Chemical equilibrium
Introduction
The subject of kinetics is normally divorced from chemical equilibrium. This is
true especially when one is considering energies. As you have seen before, the
difference in energies of the reactants and products is independent of the
reaction mechanism. In addition, you will see later that although catalysts affect
rates of reactions, they do not change chemical equilibrium. Thus, it is
appropriate to separate kinetics from thermodynamics. However, completely
divorcing kinetics from thermodynamics is not entirely correct. Principles
governing chemical equilibrium are related in one way or another to dynamics.
One should not forget that a simple dynamic equilibrium consists of two
opposing reactions occurring at the same rate. Kinetics is therefore central to
an understanding of chemical equilibrium.
In our previous discussions, we have examined the factors which govern how
fast reactions occur. For a reaction to occur, molecules need to collide, the
energy resulting from these collisions provide the energy required to break bonds
in the reactant molecules. Using our imagination, we can envision how the
energy changes as the reaction progresses. Chemical reactions occur when
molecules collide in an appropriate orientation and with sufficient energy to break
bonds. We can make a hypothesis based on this observation: The activation
energy should be related to the dissociation energy of the bonds which are about
to be broken. Energies of activation can be experimentally determined by
measuring rates at various temperatures and drawing an Arrhenius plot.
Representative data are shown in the following:
Reaction
Activation Energy
(Kcal/mole)
Bond Strengths
(Kcal/mole)
H2 +Cl2 --> 2HCl
60
H2=104, Cl2=58
2HI --> H2 + I2
44
HI=72
2NO + Br2 --> 2NOBr
1.3
NO=147, Br2=45
CN- + CH3I --> CH3CN + I- 20
CH3-I=47
Upon examination of the above data, it is clear that only the first one reveals a
relationship between the dissociation energy and the activation energy. Here the
activation energy is 60 Kcal/mole which is very close to the bond energy of a
chlorine molecule(Cl2), 58 Kcal/mole. The formation of HCl from the elements is,
in fact, a photochemical reaction where a photon is first absorbed. The energy of
this photon is used specifically to break the Cl-Cl bond in Cl2. From other
experiments, it is also known that the breaking up of the chlorine molecule is the
rate-determining step for this reaction. Thus, it is of no surprise that the energy of
activation is closely related to the bond energy of Cl2.
But what about the other reactions? Clearly, the activation energy is lower than
the bond energies of the reactants. Of course, reactions prefer to follow the route
which involves the least activation energy. A reaction can have a large number of
possible mechanisms, the reaction chooses the path of minimum energy
requirement. The route with the lowest energy of activation leads to faster rates
and, therefore, reactions would occur more often via this route. Since the energy
of activation is lower than the bond dissociation energy, it is reasonable to state
that while bonds are being broken, some of the bonds which need to be formed
are simultaneously being created. There is an intermediate between the
reactants and products in which bonds are partially broken and partially formed.
This is illustrated in the following diagram:
Thus, while the H-I bond is partially broken, H-H bond and I-I bonds have been
partially formed. Having an intermediate allows for a lower activation energy,
certainly lower than the route in which the HI bonds are fully broken first.
One thing should be very clear in the diagram shown above:
The mechanism does not alter the overall change in energy.
The overall change in energy is 4 Kcal/mole no matter what the reaction
mechanism is. This brings us to the next topic in this course, another visit in the
area of thermodynamics (Do not forget kinetics, however. In fact, we will use
kinetics in our discussion of chemical equilibrium).
In the reaction: 2HI --> H2 + I2, the energy of activation is apparently 44
Kcal/mole. This reaction occurs when 2HI molecules collide with sufficient
energy. Once the 44 Kcal/mole barrier is surmounted, the intermediate can easily
transform into the final products, H2 and I2. Consider the reverse direction. As
soon as H2 and I2 molecules are formed, the possibility of an iodine molecule
colliding against a hydrogen molecule increases. And if they do collide with
sufficient energy (in this case, 40 Kcal/mole, even less than the previous
condition), the intermediate can be formed once more and the reverse reaction
can occur, forming 2 HI molecules.
Here are some experimental data obtained at 698.6 K:
Initial Concentrations
Final Concentrations
(10-3 moles/L)
(10-3 moles/L)
Expt.
H2
I2
HI
H2
I2
HI
1
10.6673 11.9652
0
1.8313 3.1292
17.671
2
10.6673 10.7610
0
2.2423 2.3360
16.850
3
11.3540 9.0440
0
3.5600 1.2500
15.588
4
8.6737
4.8468
5.326 4.5647 0.7378
13.544
5
0
0
10.692 1.1409 1.1409
8.410
6
0
0
4.646 0.4953 0.4953
3.655
This table clearly shows that reactions can be reversible. In fact, all reactions are
reversible to some extent. Some reactions seem to be complete only because
our detection methods are sometimes not infinitely sensitive. In fact, for
preparations to be nearly complete, removal of the product from the reaction
mixture is necessary. Of course, if the products are removed as they are formed,
then the reverse reaction becomes impossible.
In the discussion below, we are looking at the reaction in the following direction:
H2(g) + I2(g) <==> 2HI (g)
The final concentrations listed in the above table no longer change with time. In
experiment 1, no HI is present initially. Collisions between H2 and I2 result in the
formation of HI molecules. As the reaction proceeds, the amounts of H2 and I2
are decreasing. Thus, collisions between these two molecules will become less
frequent. The rate of formation of HI should therefore decrease with time. On the
other hand, as HI molecules are formed, the possibility of an HI molecule
colliding with another HI molecule increases. This collision can then lead to the
reverse reaction. As more HI is produced, the rate of the reverse reaction
increases. After some time the rate of the forward reaction has decreased
enough and the rate of the reverse reaction has increased enough such that both
rates are now equal. Chemical equilibrium is achieved. The system will no longer
undergo a net change, however, opposing reactions are still occurring but at the
same rate. For this reason, no noticeable overall change is observed.
Thus, at chemical equilibrium, the following is true:
rate of forward reaction = rate of reverse reaction
kf [H2][I2] = kr [HI]2
Rearranging the above equation so that the constants are on one side and the
concentrations are on the other:
kf / kr = [HI]2/[H2][I2]
Since both kf and kr are constants, their ratio should also be a constant. We will
now introduce a new symbol representing this ratio:
K = [HI]2/[H2][I2]
This equation relates the concentration of the products to the concentration of the
reactants when dynamic equilibrium is achieved. Examining the previous table
indicates that this relationship is indeed obeyed:
Experiment
[HI]2/[H2][I2]
1
54.50
2
53.97
3
54.61
4
54.50
5
54.35
6
54.59
K is given the name, equilibrium constant.
Principle of Microscopic Reversibility - Chemical
equilibrium
One can determine the expression for K from the balanced chemical
equation:
aA + bB --> cC + dD
K = ([C]c[D]d)/([A]a[B]b
Unlike kinetics, the coefficients found in the balanced equation are
automatically the exponents even for reactions which involve more than
one step (non-elementary reactions).
For example a two-step mechanism for the reaction 2A + Z --> C:
(1) A + A--> B
(2) B + Z--> C
The reverse reaction will look like:
(1) C --> B + Z
(2) B --> A + A
This is the principle of microscopic reversibility: the forward and reverse of a
particular pathway are related as mirror images. Thus, each mechanistic step
must be at equilibrium when the whole system is at equilibrium:
(1) A + A<===> B
(2) B + Z <====> C
k1 [A]2 = k-1[B]
and
k2 [B][Z] = k-2 [C]
The equilibrium expressions for the two steps are as follows:
k1 / k-1 = [B]/[A]2
k2 / k-2 = [C]/([B][Z])
Multiplying the two equations yields:
(k1k2 / k-1k-2) = [C]/([A]2[Z])
or
K = [C]/([A]2[Z])
the equation one gets directly from the stoichiometry of the reaction: 2A + Z --> C
It should be clear how an understanding of kinetics helps in understanding
chemical equilibrium. Now, let us move a little bit off-tangent and begin asking
what a catalyst does to a system in chemical equilibrium. From the previous
topic, we know that catalysts provide an alternate mechanism or route with lower
activation energy. Using the Arrhenius equation, the rate constant for the forward
reaction can be expressed as follows for the catalyzed reaction:
kf = Af exp(-(Eaf - C)/RT)
where C is the amount the energy of activation for the forward reaction (Eaf) is
lowered. The rate constant of the reverse is likewise affected and the Energy of
activation (Ear) is lowered by the same amount, C. Thus, the rate constant for the
reverse reaction is given by:
kr = Ar exp(-(Ear - C)/RT)
The uncatalyzed K is given by:
kf / kr = (Af exp (-Eaf/RT) / Ar exp (-Ear/RT))
kf / kr = (Af/Ar) exp (-(Eaf - Ear)/RT)
For the catalyzed reaction, K is given by:
kf / kr = (Af exp (-(Eaf - C)/RT) / Ar exp (-(Ear - C)/RT))
which obviously yields the same equation:
kf / kr = (Af/Ar) exp (-(Eaf - Ear)/RT)
The equilibrium constant depends only on the difference between the energies of
the reactants and the products as reflected in the difference between Eaf and Ear.
If you go back to the energy diagram for the reaction, the difference between the
energy of activation of the forward and the reverse reaction is exactly equal to
the difference between the energy of the reactants and the energy of the
products. The introduction of a catalyst speeds up both forward and reverse
reactions by the same amount and, thus, has no effect on chemical
equilibrium.
The textbook devotes a great deal of discussion regarding the Haber process:
This process involves the production of ammonia from the elements, nitrogen
and hydrogen:
N2(g) + 3H2(g) <----> 2NH3(g)
Epilogue: Remarks on Science and the Social Order (taken from Leonard K.
Nash (Emeritus Professor of Chemistry at Harvard University) Elements of
Chemical Thermodynamics, Mass.: Addison-Wesley Publishing Company, Inc.,
1962, pp.89-90)
Haber began his studies of the synthesis of ammonia in 1904. On the basis of
calculations like that given above (equations which make use of constant of
equilibrium expressions), confirmed by a great many experiments, he came to
conceive the process for which he took the key patent in 1908. This patent
describes a continuous recirculation of the process gas --- using heat exchangers
but maintaining throughout the same high pressure --- in such fashion that even
at low equilibrium concentration of ammonia is removed continuously, by
condensation, while the remaining process gas is reheated and passed again
over the catalyst. By 1909 Haber was able to demonstrate, to a representative of
the Badische Anilin und Sodafabrik, a tiny pilot system producing about 80 grams
of liquid ammonia per hour. The many remaining problems of the synthesis were
solved in the laboratories of this concern which --- five years later, in 1914 --- had
managed to bring the Haber process into quantity production.
That date is significant. All the high explosives used in the World War of 19141918?absolutely require nitric acid which, at the outbreak of the World War I, was
derived almost exclusively from Chile saltpeter, NaNO3. A country cut off from its
supply of this mineral, as Germany was by the British blockade, could, however,
still obtain nitric acid if only it could fix atmospheric nitrogen as ammonia, since
ammonia can be converted into nitric acid by a series of steps summarized in the
equation
NH3 + 2O2 --> HNO3 + H2O
The Haber process was brought into production not a minute too early for
Germany. Because of the Haber process, Germany avoided early defeat, the war
was prolonged. Is it a blessing so to be saved? Prolongation of the war so
depleted Germany, and so embittered her foes, that the postwar situation may be
thought to have made inevitable the rise of a Hitler. If this be so, the Haber
process was no blessing for Germany, and certainly no blessing for Haber, who,
with the rise of Hitler, became "the Jew Haber" driven to his death in exile.
In the late 19th century there were prophets of disaster who foretold the early
doom of urban civilization, an early resurgence of chronic famine, consequent to
the ultimately inevitable exhaustion of the supply of Chile saltpeter, the major
source of fixed nitrogen for agricultural fertilizer. The Haber process frees us
once and for all from this threat?
?The scientist makes an ethical judgment, and assumes a moral responsibility,
when he elects to participate in the technological exploitation of science for
destructive purposes. "Social demand" may applaud, but cannot justify, such a
decision --- any more than it can the decision of the smith who turns iron into
swords rather than ploughshares. But science, scientific knowledge, is ethically
as neutral as the iron: "evil"; only when men forge it as a sword; "good" when
beaten into a ploughshare. Conceivably there is some knowledge that can lead
only to "evil"; certainly there is none that can lead only to "good"? One can never
deny the possibility of what Sophocles though a certainty, "No great thing ever
enters human life without a curse." Unlike the Greeks, however, we have an
abiding faith in the possibility of ameliorating the human condition which
emboldens us to push on in science, and elsewhere, with Whitehead's conviction
that: "Panic of error is the death of progress."
Le Chatelier's Principle - Chemical equilibrium
(Historical) Law of Mass Action
In 1864, Cato Maximilian Guldberg and Peter Waage postulated the expression
for the equilibrium constant.
aA + bB --> cC + dD
K = ([C]c[D]d)/([A]a[B]b
The system will try to obey the above equation at all times. This was illustrated in
the last lecture in the case of HI being formed from hydrogen and iodine and
vice-versa.
Important considerations:
(1) The equilibrium constant is normally dimensionless eventhough its evaluation
may produce units.
(2) The notation Kc is normally used to denote that the equilibrium constant refers
to the expression in which the amounts of materials are expressed in molar
concentrations.
(3) For a particular reaction, the value of the equilibrium constant varies with
temperature. Remember,
K = kf / kr = (Af/Ar) exp (-(Eaf - Ear)/RT)
(4) The notation Kp is used when the amounts of the materials are expressed as
gas pressures.
(5) For the value of an equilibrium constant to be meaningful, we must specify
how the equilibrium reaction is written, i.e., which species are written on the
product side and which are written on the reactant side. The equilibrium
expression for a reaction written in one direction is the reciprocal of the one for
the reaction written in the reverse direction.
Quantitative considerations:
When K >> 1, formation of products is favored
When K << 1, reactants are favored
Relating Kcand Kp:
The ideal gas law: PV = nRT
rearranges into: P = (n/V)RT
for gases (n/V) will be the molar concentration, [C]:
Thus, P = [C]RT
and Kp = Kc(RT)(ngas in products-ngas in reactants)
Question: What can affect chemical equilibirium?
Le Chatelier's Principle:
NaCl(s) <==> Na+(aq) + Cl-(aq)
In the laboratory, a saturated aqueous solution of NaCl was prepared. A
saturated aqueous solution means that the maximum
amount of NaCl which can be dissolved in a given amount of water is present.
Additional crystals of NaCl when added to a
saturated solution will not dissolve. The constant of equilibrium expression for the
dissolution of NaCl in water is given by:
Kc = [Na+][Cl-]/[NaCl(s)]
The above equation contains the concentration of NaCl in solid NaCl. This
concentration is related to the density of solid NaCl
and its molar mass. Since the density of NaCl or of any pure solid or pure liquid
does not vary with the extent of a reaction, the
concentration of any pure solid or liquid can be regarded as a constant
and, thus, can be further absorbed into the equilibrium constant:
Kc' = [Na+][Cl-]
where Kc' = Kc[NaCl(s)]. For dissolution processes, this constant is given a
special name: Constant of Solubility Product,
and the symbol Ksp.
Ksp of NaCl = [Na+][Cl-]
The constant of solubility product is, in fact, another statement of solubility. The
higher the Ksp value is the higher the solubility.
Ksp gives the highest concentration for dissolved species before crystallization or
precipitation will occur. If the product
[Na+][Cl-] is greater than Ksp of NaCl, NaCl crystals will be formed. Thus, in a
saturated solution Ksp = [Na+][Cl-]. The
dynamic equilibrium can be easily disturbed by changing any one of these
concentrations: either [Na+] or [Cl-].
How does the system respond to a disturbance, the answer is given by Le
Chatelier's Principle:
If a system at equilibrium is disturbed by a change in temperature,
pressure, or the concentration of one of the
components, the system will shift its equilibrium position so as to
counteract the effect of the disturbance.
-Le Chatelier's Principle (1888)
"It is known that in the blast furnace the reduction of iron oxide is produced by
carbon monoxide, according to the reaction:
Fe2O3(s) + 3CO(g) <==> 2Fe(s) + 3CO2(g)
but the gas leaving the chimney contains a considerable proportion of carbon
monoxide,?. Because this incomplete reaction was thought to be due to an
insufficiently prolonged contact between carbon monoxide and the iron ore
(confusing a problem with equilibrium with that of kinetics), the dimensions of the
furnaces have been increased. In England they have been made as high as thirty
meters. But the proportion of carbon monoxide escaping has not diminished, thus
demonstrating, by an experiment costing several hundred thousand francs, that
the reduction of iron oxide by carbon monoxide is a limited reaction.
Acquaintance with the laws of equilibrium would have permitted the same
conclusion to be reached more rapidly and far more economically."
Observations from the laboratory:
(1) Addition of 1 mL concentrated HCl (12M) to 4 mL of a saturated solution of
NaCl causes NaCl to crystallize.
Concentrated HCl contains water and its addition increases the volume of the
solution. However, one should keep in mind that it is 12M in Cl-. From the
experiment you performed before Le Chatelier's experiment, you were told that a
saturated solution of NaCl is only 5.4 M which means that, at the most 5.4M of
[Na+] and 5.4M of [Cl-] can exist in solution. Addition of HCl results in a dramatic
increase in [Cl-] and disturbs the equilibrium. Upon addition of concentrated HCl,
the reaction quotient [Na+][Cl-] becomes greater than Ksp of NaCl. Crystallization
of NaCl thus occur to relieve this stress.
(2) Addition of ethanol (a compound very much capable of hydrogen-bonding
with water) reduces the amount of available
water molecules needed for separating the Na+ and Cl- ions. Thus, one can look
at this situation as an increase in the effective
concentration of the dissolved ions (due to less solvent available) and, likewise,
causes precipitation of NaCl.
(3) Adding solid NaNO3 to a saturated solution of NaCl also causes crystallization
of NaCl. Increasing [Na+] definitely
disturbs the equilibrium and the system readjusts by crystallizing NaCl out until
the product [Na+][Cl-] becomes equal again to
Ksp.
(4) Adding concentrated NaNO3(aq) does not cause crystallization of NaCl. Once
again, from the experiment you performed
before this one, you were told that the concentration of a saturated NaNO 3
solution is about 7.6 M. The saturated NaCl is
about 5.4M. Mixing equal volumes of these two solutions does not increase the
product [Na+][Cl-]. Since the NaNO3 solution
is rich in Na+, [Na+] in the original NaCl solution does go up (to about 6.5M).
However, to precipitate NaCl, it is not just
[Na+] that determines whether NaCl will precipitate or not, the [Cl-] concentration
also requires our attention. Since the
concentrated solution of NaNO3 is devoid of Cl- ions, its addition actually reduces
[Cl-] since, remember, we are also adding
water in the process. The [Cl-] is essentially reduced to half of its original value, it
is now 2.7 M. Clearly, the product
[Na+][Cl-](6.5 x 2.7) is less than the original value for the saturated solution (5.4 x
5.4). Contrast this to the result when one
adds concentrated HCl, the previous experiment. Adding 1 mL of 12M HCl
effectively increases [Cl-] to 6.7 while
reducing [Na+] to about 4.7 (Concentrated HCl is about 40% HCl, so adding 1 mL
is essentially adding only 0.6 mL
of water). The product 6.7 x 4.7 is greater than 5.4 x 5.4, thus, resulting in
crystallization.
Change in reactant or product concentration
If a chemical system is at equilibrium and we add a substance (either a reactant
or a product), the reaction will shift so as to reestablish equilibrium by consuming
part of the added substance. Conversely, removal of a substance will result in the
reaction moving in the direction that forms more of the substance. Remember
that this is a change in concentration so changing the amount of a pure solid or
liquid which is already in equilibrium should not make the reaction consume the
additional solid or liquid. Example, if solid NaCl is already in equilibrium with a
saturated solution of NaCl, adding more pure NaCl solid will not cause additional
dissolving. Conversely, removal of some of the solid NaCl which is in equilibrium
with a saturated solution of NaCl will not cause further crystallization of NaCl
from the saturated solution.
Effects of volume and pressure changes
These effects are present when the two opposing reactions are of different
molecularity. In solution, volume changes can be achieved by addition of solvent.
Changing the volume effectively changes concentration. Adding more solvent is
essentially a dilution. If the number of solute species on the reactant side is not
the same as on the product side, then volume changes can cause a shift in
equilibrium. Increasing the volume favors the process with lower molecularity.
Increasing the volume then will cause the system to shift in the direction that
increases the number of solute species. This is the same for reactions involving
gases. Decreasing the volume of the container causes an equilibrium mixture of
gases to shift in the direction that reduces the number of moles of gas. For
gases, increasing the pressure by adding one of the gases participating in the
reaction, will also disturb equilibrium (This is essentially a change in reactant or
product concentration). Increasing the total pressure of a reaction vessel by
adding a spectator gas does not affect equilibrium.
Remember that volume changes can only have effect if the number of solute
species or gas species in the reactant side is not equal to the if the number of
solute species or gas species in the product side.
Effect of Temperature
One can treat heat as a reactant (for an endothermic process) or a product (for
an exothermic process and the same principle used for changes in concentration
of reactants or products can be used to deduce the effect of heat on an
equilibrium reaction.