here
The condition for a thermodynamically favorable reaction is very simple:
If ΔG < 0, the reaction will take place spontaneously, although the
rate may be extremely small.
If ΔG = 0, the reaction is at equilibrium.
If ΔG > 0, the reaction cannot take place without energy being
supplied from outside the system.
Here ΔG represents the change in Free Energy on going from reactants to
products. The condition for a negative free energy value indicates the reaction
releases energy to the environment.
Chemical Equilibrium, slide 18 here
This type of situation can be represented by an energy diagram as shown
in figure 5-1. The reaction shown is thermodynamically favorable. Energy
barrier height (E0) represents the kinetic barrier. If the kinetic barrier is high, the
reaction may be extremely slow. Substances known as catalysts work by
reducing the kinetic barrier thus allowing the reaction to speed up. High kinetic
barriers lead to metastable systems.
Figure 5-1
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Chemical Equilibrium, slide 19 here
This is a common situation in geology. Many high pressure minerals, formed
at great depths within the earth, do not immediately revert to the lower pressure
polymorphs when they reach the surface. The word "stable" should only be
applied to systems that are at equilibrium.
Chemical Equilibrium, slide 20 here
The equilibrium constant, K, may be defined for a general system as follows:
5-14
5-15
Each product's concentration is raised to the coefficient of that species in
the equation, and multiplied by the concentration of each other product species,
raised to the correct exponent. This represents the numerator, and the
denominator is given by the same type of expression for the reactants.
NOTE: Each equilibrium constant applies only to the reaction for which it is
written. Many ways of writing nearly the same reaction exist.
Chemical Equilibrium, slide 21 here
5-16
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Environmental Geochemistry, GLY 4241/5243, © David Warburton, 2016
5-17
Chemical Equilibrium, slide 22 here
5-18
5-19
Chemical Equilibrium, slide 23 here
5-20
5-21
Chemical Equilibrium, slide 24 here
Solubility products are a special type of equilibrium constant.
5-22
5-23
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Environmental Geochemistry, GLY 4241/5243, © David Warburton, 2016
What is the concentration of a solid? It is effectively a constant, so we
may set it equal to one. Then equation 5-23 reduces to:
5-24
Here Ksp represents the solubility product.
Chemical Equilibrium, slide 25 here
The solubility in moles/liter is equal to [Pb2+] or [SO42-], since one ion of
each is produced when the PbSO4 molecule dissolves. Thus,
5-25
Similarly, the solubility product for galena, PbS, is 10-27.5.
Chemical Equilibrium, slide 26 here
What happens if galena is added to a solution already saturated with
anglesite? Let X = solubility of galena.
5-26
5-27
Chemical Equilibrium, slide 27 here
This could be solved as a quadratic equation. However, X will be less than the
solubility of pure galena. Therefore,
5-28
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5-29
5-30
Thus, in the presence of anglesite the solubility of galena is greatly reduced.
Chemical Equilibrium, slide 28 here
It is important that we check our approximation. If x = 2.4 x 10-24, is equation
5-29 true? Obviously it is, so the approximation is useful.
The galena-anglesite case is a specific example of what is usually called
the common-ion effect. When two salts, which share one ion in common, are
both present in the same solution, the common-ion effect states that the
solubility of both salts is decreased. The presence of ions different from those
furnished by the salt itself often increases the solubility of the salt. This is
simply an observation and cannot be derived from equilibrium reasoning.
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References
P. Waage, and C.M. Guldberg, Forhandlinger: Videnskabs-Selskabet I
Christiana, 35, 1864.
Konrad B. Krauskopf, Introduction to Geochemistry, Second edition, McGrawHill, New York, 1979.
Henri Louis Le Chatelier, Comptes rendus, 99, 786, 1884.
Henri Louis Le Chatelier, Annales des Mines, 13 (2), 157, 1888.
P. Waage and C.M. Guldberg, Studies Concerning Affinity, Henry I. Abrash,
Translator, Last modified September 21, 2000,
http://archive.today/1ppvd#selection-31.0-37.27, (last seen August 19,
2014).
4241LN05_PP_F16.pdf
September 1, 2016
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