In beginning our study of the reactions of
gases, we will assume a knowledge of the physical properties of
gases as described by the Ideal Gas Law
and an understanding of these properties as given by the postulates
and conclusions of the Kinetic Molecular
Theory. We assume that we have developed a dynamic
model of phase equilibrium in terms of competing rates. We will
also assume an understanding of the bonding, structure, and
properties of individual molecules.
In performing stoichiometric calculations, we
assume that we can calculate the amount of product of a reaction
from the amount of the reactants we start with. For example, if we
burn methane gas,
CH4(g)CH4(g),
in excess oxygen, the reaction
CH4(g)+2O2(g)→CO2(g)+2H2O(g)CH4(g)+2O2(g)→CO2(g)+2H2O(g)
(1)
occurs, and the number of moles of
CO2(g)CO2(g)
produced is assumed to equal the number of moles of
CH4(g)CH4(g)
we start with.
From our study of phase transitions we have
learned the concept of equilibrium. We observed that, in the
transition from one phase to another for a substance, under certain
conditions both phases are found to coexist, and we refer to this
as phase equilibrium. It should not surprise us that these same
concepts of equilibrium apply to chemical reactions as well. In the
reaction,
therefore, we should examine whether the reaction actually produces
exactly one mole of
CO2CO2
for every mole of
CH4CH4
we start with or whether we wind up with an equilibrium mixture
containing both
CO2CO2
and
CH4CH4.
We will find that different reactions provide us with varying
answers. In many cases, virtually all reactants are consumed,
producing the stoichiometric amount of product. However, in many
other cases, substantial amounts of reactant are still present when
the reaction achieves equilibrium, and in other cases, almost no
product is produced at equilibrium. Our goal will be to understand,
describe and predict the reaction equilibrium.
An important corollary to this goal is to
attempt to control the equilibrium. We will find that varying the
conditions under which the reaction occurs can vary the amounts of
reactants and products present at equilibrium. We will develop a
general principle for predicting how the reaction conditions affect
the amount of product produced at equilibrium.
We begin by analyzing a significant industrial
chemical process, the synthesis of ammonia gas,
NH3NH3,
from nitrogen and hydrogen:
N2(g)+3H2(g)→2NH3(g)N2(g)+3H2(g)→2NH3(g)
(2)
If we start with 1 mole of
N2N2
and 3 moles of
H2H2,
the balanced equation predicts that we will produce 2 moles of
NH3NH3.
In fact, if we carry out this reaction starting with these
quantities of nitrogen and hydrogen at 298K in a 100.0L reaction
vessel, we observe that the number of moles of
NH3NH3
produced is 1.91 mol. This "yield" is less than
predicted by the balanced equation, but the difference is not due
to a limiting reagent factor. Recall that, in stoichiometry, the
limiting reagent is the one that is present in less than the ratio
of moles given by the balanced equation. In this case, neither
N2N2
nor
H2H2
is limiting because they are present initially in a 1:3 ratio,
exactly matching the stoichiometry. Note also that this seeming
deficit in the yield is not due to any experimental error or
imperfection, nor is it due to poor measurements or preparation.
Rather, the observation that, at 298K, 1.91 moles rather than 2
moles are produced is completely reproducible: every measurement of
this reaction at this temperature in this volume starting with 1
mole of
N2N2
and 3 moles of
H2H2
gives this result. We conclude that the reaction achieves
reaction equilibrium in which all three
gases are present in the gas mixture. We can determine the amounts
of each gas at equilibrium from the stoichiometry of the reaction.
When
nNH3=1.91molnNH31.91mol
are created, the number of moles of
N2N2
remaining at equilibrium is
nN2=0.045molnN20.045mol
and
nH2=0.135molnH20.135mol.
It is important to note that we can vary the
relative amount of
NH3NH3
produced by varying the temperature of the reaction, the volume of
the vessel in which the reaction occurs, or the relative starting
amounts of
N2N2
and
H2H2.
We shall study and analyze this observation in detail in later
sections. For now, though, we demonstrate that the concept of
reaction equilibrium is general to all reactions.
Consider the reaction
H2(g)+I2(g)→2HI(g)H2(g)+I2(g)→2HI(g)
(3)
If we begin with 1.00 mole of
H2H2
and 1.00 mole of
I2I2
at 500K in a reaction vessel of fixed volume, we observe that, at
equilibrium,
nHI=1.72molnHI1.72mol,
leaving in the equilibrium mixture
nH2=0.14molnH20.14mol
and
nI2=0.14molnI20.14mol.
Similarly, consider the decomposition
reaction
N2O4(g)→2NO2(g)N2O4(g)→2NO2(g)
(4)
At 298K in a 100.0L reaction flask, 1.00 mol
of
N2O4N2O4
partially decomposes to produce, at equilibrium,
nNO2=0.64molnNO20.64mol
and
nN2O4=0.68molnN2O40.68mol.
Some chemical reactions achieve an equilibrium
that appears to be very nearly complete reaction. For
example,
H2(g)+Cl2(g)→2HCl(g)H2(g)+Cl2(g)→2HCl(g)
(5)
If we begin with 1.00 mole of
H2H2
and 1.00 mole of
Cl2Cl2
at 298K in a reaction vessel of fixed volume, we observe that, at
equilibrium,
nHClnHCl
is almost exactly 2.00 mol, leaving virtually no
H2H2
or
Cl2Cl2.
This does not mean that the reaction has not come to equilibrium.
It means instead that, at equilibrium, there are essentially no
reactants remaining.
In each of these cases, the amounts of
reactants and products present at equilibrium vary as the
conditions are varied but are completely reproducible for fixed
conditions. Before making further observations that will lead to a
quantitative description of the reaction equilibrium, we consider a
qualitative description of equilibrium.
We begin with a dynamic equilibrium
description. We know from our studies of phase transitions that
equilibrium occurs when the rate of the forward process
(e.g. evaporation) is matched by the
rate of reverse process (e.g.
condensation). Since we have now observed that gas reactions also
come to equilibrium, we postulate that at equilibrium the forward
reaction rate is equal to the reverse reaction rate. For example,
in the reaction here, the rate of
decomposition of
N2O4N2O4
molecules at equilibrium must be exactly matched by the rate of
recombination (or dimerization) of
NO2NO2
molecules.
To show that the forward and reverse reactions
continue to happen at equilibrium, we start with the
NO2NO2
and
N2O4N2O4
mixture at equilibrium and we vary the volume of the flask
containing the mixture. We observe that, if we increase the volume
and the reaction is allowed to come to equilibrium, the amount of
NO2NO2
at equilibrium is larger at the expense of a smaller amount of
N2O4N2O4.
We can certainly conclude that the amounts of the gases at
equilibrium depend on the reaction conditions. However, if the
forward and reverse reactions stop once the equilibrium amounts of
material are achieved, the molecules would not "know"
that the volume of the container had increased. Since the reaction
equilibrium can and does respond to a change in volume, it must be
that the change in volume affects the rates of both the forward and
reverse processes. This means that both reactions must be occurring
at equilibrium, and that their rates must exactly match at
equilibrium.
This reasoning reveals that the amounts of
reactant and product present at equilibrium are determined by the
rates of the forward and reverse reactions. If the rate of the
forward reaction (e.g. decomposition
of
N2O4N2O4)
is faster than the rate of the reverse reaction, then at
equilibrium we have more product than reactant. If that difference
in rates is very large, at equilibrium there will be much more
product than reactant. Of course, the converse of these conclusions
is also true. It must also be the case that the rates of these
processes depends on, amongst other factors, the volume of the
reaction flask, since the amounts of each gas present at
equilibrium change when the volume is changed.
It was noted above that the equilibrium
partial pressures of the gases in a reaction vary depending upon a
variety of conditions. These include changes in the initial numbers
of moles of reactants and products, changes in the volume of the
reaction flask, and changes in the temperature. We now study these
variations quantitatively.
Consider first the reaction here. Following on our previous study of this
reaction, we inject an initial amount of
N2O4(g)N2O4(g)
into a 100L reaction flask at 298K. Now, however, we vary the
initial number of moles of
N2O4(g)N2O4(g)
in the flask and measure the equilibrium pressures of both the
reactant and product gases. The results of a number of such studies
are given here.
Equilibrium Partial Pressures in Decomposition
Reaction
| Initial
nN2O4nN2O4 |
PN2O4PN2O4
(atm) |
PNO2PNO2
(atm) |
| 0.1 |
0.00764 |
0.033627 |
| 0.5 |
0.071011 |
0.102517 |
| 1 |
0.166136 |
0.156806 |
| 1.5 |
0.26735 |
0.198917 |
| 2 |
0.371791 |
0.234574 |
| 2.5 |
0.478315 |
0.266065 |
| 3 |
0.586327 |
0.294578 |
| 3.5 |
0.695472 |
0.320827 |
| 4 |
0.805517 |
0.345277 |
| 4.5 |
0.916297 |
0.368255 |
| 5 |
1.027695 |
0.389998 |
We might have expected that the amount of
NO2NO2
produced at equilibrium would increase in direct proportion to
increases in the amount of
N2O4N2O4
we begin with. Table 1 shows that
this is not the case. Note that when we increase the initial amount
of
N2O4N2O4
by a factor of 10 from 0.5 moles to 5.0 moles, the pressure of
NO2NO2
at equilibrium increases by a factor of less than 4.
The relationship between the pressures at
equilibrium and the initial amount of
N2O4N2O4
is perhaps more easily seen in a graph of the data in Table 1, as shown in Figure 1. There are some interesting features
here. Note that, when the initial amount of
N2O4N2O4
is less than 1 mol, the equilibrium pressure of
NO2NO2
is greater than that of
N2O4N2O4.
These relative pressures reverse as the initial amount increases,
as the
N2O4N2O4
equilibrium pressure keeps track with the initial amount but the
NO2NO2
pressure falls short. Clearly, the equilibrium pressure of
NO2NO2
does not increase proportionally with the initial amount of
N2O4N2O4.
In fact, the increase is slower than proportionality, suggesting
perhaps a square root relationship between the pressure of
NO2NO2
and the initial amount of
N2O4N2O4.
We test this in Figure 2 by plotting
PNO2PNO2
at equilibrium versus the square root of the initial number of
moles of
N2O4N2O4.
Figure 2 makes it clear that this is
not a simple proportional relationship, but it is closer. Note in
Figure 1 that the equilibrium
pressure
PN2O4PN2O4
increases close to proportionally with the initial amount of
N2O4N2O4.
This suggests plotting
PNO2PNO2
versus the square root of
PN2O4PN2O4.
This is done in Figure 3, where we
discover that there is a very simple proportional relationship
between the variables plotted in this way. We have thus observed
that
PNO2=cPN2O4
PNO2
c
2PN2O4
(6)
where cc is the
slope of the graph. Equation 6 can be
rewritten in a standard form
Kp=PNO22PN2O4
Kp
PNO22
PN2O4
(7)
To test the accuracy of this equation and to
find the value of
KpKp,
we return to Table 1 and add
another column in which we calculate the value of
KpKp
for each of the data points. Table 2 makes it clear that the
"constant" in Equation 7
truly is independent of both the initial conditions and the
equilibrium partial pressure of either one of the reactant or
product. We thus refer to the constant
KpKp
in Equation 7 as the
reaction equilibrium constant.
Equilibrium Partial Pressures in Decomposition
Reaction
| Initial
nN2O4nN2O4 |
PN2O4PN2O4
(atm) |
PNO2PNO2
(atm) |
KpKp |
| 0.1 |
0.00764 |
0.0336 |
0.148 |
| 0.5 |
0.0710 |
0.102 |
0.148 |
| 1 |
0.166 |
0.156 |
0.148 |
| 1.5 |
0.267 |
0.198 |
0.148 |
| 2 |
0.371 |
0.234 |
0.148 |
| 2.5 |
0.478 |
0.266 |
0.148 |
| 3 |
0.586 |
0.294 |
0.148 |
| 3.5 |
0.695 |
0.320 |
0.148 |
| 4 |
0.805 |
0.345 |
0.148 |
| 4.5 |
0.916 |
0.368 |
0.148 |
| 5 |
1.027 |
0.389 |
0.148 |
It is very interesting to note the functional
form of the equilibrium constant. The product
NO2NO2
pressure appears in the numerator, and the exponent 2 on the
pressure is the stoichiometric coefficient on
NO2NO2
in the balanced chemical equation. The reactant
N2O4N2O4
pressure appears in the denominator, and the exponent 1 on the
pressure is the stoichiometric coefficient on
N2O4N2O4
in the chemical equation.
We now investigate whether other reactions
have equilibrium constants and whether the form of this equilibrium
constant is a happy coincidence or a general observation. We return
to the reaction for the synthesis of
ammonia.
In a previous section, we
considered only the equilibrium produced when 1 mole of
N2N2
is reacted with 3 moles of
H2H2.
We now consider a range of possible initial values of these
amounts, with the resultant equilibrium partial pressures given in
Table 3. In addition,
anticipating the possibility of an equilibrium constant, we have
calculated the ratio of partial pressures given by:
Kp=PNH32PN2PH23
Kp
PNH32
PN2
PH23
(8)
In Table 3,
the equilibrium partial pressures of the gases are in a very wide
variety, including whether the final pressures are greater for
reactants or products. However, from the data in Table 3, it is clear that, despite these
variations,
KpKp
in Equation 8 is essentially a
constant for all of the initial conditions examined and is thus the
reaction equilibrium constant for this reaction.
Equilibrium Partial Pressures of the Synthesis
of Ammonia
| V (L) |
nN2nN2 |
nH2nH2 |
PN2PN2 |
PH2PH2 |
PNH3PNH3 |
KpKp |
| 10 |
1 |
3 |
0.0342 |
0.1027 |
4.82 |
6.2×1056.25 |
| 10 |
0.1 |
0.3 |
0.0107 |
0.0322 |
0.467 |
6.0×1056.05 |
| 100 |
0.1 |
0.3 |
0.00323 |
0.00968 |
0.0425 |
6.1×1056.15 |
| 100 |
3 |
3 |
0.492 |
0.00880 |
0.483 |
6.1×1056.15 |
| 100 |
1 |
3 |
0.0107 |
0.0322 |
0.467 |
6.0×1056.05 |
| 1000 |
1.5 |
1.5 |
0.0255 |
0.00315 |
0.0223 |
6.2×1056.25 |
Studies of many chemical reactions of gases
result in the same observations. Each reaction equilibrium can be
described by an equilibrium constant in which the partial pressures
of the products, each raised to their corresponding stoichiometric
coefficient, are multiplied together in the numerator, and the
partial pressures of the reactants, each raised to their
corresponding stoichiometric coefficient, are multiplied together
in the denominator. For historical reasons, this general
observation is sometimes referred to as the Law of Mass
Action.
We have previously observed that phase
equilibrium, and in particular vapor pressure, depend on the
temperature, but we have not yet studied the variation of reaction
equilibrium with temperature. We focus our initial study on
this reaction
and we measure the equilibrium partial pressures at a variety of
temperatures. From these measurements, we can compile the data
showing the temperature dependence of the equilibrium constant
KpKp
for this reaction in Table 4.
Equilibrium Constant for the Synthesis of
HI
| T (K) |
KpKp |
| 500 |
6.25×10-36.25-3 |
| 550 |
8.81×10-38.81-3 |
| 650 |
1.49×10-21.49-2 |
| 700 |
1.84×10-21.84-2 |
| 720 |
1.98×10-21.98-2 |
Note that the equilibrium constant increases
dramatically with temperature. As a result, at equilibrium, the
pressure of
HIHI
must also increase dramatically as the temperature is
increased.
These data
do not seem to have a simple relationship between
KpKp
and temperature. We must appeal to arguments based on
Thermodynamics, from which it is possible to show that the
equilibrium constant should vary with temperature according to the
following equation:
lnKp=-ΔH°RT+ΔS°R
Kp
ΔH°RT
ΔS°R
(9)
If
ΔH°ΔH°
and
ΔS°ΔS°
do not depend strongly on the temperature, then this equation would
predict a simple straight line relationship between
lnKpKp
and
1T1T.
In addition, the slope of this line should be
-ΔH°RΔH°R.
We test this possibility with the graph in Figure 4.
In fact, we do observe a straight line through
the data. In this case, the line has a negative slope. Note
carefully that this means that
KpKp
is increasing with temperature.
The negative slope via Equation 9
means that
-ΔH°RΔH°R
must be negative, and indeed for this reaction in this
temperature range,
ΔH°=15.6kJmolΔH°15.6kJmol.
This value matches well with the slope of the line in Figure 4.
Given the validity of Equation 9 in describing the temperature
dependence of the equilibrium constant, we can also predict that an
exothermic reaction with
ΔH°<0ΔH°0
should have a positive slope in the graph of
lnKpKp
versus
1T1T,
and thus the equilibrium constant should
decrease with increasing
temperature. A good example of an exothermic reaction is the
synthesis of
ammonia for which
ΔH°=-99.2kJmolΔH°-99.2kJmol.
Equilibrium constant data are given in Table 5. Note that, as predicted, the
equilibrium constant for this exothermic reaction decreases rapidly
with increasing temperature. The data from Table 5 is shown in Figure 5, clearly showing the contrast
between the endothermic reaction and the exothermic reaction. The
slope of the graph is positive for the exothermic reaction and
negative for the endothermic reaction. From Equation 9, this is a general result for all
reactions.
Equilibrium Constant for the Synthesis of Ammonia
| T (K) |
KpKp
|
| 250 |
7×10878 |
| 298 |
6×10565 |
| 350 |
2×10323 |
| 400 |
36 |
One of our goals at the outset was to
determine whether it is possible to control the equilibrium which
occurs during a gas reaction. We might want to force a reaction to
produce as much of the products as possible. In the alternative, if
there are unwanted by-products of a reaction, we might want
conditions which minimize the product. We have observed that the
amount of product varies with the quantities of initial materials
and with changes in the temperature. Our goal is a systematic
understanding of these variations.
A look back at Table 1 and Table 2 shows that the equilibrium
pressure of the product of the reaction increases with increasing
the initial quantity of reaction. This seems quite intuitive. Less
intuitive is the variation of the equilibrium pressure of the
product of this
reaction with variation in the volume of the
container, as shown in Table 3.
Note that the pressure of
NH3NH3
decreases by more than a factor of ten when the volume is increased
by a factor of ten. This means that, at equilibrium, there are
fewer moles of
NH3NH3
produced when the reaction occurs in a larger volume.
To understand this effect, we rewrite the
equilibrium constant in Equation 8 to
explicit show the volume of the container. This is done by applying
Dalton's Law of Partial Pressures,
so that each partial pressure is given by the Ideal Gas Law:
Kp=nNH32RTV2nN2RTVnH23RTV3=nNH32nN2nH23RTV2
Kp
nNH32
RTV2
nN2
RTV
nH23
RTV3
nNH32
nN2
nH23
RTV2
(10)
Therefore,
KpRTV2=nNH32nN2nH23
Kp
RTV2
nNH32
nN2
nH23
(11)
This form of the equation makes it clear that,
when the volume increases, the left side of the equation decreases.
This means that the right side of the equation must decrease also,
and in turn, nNH3nNH3 must decrease while nN2nN2 and nH2nH2must
increase. The equilibrium is thus shifted from products to
reactants when the volume increases for this reaction.
The effect of changing the volume must be
considered for each specific reaction, because the effect depends
on the stoichiometry of the reaction. One way to determine the
consequence of a change in volume is to rewrite the equilibrium
constant as we have done in Equation 11.
Finally, we consider changes in temperature.
We note that
KpKp
increases with
TT for
endothermic reactions and decreases with
TT for
exothermic reactions. As such, the products are increasingly
favored with increasing temperature when the reaction is
endothermic, and the reactants are increasingly favored with
increasing temperature when the reaction is exothermic. On
reflection, we note that when the reaction is exothermic, the
reverse reaction is endothermic. Putting these statements together,
we can say that the reaction equilibrium always shifts in the
direction of the endothermic reaction when the temperature is
increased.
All of these observations can be collected
into a single unifying concept known as Le
Châtelier's Principle.
When a reaction at equilibrium is stressed by
a change in conditions, the equilibrium will be reestablished in
such a way as to counter the stress.
This statement is best understood by
reflection on the types of "stresses" we have
considered in this section. When a reactant is added to a system at
equilibrium, the reaction responds by consuming some of that added
reactant as it establishes a new equilibrium. This offsets some of
the stress of the increase in reactant. When the temperature is
raised for a reaction at equilibrium, this adds thermal energy. The
system shifts the equilibrium in the endothermic direction, thus
absorbing some of the added thermal energy, countering the
stress.
The most challenging of the three types of
stress considered in this section is the change in volume. By
increasing the volume containing a gas phase reaction at
equilibrium, we reduce the partial pressures of all gases present
and thus reduce the total pressure. Recall that the response of
this reaction to
the volume increase was to create more of the reactants at the
expense of the products. One consequence of this shift is that more
gas m
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