Chemical Equilibrium
Chemical Equilibrium
Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
A STUDY OF CHEMICAL
EQUILIBRIUM
to award degree of
Submitted by : Shweta
Contents
Part A
1. Devotion
2. Declaration
3. Certification
4. Acknowledgment
5. About the Thesis
Part B
7. What is Equilibrium?
expressions
hydrates
reactions?
DEVOTION
The Thesis “A Study of Chemical Equilibrium” is devoted to the Global Open
University, Nagaland to award degree of Master of Philosophy (M. Phil) in
Chemistry.
Submitted by : Shweta
DECLARATION
I hereby declare that the thesis titled “A Study of Chemical Equilibrium” submitted
for the Master in Philosophy (M.Phil) is my original work and the dissertation has not
formed the basis or award of any degree, associationship, fellowship, or any other
similar title.
(Shweta)
Deponent
A Study of Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
CERTIFICATE
ACKNOWLEDGEMENT
It gives me great pleasure to present the work before the esteemed university for kind
consideration and evaluation.
I thank my all teachers including Mr. K.K. Jha who have guided me in many ways in
writing the dissertation.
I specially thank Dr. Ashok Kumar Yadav, my guide and teacher without whose
encouragement and help I might have not written this dissertation.
(Shweta)
A Study of Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
The dissertation has been prepared under the guidance of Dr. Ashok Kumar Yadav. It
is my endeavor to present the work in easy and lucid language. Main topics are kept
in question answer format for better clarity. In respect of some topics summaries are
also given just following the concerned topic. All endeavors are made to present the
work without any error. However despite the best effort some error may creep in, for
this I heartily apologize.
A Study of Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
Chemical change occurs when the atoms that make up one or more substances
rearrange themselves in such a way that new substances are formed. These substances
are the components of the chemical reaction system; those components which
decrease in quantity are called reactants, while those that increase are products.
reactants → products
Chemical change is one of the two central concepts of chemical science, the other
being structure. The very origins of Chemistry itself are rooted in the observations of
transformations such as the combustion of wood, the freezing of water, and the
winning of metals from their ores that have always been a part of human experience.
It was, after all, the quest for some kind of constancy underlying change that led the
Greek thinkers of around 200 BCE to the idea of elements and later to that of the
atom. It would take almost 2000 years for the scientific study of matter to pick up
these concepts and incorporate them into what would emerge, in the latter part of the
19th century, as a modern view of chemical change.
The first thing we need to know about a chemical reaction represented by a balanced
equation is whether it can actually take place. If the reactants and products are all
substances capable of an independent existence, then the answer is always "yes", at
least in principle. This answer must always be qualified, however, by the following
two considerations:
• Does the reaction take place to a sufficient extent to produce useful (or
even detectable) quantities of products?
A Study of Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
there are often ways of getting around both kinds of limitations, and their discovery
and practical applications constitute an important area of industrial chemistry.
What is Equilibrium?
Basically, the term refers to what we might call a "balance of forces". In the case of
mechanical equilibrium, this is its literal definition. A book sitting on a table top
remains at rest because the downward force exerted by the earth's gravity acting on
the book's mass (this is what is meant by the "weight" of the book) is exactly
balanced by the repulsive force between atoms that prevents two objects from
simultaneously occupying the same space, acting in this case between the table
surface and the book. If we pick up the book and raise it above the table top, the
additional upward force exerted by our arm destroys the state of equilibrium as the
book moves upward. If we wish to hold the book at rest above the table, we adjust the
upward force to exactly balance the weight of the book, thus restoring equilibrium.
An object is in a state of mechanical equilibrium when it is either static (motionless)
or in a state of unchanging motion. From the relation f = ma , it is apparent that if the
net force on the object is zero, its acceleration must also be zero, so if we can see that
an object is not undergoing a change in its motion, we know that it is in mechanical
equilibrium.
remove heat from our hand more slowly. Thermal equilibrium is something we often
want to avoid, or at least postpone; this is why we insulate buildings, perspire in the
summer and wear heavier clothing in the winter.
When a chemical reaction takes place in a container which prevents the entry or
escape of any of the substances involved in the reaction, the quantities of these
components change as some are consumed and others are formed. Eventually this
change will come to an end, after which the composition will remain unchanged as
long as the system remains undisturbed. The system is then said to be in its
equilibrium state, or more simply, "at equilibrium".
The direction in which we write a chemical reaction (and thus which components are
considered reactants and which are products) is arbitrary. Thus the two equations
represent the same chemical reaction system in which the roles of the components are
reversed, and both yield the same mixture of components when the change is
completed.
in the quantities of the components will occur as long as the system remains
undisturbed.
The two diagrams below show how the concentrations of the three components of this
chemical reaction change with time (under the conditions in which this particular
reaction is usually carried out, the time scale is typically 10 to 100 minutes.) Examine
the two sets of plots carefully, noting which substances have zero initial
concentrations, and are thus "products" of the reaction equations shown. Satisfy
ourself that these two sets represent the same chemical reaction system, but with the
reactions occurring in opposite directions. Most importantly, note how the final
(equilibrium) concentrations of the components are the same in the two cases.
In general, both processes (forward and reverse) can be expected to occur, resulting in
an equilibrium mixture containing finite amounts of all of the components of the
reaction system. (We use the word components when we do not wish to distinguish
between reactants and products.)
If the equilibrium state is one in which significant quantities of both reactants and
products are present (as in the hydrogen iodide example given above), then the
reaction is said to incomplete or reversible.
The latter term is preferable because it avoids confusion with "complete" in its other
sense of being finished, implying that the reaction has run its course and is now at
equilibrium.
In principle, all chemical reactions are reversible, but this reversibility may not be
observable if the fraction of products in the equilibrium mixture is very small, or if
the reverse reaction is kinetically inhibited (very slow.)
A Study of Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
Guldberg and Waage showed that the rate of the reaction in either direction is
proportional to what they called the "active masses" of the various components:
in which the proportionality constants k are called rate constants and the quantities in
square brackets represent concentrations. If we combine the two reactants A and B,
the forward reaction starts immediately; then, as the products C and D begin to build
up, the reverse process gets underway. As the reaction proceeds, the rate of the
forward reaction diminishes while that of the reverse reaction increases. Eventually
the two processes are proceeding at the same rate, and the reaction is at equilibrium:
Clearly, if we observe some change taking place— a change in color, the emission of
gas bubbles, the appearance of a precipitate, or the release of heat, we know the
reaction is not at equilibrium. However, the absence of any sign of change does not
by itself establish that the reaction is at equilibrium, which is defined above as the
lack of any tendency for change to occur; "tendency" is not a property that is directly
observable! Consider, for example, the reaction representing the synthesis of water
from its elements:
2 H2 + O2 → 2 H2O
We can store the two gaseous reactants in the same container indefinitely without any
observable change occurring. But if we create an electrical spark in the container or
introduce a flame, bang! After we pick ourselves up off the floor and remove the
shrapnel from what's left of our body, we will know very well that the system was not
initially at equilibrium! It happens that this particular reaction has a tremendous
tendency to take place, but for reasons that we will discuss in a later chapter, nothing
can happen until we "set it off" in some way- in this case by exposing the mixture to a
flame, or (in a more gentle way) by introducing a platinum wire, which acts as a
catalyst. A reaction of this kind is said to be highly favored thermodynamically,
but inhibited kinetically. The hydrogen iodide reaction, by contrast, is only
moderately favored thermodynamically (that's why it is incomplete), but its kinetics
are reasonably facile.
It is almost always the case, however, that once a reaction actually starts, it will
continue on its own until it reaches equilibrium, so if we can observe the change as it
occurs and see it slow down and stop, we can be reasonably certain that the system is
in equilibrium. This is by far the chemist's most common criterion.
A Study of Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
There is one other experimental test for equilibrium in a chemical reaction, although
it is really only applicable to the kind of reactions we described above as being
reversible. As we shall see later, the equilibrium state of a system is always sensitive
to the temperature, and often to the pressure, so any changes in these variables,
however, small, will temporarily disrupt the equilibrium, resulting in an observable
change in the composition of the system as it moves toward its new equilibrium state.
Similarly, addition or removal of one component of the reaction will affect the
amounts of all the others. If carrying out any of these operations fails to produce an
observable change, then it is likely that the reaction is kinetically inhibited and that
the system is not at equilibrium.
Make sure we thoroughly understand the following essential ideas which have been
presented above. It is especially important that we know the precise meanings of all
the highlighted terms in the context of this topic.
• Any reaction that can be represented by a balanced chemical equation can take
place, at least
• The tendency for the change to occur may be so small that the quantity of
products formed may be very low, and perhaps negligible. A reaction of this
kind is said to be thermodynamically inhibited. The tendency for chemical
change is governed solely by the properties of the reactants and products, and
can be predicted by applying the laws of thermodynamics.
• The rate at which the reaction proceeds may be very small or even zero, in
which case we say the reaction is kinetically inhibited. Reaction rates depend
on the mechanism of the reaction— that is, on what actually happens to the
atoms as reactants are transformed into products. Reaction mechanisms cannot
A Study of Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
The law of mass action states that any chemical change is a competition between a
forward reaction (left-to-right in the chemical equation) and a reverse reaction. The
rate of each of these processes is governed by the concentrations of the substances
reacting; as the reaction proceeds, these rates approach each other and at equilibrium
they become identical.
A Study of Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
equilibrium state, then the composition of the system will tend to change until that
new equilibrium state is attained. (We say "tend to change" because if the reaction is
kinetically inhibited, the change may be too slow to observe or it may never take
place.) In 1884, the French chemical engineer and teacher Henri le Châtelier (1850-
1936) showed that in every such case, the new equilibrium state is one that partially
reduces the effect of the change that brought it about.
This law is known to every Chemistry student as the le Châtelier principle. His
original formulation was somewhat complicated, but a reasonably useful paraphrase
of it reads as follows:
To see how this works (and we must do so, as this is of such fundamental importance
that we simply cannot do any meaningful Chemistry without a thorough working
understanding of this principle), look again at the hydrogen iodide dissociation
reaction
2 HI H2 + I2
principle states that the net reaction will be in a direction that tends to reduce the
effect of the added H2. This can occur if some of the H2 is consumed by reacting with
I2 to form more HI; in other words, a net reaction occurs in the reverse direction.
Chemists usually simply say that "the equilibrium shifts to the left".
To get a better idea of how this works, carefully examine the diagram below
which follows the concentrations of the three components of this reaction as they
might change in time (the time scale here will typically be about an hour):
The following table contains several examples showing how changing the quantity of
a reaction component can shift an established equilibrium.
A Study of Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
hemoglobin + O2 oxyhemoglobin
The partial pressure of O2 in the air is 0.2 atm, sufficient to allow these molecules
to be taken up by hemoglobin (the red pigment of blood) in which it becomes
loosely bound in a complex known as oxyhemoglobin. At the ends of the
capillaries which deliver the blood to the tissues, the O2 concentration is reduced
by about 50% owing to its consumption by the cells. This shifts the equilibrium to
the left, releasing the oxygen so it can diffuse into the cells.
Carbon dioxide reacts with water to form a weak acid H2CO3 which would cause
the blood pH to fall to dangerous levels if it were not promptly removed as it is
excreted by the cells. This is accomplished by combining it with carbonate ion
through the reaction
which is forced to the right by the high local CO2 concentration within the tissues.
Once the hydrogen carbonate (bicarbonate) ions reach the lung tissues where the
CO2 partial pressure is much smaller, the reaction reverses and the CO2 is
expelled.
Air that contains as little as 0.1 percent carbon monoxide can tie up about half of
the hemoglobin binding sites, reducing the amount of O2 reaching the tissues to
fatal levels. Carbon monoxide poisoning is treated by administration of pure O2
which promotes the shift of the above equilibrium to the left. This can be made
even more effective by placing the victim in a hyperbaric chamber in which the
pressure of O2 can be made greater than 1 atm.
A Study of Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
Virtually all chemical reactions are accompanied by the liberation or uptake of heat.
If we regard heat as a "reactant" or "product" in an endothermic or exothermic
reaction respectively, we can use the le Châtelier principle to predict the direction in
which an increase or decrease in temperature will shift the equilibrium state. Thus for
the oxidation of nitrogen, an endothermic process, we can write
[heat] + N2 + O2 2 NO
Nitric oxide, the product of this reaction, is a major air pollutant which initiates a
sequence of steps leading to the formation of atmospheric smog. Its formation is an
unwanted side reaction which occurs when the air (which is introduced into the
combustion chamber of an engine to supply oxygen) gets heated to a high
temperature. Designers of internal combustion engines now try, by various means, to
limit the temperature in the combustion region, or to restrict its highest-temperature
part to a small volume within the combustion chamber.
A Study of Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
We will recall that if the pressure of a gas is reduced, its volume will increase;
pressure and volume are inversely proportional. With this in mind, suppose that the
reaction
2 NO2(g) N2O4(g)
is in equilibrium at some arbitrary temperature and pressure, and that we double the
pressure, perhaps by compressing the mixture to a smaller volume. From the le
Châtelier principle we know that the equilibrium state will change to one that tends to
counteract the increase in pressure. This can occur if some of the NO2 reacts to form
more of the dinitrogen tetroxide, since two moles of gas is being removed from the
system for every mole of N2O4 formed, thereby decreasing the total volume of the
system. Thus increasing the pressure will shift this equilibrium to the right.
It is important to understand that the changing the pressure will have a significant
effect only on reactions in which there is a change in the number of moles of gas. For
the above reaction, this change
In the case of the nitrogen oxidation reaction described previously, –ng = 0 and
pressure will have no effect.
The volumes of solids and liquids are hardly affected by the pressure at all, so for
reactions that do not involve gaseous substances, the effects of pressure changes are
ordinarily negligible. Exceptions arise under conditions of very high pressure such as
exist in the interior of the Earth or near the bottom of the ocean. A good example is
A Study of Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
The skeletons of several varieties of microscopic organisms that inhabit the top of the
ocean are made of CaCO3, so there is a continual rain of this substance toward the
bottom of the ocean as these organisms die. As a consequence, the floor of the
Atlantic Ocean is covered with a blanket of calcium carbonate. This is not true for the
Pacific Ocean, which is deeper; once the skeletons fall below a certain depth, the
higher pressure causes them to dissolve. Some of the seamounts (undersea mountains)
in the Pacific extend above the solubility boundary so that their upper parts are
covered with CaCO3 sediments.
The effect of pressure on a reaction involving substances whose boiling points fall
within the range of commonly encountered temperature will be sensitive to the states
of these substances at the temperature of interest.
For reactions involving gases, only changes in the partial pressures of those gases
directly involved in the reaction are important; the presence of other gases has no
effect.
Example 1
The commercial production of hydrogen is carried out by treating natural gas with
steam at high temperatures and in the presence of a catalyst (“steam reforming of
methane”):
Given the following boiling points: CH4 (methane) = –161°C, H2O = 100°C, CH3OH
= 65°, H2 = –253°C, predict the effects of an increase in the total pressure on this
equilibrium at 50°, 75° and 120°C.
Solution: Calculate the change in the moles of gas for each process:
The le Châtelier principle tells us that in order to maximize the amount of product in
the reaction mixture, it should be carried out at high pressure and low temperature.
However, the lower the temperature, the slower the reaction (this is true of virtually
all chemical reactions.) As long as the choice had to be made between a low yield of
ammonia quickly or a high yield over a long period of time, this reaction was
infeasible economically.
Nitrogen is available for free, being the major component of air, but the strong triple
bond in N2 makes it extremely difficult to incorporate this element into species such
as NO3– and NH4+ which serve as the starting points for the wide variety of nitrogen-
containing compounds that are essential for modern industry. This conversion is
known as nitrogen fixation, and because nitrogen is an essential plant nutrient,
modern intensive agriculture is utterly dependent on huge amounts of fixed nitrogen
in the form of fertilizer. Until around 1900, the major source of fixed nitrogen was the
NaNO3 (Chile saltpeter) found in extensive deposits in South America. Several
chemical processes for obtaining nitrogen compounds were developed in the early
1900's, but they proved too inefficient to meet the increasing demand.
Although the direct synthesis of ammonia from its elements had been known for
some time, the yield of product was found to be negligible. In 1905, Fritz Haber
(1868-1934) began to study this reaction, employing the thinking initiated by le
Châtelier and others, and the newly-developing field of thermodynamics that served
as the basis of these principles. From the le Châtelier law alone, it is apparent that this
exothermic reaction is favored by low temperature and high pressure. However, it
was not as simple as that: the rate of any reaction increases with the temperature, so
A Study of Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
working with temperature alone, one has the choice between a high product yield
achieved only very slowly, or a very low yield quickly. Further, the equipment and
the high-strength alloy steels need to build it did not exist at the time. Haber solved
the first problem by developing a catalyst that would greatly speed up the reaction at
lower temperatures.
The second problem, and the development of an efficient way of producing hydrogen,
would delay the practical implementation of the process until 1913, when the first
plant based on the Haber-Bosch process (as it is more properly known, Carl Bosch
being the person who solved the major engineering problems) came into operation.
The timing could not have been better for Germany, since this country was about to
enter the First World War, and the Allies had established a naval blockade of South
America, cutting off the supply of nitrate for the German munitions industry.
Bosch's plant operated the ammonia reactor at 200 atm and 550°C. Later, when
stronger alloy steels had been developed, pressures of 800-1000 atm became
common. The source of hydrogen in modern plants is usually natural gas, which is
mostly methane:
Much has been written about some of the more controversial and tragic aspects of
Haber's life. Shown here are the major figures in the cast in the 2005 Vancouver
production of Vern Thiessen's play Einstein's Gift which centers on the conflict and
tragedy in Fritz Haber's life. His first wife, Clara, was his former student and a
chemist; Haber's decision to help Germany develop chlorine as a war gas led to her
suicide.
Summary
Make sure we thoroughly understand the following essential ideas which have been
presented above. It is especially important that we know the precise meanings of all
the highlighted terms in the context of this topic.
• A system in its equilibrium state will remain in that state indefinitely as long
as it is undisturbed. If the equilibrium is destroyed by subjecting the system to
a change of pressure, temperature, or the number of moles of a substance, then
a net reaction will tend to take place that moves the system to a new
equilibrium state. le Châtelier's principle says that this net reaction will occur
in a direction that partially offsets the change.
• The le Châtelier Principle has practical effect only for reactions in which
significant quantities of both reactants and products are present at
equilibrium— that is, for reactions that are thermodynamically reversible.
• Addition of more product substances to an equilibrium mixture will shift the
equilibrium to the left; addition of more reactant substances will shift it to the
right. These effects are easily explained in terms of competing forward- and
reverse reactions— that is, by the law of mass action.
A Study of Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
In the previous section we defined the equilibrium expression for the reaction
aA+bB→cC+dD as
In the general case in which the concentrations can have any arbitrary values
(including zero), this expression is called the equilibrium quotient and its value is
denoted by Q (or Qc if we wish to emphasize that the terms represent molar
concentrations.) If the terms correspond to equilibrium concentrations, then the above
expression is called the equilibrium constant and its value is denoted by K (or Kc .)
K is thus the special value that Q has when the reaction is at equilibrium
The value of Q in relation to K serves as an index how the composition of the reaction
system compares to that of the equilibrium state, and thus it indicates the direction in
which any net reaction must proceed. For example, if we combine the two reactants A
and B at concentrations of 1 mol L–1 each, the value of Q will be 0÷1=0. If instead
our mixture consists only of the two products C and D, Q will be indeterminately
large (1÷0). It is easy to see (by simple application of the Le Châtelier principle) that
the ratio of Q/K immediately tells us whether, and in which direction, a net reaction
will occur as the system moves toward its equilibrium state. The three possibilities are
shown in the table below.
Q/K
It is very important that we be able to work out these relations for ourself, not by
memorizing them, but from the definitions of Q and K.
The formal definitions of Q and K are quite simple, but they are of limited usefulness
unless we are able to relate them to real chemical situations. The following diagrams
illustrate the relation between Q and K from various standpoints. Take some time to
study each one carefully, making sure that we are able to relate the description to the
illustration.
Each tiny dot on the graph represents a possible combination of NO2 and N2O4
concentrations that produce a certain value of Q for the chemical reaction system
N2O4 2 NO2. (There are of course an infinite number of possible Q's of this system
within the concentration boundaries shown on the plot.) Only those dots that fall on
the red line correspond to equilibrium states of this system (those for which Q = K ).
The line itself is a plot of [NO2] = ([N2O4]K)0.5. If the system is initially in a non-
equilibrium state, its composition will change in a direction that moves it to one on
the line.
One of the simplest equilibria we can write is that between a solid and its vapor.
Using the sublimation of iodine I2(s) I2 (g) as an example, we see that the possible
equilibrium states of the system (shaded area in the diagram) are limited to those in
which at least some solid is present, but that within this region, the quantity of iodine
A Study of Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
vapor is constant (red line) as long as the temperature is unchanged. The arrow shows
the succession of states the system passes through when 0.29 mole of solid iodine is
placed in a 1-L sealed container; the unit slope of this line reflects the fact that each
mole of I2 removed from the solid ends up in the vapor.
The expressions given in the rightmost column above simply reflect the fact that the
rate at which a substance undergoes change should be proportional to its
concentration; this is just another statement of the Law of Mass Action. The
proportionality constants kf and kr are the forward and reverse rate constants. If we
start with substance A alone, the absence of B means that the forward reaction alone
is proceeding. Then, as the concentration of B begins to build up, the reverse reaction
comes into operation, the rate of the forward reaction diminishes due to the reduction
in the concentration of B. At some point these two processes will come into exact
balance so that the forward and reverse rates are the same, at which point we can
write
thus showing that the equilibrium constant can in a sense be regarded as the resultant
of the two opposing rate constants. If the rate constant of the forward reaction
exceeds that of the reverse step, then the equilibrium state will be one in which the
product dominates. (Note carefully that although the two rate constants will generally
be different, the forward and reverse rates themselves will always be identical at
equilibrium.)
The preceding paragraph shows how the concept of the equilibrium constant follows
from the Law of Mass Action, but it is not a proper derivation of the formula for
equilibrium constants in general, which must be done through the laws of
thermodynamics.
The single most important idea for we to carry along with we from this section is that
equilibrium is a dynamic process in which the forward and reverse reactions are
continually opposing each other in a dead heat.
To see if we really understand this, try explaining to ourselves how the Le Châtelier
Principle as it applies to concentrations of reaction components follows directly from
the idea of opposing reaction steps.
Although it is the ratio of kf/kr that determines K, the magnitudes of these rate
constants also make a difference; if they are small (or as is often the case, zero) then
the reaction is kinetically inhibited and equilibrium will be achieved slowly or not at
all. When their values are large, equilibrium is achieved quickly; the equilibrium is
said to be facile and the reaction is fast. Sometimes a very slow equilibrium can be
made more facile in the presence of a suitable catalyst.
A Study of Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
Summary
Make sure we thoroughly understand the following essential ideas which have been
presented above. It is especially important that we know the precise meanings of all
the highlighted terms in the context of this topic.
Most of the equilibria we deal with in this course occur in liquid solutions and
gaseous mixtures. We can express Kc values in terms of moles per liter for both, but
when dealing with gases it is often more convenient to use partial pressures. These
two measures of concentration are of course directly proportional:
All of these forms of the equilibrium constant are only approximately correct,
working best at low concentrations or pressures. The only equilibrium constant that is
truly “constant” (except that it still varies with the temperature!) is expressed in terms
of activities, which we can think of as “effective concentrations” that allow for
interactions between molecules. In practice, this distinction only becomes important
A Study of Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
for equilibria involving gases at very high pressures (such as are often encountered in
chemical engineering) and in ionic solutions more concentrated than about 0.001 M.
We will not deal much with activities in this course.
For a reaction such as CO2(g) + OH–(aq) → HCO3–(aq) that involves both gaseous
and dissolved components, a “hybrid” equilibrium constant is commonly used:
How can the concentration of a reactant or product not change when a reaction
involving that substance takes place? There are two general cases to consider.
This happens all the time in acid-base chemistry. Thus for the hydrolysis of the
cyanide- ion
CN–+ H2O → HCN + OH–, we write
A Study of Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
in which no [H2O] term appears. The justification for this omission is that water is
both the solvent and reactant, but only the tiny portion that acts as a reactant would
ordinarily go in the equilibrium expression. The amount of water consumed in the
reaction is so minute (because K is very small) that any change in the concentration of
H2O from that of pure water (55.6 mol L–1) will be negligible.
Be careful about throwing away H2O whenever we see it. In the etherification
reaction
This is most frequently seen in solubility equilibria, but there are many other
reactions in which solids are directly involved:
These are heterogeneous reactions (meaning reactions in which some components are
in different phases), and the argument here is that concentration is only meaningful
when applied to a substance within a single phase. Thus the term [CaF2] would refer
to the “concentration of calcium fluoride within the solid CaF2", which is a constant
A Study of Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
depending on the molar mass of CaF2 and the density of that solid. The
concentrations of the two ions will be independent of the quantity of solid CaF2 in
contact with the water; in other words, the system can be in equilibrium as long as
any CaF2 at all is present.
Throwing out the constant-concentration terms can lead to some rather sparse-looking
equilibrium expressions. For example, the equilibrium expression for each of the
processes shown in the following table consists solely of a single term involving the
partial pressure of a gas:
The last two processes represent changes of state or phase equilibria which can be
treated exactly the same as chemical reactions. In each of the heterogeneous
processes shown in the table, the reactants and products can be in equilibrium (that is,
permanently coexist) only when the partial pressure of the gaseous product has the
value consistent with the indicated Kp. Bear in mind also that these Kp s all increase
with the temperature.
A Study of Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
Example
What are the values of Kp for the equilibrium between liquid water and its vapor at
25°C, 100°C, and 120°C? The vapor pressure of water at these three temperatures is
23.8 torr, 760 torr (1 atm), and 1489 torr, respectively.
Comment: These vapor pressures are the partial pressures of water vapor in
equilibrium with the liquid, so they are identical with the Kp's when expressed in units
of atmospheres.
Solution:
Our ability to interpret the numerical value of a quantity in terms of what it means in
a practical sense is an essential part of developing a working understanding of
Chemistry. This is particularly the case for equilibrium constants, whose values span
the entire range of the positive numbers. Although there is no explicit rule, for most
practical purposes we can say that equilibrium constants within the range of roughly
A Study of Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
0.01 to 100 indicate that a chemically significant amount of all components of the
reaction system will be present in an equilibrium mixture and that the reaction will be
incomplete or “reversible”. As an equilibrium constant approaches the limits of zero
or infinity, the reaction can be increasingly characterized as a one-way process; we
say it is “complete” or “irreversible”. The latter term must of course not be taken
literally; the Le Châtelier principle still applies (especially insofar as temperature is
concerned), but addition or removal of reactants or products will have less effect.
Although it is by no means a general rule, it frequently happens that reactions having
very large or very small equilibrium constants are kinetically hindered, often to the
extent that the reaction essentially does not take place.
The examples in the following table are intended to show that numbers (values of K),
no matter how dull they may look, do have practical consequences!
reaction K remarks
5 10–31 at These two very different values of K illustrate very
N2(g) + O2(g) 25°C, nicely why reducing combustion-chamber
→ 2 NO(g) 0.0013 at temperatures in automobile engines is
2100°C environmentally friendly.
3 H2(g) +
7 105 at 25°C, See the discussion of this reaction in the section on
N2(g) → 2
56 at 1300°C the Haber process.
NH3(g)
10–36 at 25°C, Dissociation of any stable molecule into its atoms
H2(g) → 2
6 10–5 at is endothermic. This means that all molecules will
H(g)
5000° decompose at sufficiently high temperatures.
H2O(g) →
8 10–41 at we won’t find water a very good source of oxygen
H2(g) + ½
25°C gas at ordinary temperatures!
O2(g)
This tells us that acetic acid has a great tendency to
CH3COOH(l) decompose to carbon, but nobody has ever found
Kc = 1013 at
→ 2 H2O(l) + graphite (or diamonds!) forming in a bottle of
25°C
2 C(s) vinegar. A good example of a super kinetically-
hindered reaction!
A Study of Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
so Kp and Qp for this process would appear to have units of atm–1, and Kc and Qc
would be expressed in mol–2 L2. And yet these quantities are often represented as
being dimensionless. Which is correct? The answer is that both forms are acceptable.
There are some situations (which we will encounter later) in which K’s must be
considered dimensionless, but in simply quoting the value of an equilibrium constant
it is permissible to include the units, and this may even be useful in order to remove
any doubt about the units of the individual terms in equilibrium expressions
containing both pressure and concentration terms. In carrying out our own
calculations, however, there is rarely any real need to show the units.
concentration or pressure terms that go into them are really ratios having the forms(n
mol L–1)/(1 mol L–1) or (n atm)/(1 atm) in which the unit quantity in the denominator
refers to the standard state of the substance; thus the units always cancel out. For
substances that are liquids or solids, the standard state is just the concentration of the
substance within the liquid or solid, so for something like CaF2(s), the term going into
the equilibrium expression is [CaF2]/[CaF2] which cancels to unity; this is the reason
A Study of Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
we don’t need to include terms for solid or liquid phases in equilibrium expressions.
The subject of standard states would take us somewhat beyond where we need to be
at this point in the course, so we will simply say that the concept is made necessary
by the fact that energy, which ultimately governs chemical change, is always relative
to some arbitrarily defined zero value which, for chemical substances, is the standard
state.
Here are some of the possibilities for the reaction involving the equilibrium between
gaseous water and its elements:
baking powder.)
Many common inorganic salts form solids which incorporate water molecules into
their crystal structures. These water molecules are usually held rather loosely and can
escape as water vapor. Copper(II) sulfate, for example forms a pentahydrate in which
four of the water molecules are coordinated to the Cu2+ ion while the fifth is
hydrogen-bonded to SO42–. This latter water is more tightly bound, so that the
pentahydrate loses water in two stages on heating:
These dehydration steps are carried out at the temperatures indicated above, but at
any temperature, some moisture can escape from a hydrate. For the complete
dehydration of the pentahydrate we can define an equilibrium constant
The vapor pressure of the hydrate (for this reaction) is the partial pressure of water
vapor at which the two solids can coexist indefinitely; its value is Kp1/5 atm. If a
hydrate is exposed to air in which the partial pressure of water vapor is less than its
vapor pressure, the reaction will proceed to the right and the hydrate will lose
moisture. Vapor pressures always increase with temperature, so any of these
compounds can be dehydrated by heating.
Loss of water usually causes a breakdown in the structure of the crystal; this is
commonly seen with sodium sulfate, whose vapor pressure is sufficiently large that it
can exceed the partial pressure of water vapor in the air when the relative humidity is
low. What one sees is that the well-formed crystals of the decahydrate undergo
deterioration into a powdery form, a phenomenon known as efflorescence. When a
solid is able to take up moisture from the air, it is described as hygroscopic. A small
number of anhydrous solids that have low vapor pressures not only take up
atmospheric moisture on even the driest of days, but will become wet as water
molecules are adsorbed onto their surfaces; this is most commonly observed with
sodium hydroxide and calcium chloride. With these solids, the concentrated solution
that results continues to draw in water from the air so that the entire crystal eventually
dissolves into a puddle of its own making; solids exhibiting this behavior are said to
be deliquescent.
Example
A Study of Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
At what relative humidity will copper sulfate pentahydrate lose its waters of hydration
when the air temperature is 30°C? What is Kp for this process at this temperature?
Solution: From the table above, we see that the vapor pressure of the hydrate is 12.5
torr, which corresponds to a relative humidity (we remember what this is, don’t we?)
of 12.5/31.6 = 0.40 or 40%. This is the humidity that will be maintained if the hydrate
is placed in a closed container of dry air
For this hydrate, Kp = P(H2O)0.5, so the partial pressure of water vapor that will be in
equilibrium with the hydrate and the dehydrated solid (remember that both solids
must be present to have equilibrium!), expressed in atmospheres, will be (12.5/760)5
= 1.20E-9.
One of the first hydrates investigated in detail was calcium sulfate hemihydrate
(CaSO4·½ H2O) which LeChâtelier (he of the “principle”) showed to be the hardened
form of CaSO4 known as plaster of Paris. Anhydrous CaSO4 forms compact,
powdery crystals, whereas the elongated crystals of the hydrate bind themselves into
a cement-like mass.
A Study of Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
Example
Calculate the value of K for the reaction CaCO3(s) + H+(aq) → Ca2+(aq) + HCO3–(aq)
Solution: The net reaction is the sum of reaction 1 and the reverse of reaction 2:
2+ 2– –6.3
CaCO3(s) → Ca (aq) + CO3 (aq) K1 = 10
+ 2– – –(–10.3)
H (aq) + CO3 (aq) → HCO3 (aq) K–2 = 10
+ 2+ – (-8.4+10.3) +1.9
CaCO3(s) + H (aq) → Ca (aq) + HCO3 (aq) K = K1/K2 = 10 = 10
The individual coupled reactions do not need to actually take place in the overall
process; very frequently we use their equilibrium constants in order to calculate the
equilibrium constant of another reaction.
Problem Example 4
The synthesis of HBr from hydrogen and liquid bromine has an equilibrium constant
Kp = 4.5¥1015 at 25°C. Given that the vapor pressure of liquid bromine is 0.28 atm,
find Kp for the homogeneous gas-phase reaction at the same temperature.
Solution: The net reaction we seek is the sum of the heterogeneous synthesis of HBr
and the reverse of the vaporization of liquid bromine:
Summary
Example
In an experiment carried out by Taylor and Krist (J. Am. Chem. Soc. 1941: 1377),
hydrogen iodide was found to be 22.3% dissociated at 730.8°K.
Calculate Kc for 2 HI(g) H2(g) + I2(g).
Solution:
No explicit molar concentrations are given, but we do know that for every n moles of
HI, 0.223n moles of each product is formed and (1–0.233)n = 0.777n moles of HI
remains. For simplicity, we assume that n=1 and that the reaction is carried out in a
1.00-L vessel, so that we can substitute the required concentration terms directly into
the equilibrium expression.
A Study of Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
Example
Ordinary white phosphorus, P4, forms a vapor which dissociates into diatomic
molecules at high temperatures:
P4(g) 2 P2(g)
A sample of white phosphorus, when heated to 1000°C, formed a vapor having a total
pressure of 0.20 atm and a density of 0.152 g L–1. Use this information to evaluate the
equilibrium constant Kp for this reaction.
Solution: Before worrying about what the density of the gas mixture has to do with
Kp , start out in the usual way by laying out the information required to express Kp in
terms of an unknown x.
P4 2 P2
Since K is independent of the number of
initial moles: 1 1-x
moles, assume the simplest case.
moles at
1-x 2x x is the fraction of P4 that dissociates.
equilibrium:
The denominator is the total number of
eq. mole
moles:
fractions:
(1-x) + 2x = 1+x.
eq. partial Partial pressure is the mole fraction times
pressures: the total pressure.
Now we need to find the dissociation fraction x of P4, and at this point we hope we
remember those gas laws that we were told we would be needing later in the course!
The density of a gas is directly proportional to its molecular weight, so we need to
calculate the densities of pure P4 and pure P2 vapors under the conditions of the
experiment. One of these densities will be greater than 0.152 gL–1 and the other will
be smaller; all we need to do is to find where the measured density falls in between
the two limits, and wef will have the dissociation fraction.
The molecular weight of phosphorus is 31.97, giving a molar mass of 127.9 g for P4.
This mass must be divided by the volume to find the density; assuming ideal gas
behavior, the volume of 127.9 g (1 mole) of P4 is given by RT/P, which works out to
522 L (remember to use the absolute temperature here.) The density of pure P4 vapor
under the conditions of the experiment is then
The density of pure P2 would be half this, or 0.122 g L–1. The difference between
these two limiting densities is 0.123 g L–1, and the difference between the density of
pure P4 and that of the equilibrium mixture is (.245 –.152) g L–1 or 0.093 g L–1. The
ratio 0.093 0.123 = 0.76 is therefore the fraction of P4 that remains and its fractional
dissociation is (1 – 0.76) = 0.24. Substituting into the equilibrium expression above
gives Kp = 1.2.
A Study of Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
This is by far the most common kind of equilibrium problem we will encounter:
starting with an arbitrary number of moles of each component, how many moles of
each will be present when the system comes to equilibrium? The principal source of
confusion and error for beginners at this relates to the need to determine the values of
several unknowns (a concentration or pressure for each component) from a single
equation, the equilibrium expression. The key to this is to make use of the
stoichiometric relationships between the various components, which usually allow us
to express the equilibrium composition in terms of a single variable. The easiest and
most error-free way of doing this is adopt a systematic approach in which we create
and fill in a small table as shown in the following problem example. We then
substitute the equilibrium values into the equilibrium constant expression, and solve it
for the unknown. This very often involves solving a quadratic or higher-order
equation. Quadratics can of course be solved by using the familiar quadratic formula,
but it is often easier to use an algebraic or graphical approximation, and for higher-
order equations this is the only practical approach. There is almost never any need to
get an exact answer, since the equilibrium constants we start with are rarely known all
that precisely.
Example
Phosgene (COCl2) is a poisonous gas that dissociates at high temperature into two
other poisonous gases, carbon monoxide and chlorine. The equilibrium constant Kp =
0.0041 at 600°K. Find the equilibrium composition of the system after 0.124 atm of
COCl2 is allowed to reach equilibrium at this temperature.
A Study of Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
COCl2 CO Cl2
initial pressures: 0.124 atm 0 0
change: –x +x +x
equilibrium pressures: 0.124 – x x x
which gives x = 0.0225. To see how good this is, substitute this value of x into the
denominator of the original equation and solve again:
This time, solving for x gives 0.0204. Iterating once more, we get
A Study of Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
and x = 0.0206 which is sufficiently close to the previous to be considered the final
result. The final partial pressures are then 0.104 atm for COCl2, and 0.0206 atm each
for CO and Cl2.
Note: using the quadratic formula to find the exact solution yields the two roots –
0.0247 (which we ignore) and 0.0206, which show that our approximation is quite
good.
Problem Example 4
What will be the partial pressures of all three components if 0.200 mole of PCl5 and
3.00 moles of PCl3 are combined and brought to equilibrium at this temperature and
at a total pressure of 1.00 atm?
Solution:
The partial pressures in the bottom row were found by multiplying the mole fraction
of each gas by the total pressure: Pi = XiPt. The term in the denominator of each mole
fraction is the total number of moles of gas present at equilibrium: (0.200 – x) + (3.00
+ x) + x = 3.20 + x.
Substituting the equilibrium partial pressures into the equilibrium expression, we
have
Substitution of this root into the expressions for the equilibrium partial pressures in
the table yields the following values: P(PCl5) = 0.012 atm, P(PCl3) = 0.94 atm, P(Cl2)
= 0.047 atm.
In the section that introduced the LeChâtelier principle, it was mentioned that diluting
a weak acid such as acetic acid CH3COOH (“HAc”) will shift the dissociation
equilibrium to the right:
A Study of Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
Thus a 0.10M solution of acetic acid is 1.3% ionized, while in a 0.01M solution, 4.3%
of the HAc molecules will be dissociated. This is because as the solution becomes
more dilute, the product [H3O+][Ac–] decreases more rapidly than does the [HAc]
term. At the same time the concentration of H2O becomes greater, but because it is so
large to start with (about 55.5M), any effect this might have is negligible, which is
why no [H2O] term appears in the equilibrium expression.
Example
(Sucrose is ordinary table sugar.) To what volume should a solution containing 0.050
mol of each monosaccharide be diluted in order to bring about 5% conversion to
sucrose phosphate?
Solution:
Substituting into the expression for Kc in which the solution volume is the unknown,
we have
It often happens that two immiscible liquid phases are in contact, one of which
contains a solute. How will the solute tend to distribute itself between the two phases?
One’s first thought might be that some of the solute will migrate from one phase into
the other until it is distributed equally between the two phases, since this would
correspond to the maximum dispersion (randomness) of the solute. This, however,
does not take into the account the differing solubilities the solute might have in the
two liquids; if such a difference does exist, the solute will preferentially migrate into
the phase in which it is more soluble. For a solute S distributed between two phases a
and b the process Sa = Sb is defined by the distribution law
in which Ka,b is the distribution ratio (also called the distribution coefficient) and [S]i
is the solubility of the solute in the phase.
A Study of Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
This separatory funnel is used to extract a substance from one liquid into another in
which it is more soluble. The two immiscible liquids are poured into the funnel
through the opening at the top. The funnel is then shaken to bring the two phases into
intimate contact, and then set aside to allow the two liquids to separate into layers,
A Study of Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
which are then separated by allowing the more dense liquid to exit through the
stopcock at the bottom
Example
The distribution ratio for iodine between water and carbon disulfide is 650. Calculate
the concentration of I2 remaining in the aqueous phase after 50.0 mL of 0.10M I2 in
water is shaken with 10.0 mL of CS2.
Solution:
Let m1 and m2 represent the numbers of millimoles of solute in the water and CS2
layers, respectively. Kd can then be written as (m2/10 mL) ÷ (m1/50 mL) = 650. The
number of moles of solute is (50 mL) × (0.10 mmol mL–1) = 5.00 mmol, and mass
conservation requires that m1 + m2 = 5.00 mmol, so m2 = (5.00 – m1) mmol and we
now have only the single unknown m1. The equilibrium constant then becomes
Simplifying and solving for m1 yields (0.50 – 0.1)m1 / (0.02 m1) = 650, with m1 =
0.0382 mmol. The concentration of solute in the water layer is (0.0382 mmol) /
(50 mL) = 0.000763 M, showing that almost all of the iodine has moved into the CS2
layer.
A Study of Chemical Equilibrium
(By Shweta)
Mphil/4148/CHE/2008J
Summary
The six Problem Examples presented above were carefully selected to span the range
of problem types that students enrolled in first-year college chemistry courses are
expected to be able to deal with. If we are able to reproduce these solutions on our
own, we should be well prepared for an examination on this topic.
• The first step in the solution of all but the simplest equilibrium problems is to
sketch out a table showing for each component the initial concentration or
pressure, the change in this quantity (for example, +2x), and the equilibrium
values (for example, .0036 + 2x). In doing so, the sequence of calculations
required to get to the answer usually becomes apparent.
• Equilibrium calculations often involve quadratic- or higher-order equations.
Because concentrations, pressures, and equilibrium constants are seldom
known to a precision of more than a few significant figures, there is no need to
seek exact solutions. Iterative approximations or use of a graphical calculator
are adequate and convenient.
• Phase distribution equilibria play an important role in chemical separation
processes on both laboratory and industrial scales. They are also involved in
the movement of chemicals between different parts of the environment, and in
the bioconcentration of pollutants in the food chain.