Chapter 6
Chemical Equilibrium
Chemical Principles 7th Edition
Steven S. Zumdahl
6.1 The Equilibrium Condition
6.2 The Equilibrium Constant
6.3 Equilibrium Expressions Involving Pressures
6.4 The Concept of Activity
6.5 Heterogeneous Equilibria
6.6 Applications of the Equilibrium Constant
6.7 Solving Equilibrium Problems
6.8 Le Châtelier’s Principle
6.9 Equilibria Involving Real Gases
Black Sea Nettle in the Monterey Bay Aquarium in California. Jellyfish
use statocysts to maintain their physical equilibrium. A statocyst
consists of a fluid-filled sac containing statoliths that stimulate
sensory cells and help indicate position when the animal moves.
– In doing stoichiometry calculations, we assume that reactions
proceed to completion – that is, until one of the reactants is
consumed.
– There are many chemical reactions that stop far short
completion. An example is the dimerization of nitrogen
dioxide:
2NO2(g) → N2O4(g)
– The reactant NO2 is a reddish brown gas, and the product
N2O4 is a colorless gas.
– This is illustrated on the molecular level in Fig. 6.1.
– This observation is a clear indication that the reaction has
stopped short of completion.
– In fact, the system has reached chemical equilibrium, the
state in which the concentrations of all reactants and products
remain constant with time.
FIGURE 6.1
A molecular representation of the reaction 2NO2(g) → N2O4(g) over
time in a closed vessel. Note that the numbers of NO2 and N2O4 in the
container become constant (c) and (d) after sufficient time has
passed.
– Any chemical reaction carried out in a closed vessel will reach
equilibrium.
6.1 The Equilibrium Condition
– Equilibrium is not static; it is a highly dynamic situation.
– To see how this concept applies to chemical reactions,
consider the reaction between steam and carbon monoxide in
a closed vessel at a high temperature, where the reaction
takes place rapidly:
H2O(g) + CO(g) ↔ H2(g) + CO2(g)
– The plots of the concentrations of reactants and products
versus time are shown in Fig. 6.2. Note that since CO and H2O
were originally present in equal molar quantities, and since
they react in a 1:1 ratio, the concentrations of the two gases
are always equal. Also, since H2 and CO2 are formed in equal
amount, they are always present at the same concentration.
FIGURE 6.2
The changes in concentrations with time for the reaction H2O(g) +
CO(g) ↔ H2(g) + CO2(g) when equal molar quantities of H2O(g) and
CO(g) are mixed.
– Figure 6.2 is a profile of the progress of the reaction. When CO
and H2O are mixed, they immediately begin reacting to form
H2 and CO2.
– Equilibrium is a dynamic situation.
– Unless the system is somehow disturbed, on further changes
in concentrations will occur. Note that although the
equilibrium position lies far to the right, the concentrations of
reactants never reach zero; the reactants will always be
present in small but constant concentrations.
– This is shown on the microscopic level in Fig. 6.3.
– Thus, as reactant collide and react to form products
H2O(g) + CO(g) ↔ H2(g) + CO2(g)
– the concentrations of the reactants decrease, causing the rate
of this reaction (the forward reaction) to decrease – that is,
the reaction slows down. (See Fig. 6.4.)
Reddish brown nitrogen dioxide gas streaming from a flask where
copper is reacting with concentrated nitric acid.
FIGURE 6.3
(a) H2O and CO are mixed in equal numbers and begin to react (b) to
form CO2 and H2. After time has passed equilibrium is reached (c) and
the numbers of reactant and product molecules then remain
constant over time (d).
– As in the traffic analogy, there is also a reverse direction:
H2(g) + CO2(g) ↔H2O(g) + CO(g)
– Initially in this experiment no H2 and CO2 are present, and this
reverse reaction could not occur.
– The equilibrium position of a reaction-left, right, or
somewhere in between − is determined by many factors: the
initial concentrations, the relative energies of the reactants
and products, and the relative “degree of organization” of the
reactants and products.
FIGURE 6.4
The changes with time in the rates of forward and reverse reactions
for H2O(g) + CO(g) ↔ H2(g) + CO2(g) when equal molar quantities of
H2O(g) and CO(g) are mixed. Note that rates for the forward and
reverse reactions do not change in the same way with time. We will
not be concerned with the reasons for this difference at this point.
The Characteristics of Chemical Equilibrium
– To explore the important characteristics of chemical
equilibrium, we will consider the synthesis of ammonia from
elemental nitrogen and hydrogen:
N2(g) + 3H2(g) ↔ 2NH3(g)
– This process (called the Haber process) is of great commercial
value because ammonia is an important fertilizer for the
growth of corn and other crops.
– The Unite States produces almost 20 million tones of
ammonia annually.
– When gaseous nitrogen, hydrogen, and ammonia are mixed in
a closed vessel at 25°C, no apparent change in concentrations
occurs over time, regardless of the original amounts of the
gases.
– There are two possible reasons why the concentrations of the
reactants and products of a given chemical reaction remain
unchanged when mixed:
1. The system is at chemical equilibrium.
2. The forward and reverse reactions are slow that the system
moves toward equilibrium at an undetectable rate.
– The second reason applies to the nitrogen, hydrogen, and
ammonia mixture at 25°C. Because the molecules involved
have strong chemical bonds, mixtures of N2, H2, and NH3 at
25°C can exist with no apparent change over long periods of
time. However, under appropriate conditions the system does
reach equilibrium, as shown in Fig. 6.5. Note that because of
the reaction stoichiometry H2 disappears three times as fast as
N2 does, and NH3 forms twice as fast as N2 disappears.
FIGURE 6.5
A concentration profile for the reaction N2(g) + 3H2(g) ↔ 2NH3(g)
when only N2(g) and H2(g) are mixed initially.
6.2 The Equilibrium Constant
– Cato Maximilian Guldberg (1836-1902) and Peter Waage
(1833-1900), proposed the law of mass action in 1864 as a
general description of the equilibrium condition. For a
reaction of the type
jA + kB ↔ lC + mD
– where A, B, C, and D represent chemical species and j, k, l,
and m are their coefficients in the balanced equation, the law
of mass action is represented by the following equilibrium
expression:
K = [C]l[D]m/[A]j[B]k
– The law of mass action is based on experimental observations.
– The square brackets indicate the concentrations of the
chemical species at equilibrium, and K is a constant called the
equilibrium constant.
– When the observed equilibrium concentrations are inserted
into the equilibrium expression concentration from the law of
mass action for a given reaction, the result is a constant.
Example 6.1
– The following equilibrium concentrations were observed for
the Haber process at 127°C:
[NH3] = 3.1 × 10–2 mol/L
[N2] = 8.5 × 10–1 mol/L
[H2] = 3.1 × 10–3 mol/L
– a. Calculate the value of K at 127°C for the reaction.
N2(g) + 3H2(g) ↔ 2NH3(g)
– b. Calculate the value of the equilibrium constant at 127°C for
the reaction
2NH3(g) ↔ N2(g) + 3H2(g)
– c. Calculate the value of the equilibrium constant at 127°C for
the reaction given by the equation
1/2N2(g) + 3/2H2(g) ↔ NH3(g)
Solution
– a. The balanced equation for the Haber process is
N2(g) + 3H2(g) ↔ 2NH3(g)
– Thus, using the law of mass action to construct the expression
for K, we have
K = [NH3]2/[N2][H2]3
= (3.1 × 10–2 mol/L)2/(8.5 × 10–1 mol/L)(3.1 × 10–3 mol/L)3
= 3.8 × 104 L2/mol2
– b. This reaction is written in the reverse order of the equation
given in part a.
– This leads to the equilibrium expression
K’ = [N2][H2]3/[NH3]2
– which is the reciprocal of the expression used in part a. So
K’ = [N2][H2]3/[NH3]2
= 1/K = 1/(3.8 × 104 L2/mol2) = 2.6 × 10–5 mol2/L2
– c. We use the law of mass action:
K’’ = [NH3]/[N2]1/2[H2]3/2
– If we compare this expression with the one obtained in part a,
we see that since
[NH3]/[N2]1/2[H2]3/2 = ([NH3]2/[N2][H2]3)1/2
– Then
K’’ = K1/2
– Thus
K’’ = K1/2 = (3.8 × 104 L2/mol2)1/2 = 1.9 × 102 L/mol
– The K’s used in this section might best be called Kobserved since
they are calculated from “observed” concentrations that are
not corrected for the effects of nonideal. Only Kobs values has
units.
– We can draw some important conclusion from the results of
Example 6.1. For a reaction of the form
jA + kB ↔ lC + mD
– the equilibrium expression is
K = [C]l[D]m/[A]j[B]k
– If this reaction is reversed, the new equilibrium expression is
K’ = [A]j[B]k/[C]l[D]m= 1/K
– If the original reaction is multiplied by some factor n to give
njA + nkB ↔ nlC + nmD
– the equilibrium expression becomes
K’’ = [C]nl[D]nm/[A]nj[B]nk = Kn
Some Characteristics of the Equilibrium Expression
1. The equilibrium expression for a reaction written in reverse is
the reciprocal of that for the original reaction.
2. When the balanced equation for a reaction is multiplied by a
factor n, the equilibrium expression for the new reaction is
the original expression raised to the nth power. Thus Knew =
(Koriginal)n.
3. The apparent units for K are determined by the powers of the
various concentration terms. The (apparent) units for K
therefore depend on the reaction being considered. We will
have more to say about the units for K in Section 6.4.
– The law of mass action applies to solution and gaseous
equilibria.
Applying anhydrous ammonia to soybean stubble prior to planting
corn.
– The law of mass action is widely applicable. It correctly
describes the equilibrium behavior of all chemical reaction
systems whether they occur in solution or in the gas phase.
– For example, consider again the ammonia synthesis reaction.
At 500°C the value of K for this reaction is 6.0 × 10–2 L2/mol2.
Whenever N2, H2, and NH3 are mixed together at this
temperature, the system will always come to an equilibrium
position such that
[NH3]2/[N2][H2]3 = 6.0 × 10–2 L2/mol2
– This expression has the same value at 500°C, regardless of the
amounts of the gases that are mixed together initially.
– Although the special ratio of products to reactants defined by
the equilibrium expression is constant for a given reaction
system at a given temperature, the equilibrium concentrations
will not always be the same.
– Table 6.1 gives three sets of data for the synthesis of
ammonia, showing that even though the individual sets of
equilibrium concentrations are quite different for the
different situations, the equilibrium constant, which depends
on the ratio of the concentrations, remains the same (within
experimental error). Note that subscripts of zero indicate
initial concentrations.
– Each set of equilibrium concentrations is called an equilibrium
position. It is essential to distinguish between the equilibrium
constant and the equilibrium positions for a given reaction
system.
– For a reaction at a given temperature, there are many
equilibrium position but only one value for K.
6.3 Equilibrium Expressions Involving Pressures
– So far we have been describing equiliria involving gases in
terms of concentrations. Equilibria involving gases can also be
described in terms of pressures.
– The relationship between the pressure and the concentration
of a gas can be seen from the ideal gas equation:
PV = nRT
or
P = (n/V)RT = CRT
– where C equals n/V, or the number of moles of gas n per unit
volume V. Thus C represents the molar concentration of the
gas.
– For a ammonia synthesis reaction, the equilibrium expression
can be written in terms of concentrations,
K = [NH3]2/[N2][H2]3 = CNH 2/[(CN )(CH )3]
3
2
2
– or in terms of the equilibrium partial pressures of the gases,
Kp = PNH 2/[(PN )(PH )3]
3
2
2
– In this book K denotes an equilibrium constant in terms of
concentrations, and Kp represents an equilibrium constant in
terms of partial pressures.
– K involves concentrations; Kp involves pressures.
– The relationship between K and Kp for a particular reaction
follows from the fact that for an ideal gas, C = P/RT. For
example, for the ammonia synthesis reaction,
K = [NH3]2/[N2][H2]3 = CNH 2/(CN )(CH )3
3
2
2
2
3
= (PNH /RT) /(PN /RT)(PH /RT) = PNH 2/(PN )(PH )3 × (1/RT)2/(1/RT)4
3
2
2
3
2
2
3
2
2
2
= [PNH /(PN )(PH ) ](RT) = Kp(RT)
3
2
2
– However, for the synthesis of hydrogen fluoride from its
elements,
H2(g) + F2(g) ↔ 2HF(g)
– the relationship between K and Kp is
K = [HF]2/[H2][F2] = CHF2/[(CH )(CF )] = (PHF/RT)2/[(PH /RT)(PF /RT)]
2
2
2
2
2
= PHF /[(PH )(PF )] = Kp
2
2
– For the general reaction
jA + kB ↔ lC + mD
– The relationship between K and Kp is
Kp = K(RT)Δn
– For the proceeding general reaction
Kp = (PCl)(PDm)/(PAj)(PBk) = (CC × RT)l(CD × RT)m/(CA × RT)j(CB × RT)k
= [(CC)l(CD)m/(CA )j(CB)k] × [(RT)l+m/(RT)j+k] = K(RT)(l+m)–(j+K) = K(RT)Δn
– where Δn = (l + m) – (j + k), the difference in the sums of the
coefficients for the gaseous products and reactants.
– Δn always involves products minus reactants.
– We have seen that the (apparent) units of the equilibrium
constant depend on the specific reaction. For example, for the
reaction
H2(g) + F2(g) ↔ 2HF(g)
– the units of K and Kp can be found as follows:
K = CHF2/[(CH )(CF )] = (mol/L)2/[(mol/L)(mol/L)] ⇒ no units
2
2
2
Kp = PHF /(PH )(PF ) = (atm)2/[(atm)(atm)] ⇒ no units
2
2
– Note that in the above discussion we used the term “apparent
units” when referring to equilibrium constants. This term was
used because the theoretical foundation for the concept of
equilibrium based on thermodynamics includes a reference
state for each substance, which always cause the units of
concentration or pressure to cancel.
6.4 The Concept of Activity
– The true equilibrium constant expression dose not simply
involve the observed equilibrium pressure or the
concentration for a substance but involves the ratio of the
equilibrium pressure (or concentration) for a given substance
to a reference pressure (or concentration) for that substance.
– This ratio is defined as the activity of the substance, which in
terms of pressures is defined as
Activity (ith component) = ai = Pi/Preference
– where Pi = partial pressure of the ith gaseous component
– Preference = 1 atm (exactly)
– and where ideal behavior is assumed.
– Using the concept of activities, the equilibrium expression for
the reaction
jA(g) + kB(g) ↔ lC(g) + mD(g)
– is written as
K = (aC)l(aD)m/(aA)j(aB)k = (PC/Pref)l(PD/Pref)m/(PA/Pref)j(PB/Pref)k
– With all the pressure expressed in atmosphere, we have
Kp = (PC(atm)/1 atm)l(PD(atm)/1 atm)m/(PA(atm)/1 atm)j(PB(atm)/1 atm)k
= PClPDm/PAjPBk
– where Kp is unitless as shown.
– Because of the difference in reference states, in general,
K ≠ Kp
Equilibrium
composition
expressed in
concentration
units
Equilibrium
composition
expressed as
pressures
– The only exception to this principle occurs when the sum of
the powers in the numerator and denominator are the same
(as discussed previously for H2 + F2 ↔ 2HF). In such a case, K =
Kp.
6.5 Heterogeneous Equilibria
– So far we have discussed equilibria only for systems in the gas
phase, where all reactants and products are gases. These
situations represent homogeneous equilibria.
– However, many equilibria involve more than one phase and
are called heterogeneous equilibria.
– The thermal decomposition of calcium carbonate in the
commercial preparation of lime occurs by a reaction involving
both solid and gas phase:
CaCO3(g) ↔ CaO(s) + CO2(g)
Lime
– Straightforward application of the law of mass action leads to
the equilibrium expression:
K’ = [CO2][CaO]/[CaCO3]
Helicopter liming acidic wetland in Sweden.
– However, experimental results show that the position of a
heterogeneous equilibrium does not depend on the amounts
of pure solids or liquids present (see Fig. 6.6).
– This result makes sense when the meaning of an activity for a
pure liquid or solid is understood.
aCaCO = [CaCO3]/[CaCO3] = 1
3
Pure solid
Pure solid
(reference state)
and aCaO = [CaO]/[CaO] = 1
– Thus the equilibrium expression for the decomposition of
solid CaCO3 are
K = [CO2](1)/1 = [CO2]
and
Kp = PCO (1)/1 = PCO
2
2
– In summary, we can make the following general statement:
The activity of a pure solid or liquid is always 1.
FIGURE 6.6
The position of the equilibrium CaCO3(s) ↔ CaO(s) + CO2(g) does not
depend on the amounts of CaCO3(s) and CaO(s) present.
– Lime is among the top six chemicals manufactured in the
Unite States in terms of amount produced.
– If pure solids or pure liquids are involved in a chemical
reaction, their concentrations are not included in the
quilibrium expression for the reaction.
– This simplification occurs only with pure solids or liquids, not
with solutions or gases, because in these last two cases the
activity cannot be assumed to be 1.
– In the decomposition of liquid water to gaseous hydrogen and
oxygen,
where
2H2O(l) ↔ 2H2(g) + O2(g)
K = [H2]2[O2]
and
Kp = (PH )2(PO )
2
2
– Water is not included in either equilibrium expression because
it is present as a pure liquid (aH2O(l) = 1). However, if the
reaction were carried out under conditions in which the water
is a gas rather than a liquid,
where
2H2O(g) ↔ 2H2(g) + O2(g)
K = [H2]2[O2]/[H2O]2
and
Kp = (PH )2(PO )/PH
2
2
2O
2
6.6 Applications of the Equilibrium Constant
– Knowing the equilibrium constant for a reaction allows us to
predict several important features of the reaction: the
tendency of the reaction to occur (but not the speed of the
reaction), whether a given set of concentrations represents an
equilibrium condition, and the equilibrium position that will
be achieved from a given set of initial concentration.
The Extent of a Reaction
– The inherent tendency for a reaction to occur is indicated by
the magnitude of the equilibrium constant. A value of K that is
much larger than 1 means that at equilibrium the reaction
system will consist of mostly products – the equilibrium lies to
the right.
– It is important to understand that the size of K and the time
required to reach equilibrium are not directly related.
– The time required to achieve equilibrium depends on the
reaction rate.
Reaction Quotient
– When the reactants and products of a given chemical reaction
are mixed, it is useful to know whether the mixture is at
equilibrium and, if it is not, in which direction the system will
shift to reach equilibrium.
– To determine the shift in such cases, we use the reaction
quotient Q. The reaction quotient is obtained by applying the
law of mass action, but using initial concentrations instead of
equilibrium concentrations.
– For the synthesis of ammonia,
N2(g) + 3H2(g) ↔ 2NH3(g)
– the expression for the reaction quotient is
Q = [NH3]02/[N2]0[H2]03
– where the zero subscripts indicate initial concentrations.
– To determine in which direction a system will shift to reach
equilibrium, we compare the values of Q and K (Fig. 6.7).
There are three possible situations:
1. Q is equal to K. The system is at equilibrium; no shift will
occur.
2. Q is greater than K. In this case the ratio of initial
concentrations of products to initial concentrations of
reactants is too larger. For the system to reach equilibrium, a
net change of products to reactants must occur. The system
shifts to the left, consuming products and forming reactants,
until equilibrium is achieved.
3. Q is less than K. In this case the ratio of initial concentrations
of products to initial concentrations of reactants is too small.
The system must shift to the right, consuming reactants and
forming products, to attain equilibrium.
FIGURE 6.7
Comparing the values of Q and K allows us to determine the direction
of the system will shift to reach equilibrium.
Example 6.2
– For the synthesis of ammonia at 500°C, the equilibrium
constant is 6.0 × 10–2 L2/mol2. Predict the direction in which
the system will shift to reach equilibrium in each of the
following cases.
– a. [NH3]0 = 1.0 × 10–3 M; [N2]0 = 1.0 × 10–5 M; [H2]0 = 2.0 × 10–3
M
– b. [NH3]0 = 2.00 × 10–4 M; [N2]0 = 1.50 × 10–5 M; [H2]0 = 3.54 ×
10–1 M
– c. [NH3]0 = 1.0 × 10–4 M; [N2]0 = 5.0 M; [H2]0 = 1.0 × 10–2 M
Solution
– a. First we calculate the value of Q:
Q = [NH3]02/[N2]0[H2]03
= (1.0 × 10–3 mol/L)2/(1.0 × 10–5 mol/L)(2.0 × 10–3 mol/L)3
= 1.3 × 107 L2/mol2
– Since K = 6.0 × 10–2 L2/mol2, Q is much greater than K. For the
system to attain equilibrium, the concentrations of the
products must be decreased and the concentrations of the
reactants increased. The system will shift to the left:
N2(g) + 3H2(g) ← 2NH3(g)
– b. We calculate the value of Q:
Q = [NH3]02/[N2]0[H2]03
= (2.00 × 10–4 mol/L)2/(1.50 × 10–5 mol/L)(3.54 × 10–1 mol/L)3
= 6.01 × 10–2 L2/mol2
– In this case Q = K, so the system is at equilibrium. No shift will
occur.
– c. The value of Q is
Q = [NH3]02/[N2]0[H2]03
= (1.0 × 10–4 mol/L)2/(5.0 mol/L)(1.0 × 10–2 mol/L)3
= 2.0 × 10–3 L2/mol2
– Here Q is less than K, so the system will shift to the right,
attaining equilibrium by increasing the concentration of the
product and decreasing the concentrations of the reactants:
N2(g) + 3H2(g) → 2NH3(g)
Calculating Equilibrium Pressure and Concentrations
– A typical equilibrium problem involves finding the equilibrium
concentration (or pressures) of reactants and products given
the value of the equilibrium constant and the initial
concentrations (or pressures).
Example 6.3
– Assume that the reaction for the formation of gaseous
hydrogen fluoride from hydrogen and fluorine has an
equilibrium constant of 1.15 × 102 at a certain temperature. In
a particular experiment at this temperature 3.000 mol of each
component was added to a 1.500-L flask. Calculate the
equilibrium concentrations of all species.
Solution
– The balanced equation for the reaction is
H2(g) + F2(g) ↔ 2HF(g)
– The equilibrium expression is
K = 1.15 × 102 = [HF]2/[H2][F2]
– We first calculate the initial concentrations:
[HF]0 = [H2]0 = [F2]0 = 3.000 mol/1.500 L = 2.000 M
– From the value of Q,
Q = [HF]02/[H2]0[F2]0 = (2.000)2/[(2.000)(2.000)] = 1.000
– which is much less than K, we know that the system must shift
to the right to each equilibrium.
– What change in the concentrations is necessary? Since the
answer to this question is presently unknown, we will define
the change needed in terms of x. Let x equal the number of
moles per liter of H2 consumed to reach equilibrium. The
stoichiometry of the reaction shows that x mol/L of F2 will also
be consumed and that 2x mol/L of HF will be formed:
H2(g) + F2(g) → 2HF(g)
x mol/L x mol/L → 2x mol/L
– Now the equilibrium concentrations can be expressed in
terms of x:
Initial
Concentration (mol/L)
[H2]0 = 2.000
[F2]0 = 2.000
[HF]0 = 2.000
Change (mol/L)
–x
–x
+2x
Equilibrium
Concentration (mol/L)
[H2] = 2.000 – x
[F2] = 2.000 – x
[HF] = 2.000 + 2x
– These concentrations can be represented in a shorthand table
as follows:
H2(g)
+
Initial:
2.000
Change:
–x
Equilibrium: 2.000 – x
F2(g) ↔
2.000
–x
2.000 – x
2HF(g)
2.000
+2x
2.000 + 2x
– To solve for x, we substitute the equilibrium concentrations
into the equilibrium expression:
K = 1.15 × 102 = [HF]2/[H2][F2] = (2.000 + 2x)2/(2.000 – x)2
– The right side of this equation is a perfect square, so taking
the square root of both sides gives
√ 1.15 × 102 = (2.000 + 2x)/(2.000 – x)
– which yields x = 1.528. The equilibrium concentrations can
now be calculated:
[H2] = [F2] = 2.000 M – x = 0.472 M
[HF] = 2.000 M + 2x = 5.056 M
– Check: Checking these values by substituting them into the
equilibrium expression gives
[HF]2/[H2][F2] = (5.056)2/(0.472)2 = 1.15 × 102
– which agrees with the given value of K.
6.7 Solving Equilibrium Problems
Solving Equilibrium Problems
1. Write the balanced equation for the reaction.
2. Write the equilibrium expression using the law of mass action.
3. List the initial concentrations.
4. Calculate Q and determine the direction of the shift to
equilibrium.
5. Define the change needed to reach equilibrium, and define
the equilibrium concentrations by applying the change to the
initial concentrations.
6. Substitute the equilibrium concentrations into the equilibrium
expression, and solve for the unknown.
– Check your calculated equilibrium concentrations by making
sure that they give the correct value of K.
– Suppose that for a synthesis of hydrogen fluoride from
hydrogen and fluorine, 3.000 moles of H2 and 6.000 moles of
F2 are mixed in a 3.000-liter flask. The equilibrium constant for
the synthesis reaction at this temperature is 1.15 × 102. We
can calculate the equilibrium concentration of each
component as follows:
– We begin as usual by writing the balanced equation for the
reaction:
H2(g) + F2(g) ↔ 2HF(g)
– The equilibrium expression is
K = 1.15 × 102 = [HF]2/[H2][F2]
– The initial concentrations are
[H2]0 = 3.000 mol/3.000 L = 1.000 M
[F2]0 = 6.000 mol/3.000 L = 2.000 M
[HF]0 = 0
– There is no need to calculate Q; since no HF is initially present,
we know that the system must shift to the right to reach
equilibrium.
– If we let x represent the number of moles per liter of H2
consumed to reach equilibrium, we can represent the
concentrations as follows:
Initial
Concentration (mol/L)
[H2]0 = 1.000
[F2]0 = 2.000
[HF]0 = 0
Change (mol/L)
–x
–x
+2x
Equilibrium
Concentration (mol/L)
[H2] = 1.000 – x
[F2] = 2.000 – x
[HF] = 0 + 2x
– Or in shorthand form:
H2(g)
Initial:
1.000
Change:
–x
Equilibrium: 1.000 – x
+
F2(g) ↔
2.000
–x
2.000 – x
2HF(g)
0
+2x
2x
– Substituting the equilibrium concentrations into the
equilibrium expression gives
K = 1.15 × 102 = [HF]2/[H2][F2] = (2x)2/[(1.000 – x)(2.000 – x)]
– To solve for x, we performed the indicated multiplication,
(1.000 – x)(2.000 – x)(1.15 × 102) = (2x)2
– to give
(1.15 × 102)x2 – 3.000(1.15 × 102)x + 2.000(1.15 × 102) = 4x2
– and collect terms
(1.11 × 102)x2 – (3.45 × 102)x + 2.30 × 102 = 0
– This expression is a quadratic equation for the general form
ax2 + bx + c = 0
– where the roots can be obtained from the quadratic formula:
x = –b ± √(b2 – 4ac) / 2a
– In this example a = 1.11 × 102, b = –3.45 × 102, and c = 2.30 ×
102.
– Substituting these values into the quadratic formula gives two
values for x:
x = 2.14 mol/L
and
x = 0.968 mol/L
– Both of these values cannot be valid. How can we choose
between them? Since the expression for the equilibrium
concentration of H2 is
[H2] = 1.000 M – x
– The value of x cannot be 2.14 mol/L. Thus the correct value
for x is 0.968 mol/L, and the equilibrium concentrations are as
follows:
[H2] = 1.000 M – 0.968 M = 3.2 × 10–2 M
[F2] = 2.000 M – 0.968 M = 1.032 M
[HF] = 2(0.968 M) = 1.936 M
– We can check these concentrations by substituting them into
the equilibrium expression
[HF]2/[H2][F2] = (1.936)2/[(3.2 × 10–2 )(1.032)] = 1.13 × 102
– This value is in close agreement with the given value for K
(1.15 × 102), so the calculated equilibrium concentrations are
correct.
– Note that although we used the quadratic formula to solve for
x in this problem, other methods are also available. However,
use of successive approximations is often preferable.
– Starting with the quadratic equation
(1.11 × 102)x2 – (3.45 × 102)x + 2.30 × 102 = 0
– and dividing it by 1.11 × 102 to give
x2 – 3.11x + 2.07 = 0
– which can then rearranged to
or
x2 = 3.11x – 2.07
x = √3.11x – 2.07
– Now we can proceed by guessing a value of x, which is then
inserted into the square root expression. Next, we calculate a
“new” value of x from the expression
x = √3.11x – 2.07
“New” value Guessed value
calculated
of x inserted
– When the calculated value (the new value) of x agrees with
the guessed value, the equation has been solved.
Treating Systems That Have Small Equilibrium Constants
– Consider gaseous NOCl, which decomposes to form the gas
NO and Cl2. At 35°C the equilibrium constant is 1.6 × 10–5
mol/L. In an experiment in which 1.0 mole of NOCl is placed in
a 2.0-liter flask, what are the equilibrium concentrations?
– The balanced equation is
2NOCl(g) ↔ 2NO(g) + Cl2(g)
K = [NO]2[Cl2]/[NOCl]2 = 1.6 × 10–5 mol/L
– The initial concentrations are
[NOCl]0 = 1.0 mol/2.0 L = 0.50 M, [NO]0 = 0, [Cl2]0 = 0
– The changes in the concentration of NOCl and NO can then be
obtained from the balanced equation:
2NOCl(g) → 2NO(g) + Cl2(g)
2x
→ 2x
+ x
– The concentrations can be summarized as follows:
Initial
Concentration (mol/L)
[NOCl]0 = 0.5
[NO]0 = 0
[Cl2]0 = 0
Change (mol/L)
–2x
+2x
+x
Equilibrium
Concentration (mol/L)
[NOCl] = 0.5 – 2x
[NO] = 0 + 2x = 2x
[Cl2] = 0 + x = x
– Or in shorthand form:
2NOCl(g)
Initial:
0.5
Change:
–2x
Equilibrium: 0.5 – 2x
↔
2NO(g)
0
+2x
2x
+
Cl2(g)
0
+x
x
– The equilibrium concentrations must satisfy the equilibrium
expression:
K = 1.6 × 10–5 = [NO]2[Cl2]/[NOCl]2 = [(2x)2(x)]/(0.5 – 2x)2
– Since K is so small (1.6 × 10–5 mol/L), the system will not
proceed far to the right to reach equilibrium. That is, x
represents a relatively small number.
– Consequently, the term (0.5 – 2x) can be approximated by 0.5.
That is, when x is small,
0.5 – 2x ≈ 0.5
– Approximation can simplify complicated math, but their
validity should be carefully checked.
– Making this approximation allows us to simplify the
equilibrium expression:
1.6 × 10–5 = [(2x)2(x)]/(0.5 – 2x)2 ≈ [(2x)2(x)]/(0.5)2 = 4x3/(0.5)2
– Solving for x3 gives
x3 = (1.6 × 10–5)(0.5)2/4 = 1.0 × 10–6
– and
x = 1.0 × 10–2 mol/L.
– Next, we must check the validity of the approximation. If x =
1.0 × 10–2, then
0.50 – 2x = 0.50 – (1.0 × 10–2) = 0.48
– We use this approximate value of x to calculate the
equilibrium concentrations:
[NOCl] = 0.50 – 2x = 0.48 ≈ 0.50
[NO] = 2x = 2 × (1.0 × 10–2 M) = 2.0 × 10–2 M
[Cl2] = x = 1.0 × 10–2 M
Check: [NO]2[Cl2]/[NOCl]2 = [(2.0 × 10–2 )2(1.0 × 10–2 )]/(0.5)2 = 1.6 × 10–5
– This problem turned out to be relatively easy to solve because
the small value of K and the resulting small shift to the reach
equilibrium allowed simplification.
– A good way to assess whether a 4% error is acceptable here is
to examine the precision of the data given. For example, note
that the value K is 1.6 × 10–5, which can be interpreted as (1.6
± 0.1) × 10–5. Thus the uncertainty in K is at least 1 part in 16
or about 6%. Therefore, a 4% error is [NOCl] is acceptable.
6.8 Le Châtelier’s Principle
– It is important to understand the factors that control the
position of a chemical equilibrium.
– Some of his results are given in Table 6.2.
– Note that the amount of NH3 at equilibrium increase with an
increase in pressure but decreases with an increase in
temperature.
– Thus the amount of NH3 present at equilibrium is favored by
conditions of low temperature and high pressure.
– We can qualitatively predict the effects of changes in
concentration, pressure, and temperature on a system at
equilibrium by using Le Châtelier’s principle, which states that
if a change in conditions (a “stress”) is imposed on a system at
equilibrium, the equilibrium position will shift in a direction
that tends to reduce that change in conditions.
Henri Louis Le Châtelier (1850-1936), the French physical chemist and
metallurgist, seen here while a student at the École Polytechnique.
The Effect of a Change in Concentration
– Suppose there is an equilibrium position described by these
concentrations:
[N2] = 0.399 M, [H2] = 1.197 M, and [NH3] = 0.202 M
– What will happen if 1.000 mole per liter of N2 is suddenly
injected into the system at constant volume? We can answer
this question by calculating the value of Q. The concentrations
before the system adjusts are
[N2]0 = 0.399 M + 1.000 M = 1.399 M
[H2]0 = 1.197 M
[NH3]0 = 0.202 M
Added N2
– Note that we label this as “initial concentrations” because the
system in no longer at equilibrium. Then
Q = [NH3]02/[N2]0[H2]03 = (0.202)2/(1.399)(1.197)3 = 1.70 × 10–2
– Since we are not give the value of K, we must calculate it from
the first set of equilibrium concentrations:
K = [NH3]2/[N2][H2]3 = (0.202)2/(0.399)(1.197)3 = 5.96 × 10–2
– Because the concentration of N2 was increased, Q is less than
K. The system will shift to the right to arrive at the new
equilibrium position.
– We simply summarize the results:
Equilibrium Position I
[N2] = 0.399 M
[H2] = 1.197 M
[NH3] = 0.202 M
Equilibrium Position II
[N2] = 1.399 M
1.000 mol/L
⎯⎯⎯⎯⎯→
[H2] = 1.044 M
of N2 added
[NH3] = 0.304 M
– These data reveal that the equilibrium position dose in fact
shift to the right: The concentration of H2 decrease; the
concentration of NH3 increase; and, of course, since nitrogen
is added, the concentration of N2 shows an increase relative to
the amount presented at the equilibrium position.
– We can predict the shift qualitatively by using Le Châtelier’s
principle.
– Thus Le Châtelier’s principle correctly predicts that adding
nitrogen causes the equilibrium to shift to the right (see Fig.
6.7).
– If ammonia had been added instead of nitrogen, the system
would have shifted to the left to consume ammonia. So we
can paraphrase Le Châtelier’s principle for this case as follows:
If a gaseous reactant or product is added to a system at
equilibrium, the system will shift away from the added
component. If a gaseous reactant or product is removed, the
system will shift toward the removed component.
– The system shifts in the direction that compensates for the
imposed change in conditions.
Blue anhydrous cobalt(II) chloride and pink hydrated cobalt(II)
chloride. Since the reaction CoCl2(s) + 6H2O(g) → CoCl2·6H2O(s) is
shifted to the right by water vapor, CoCl2 is often used in novelty
devices to detect humidity.
FIGURE 6.8
(a) The initial equilibrium mixture of N2, H2, and NH3. (b) Addition of
N2. (c) The new equilibrium position for the system containing more
N2 (due to addition of N2), less H2 and more NH3 than the mixture in
(a).
Example 6.4
– Arsenic can be extracted from its ores by first reacting the ore
with oxygen (a process called roasting) to form solid As4O6,
which is then reduced with carbon:
As4O6(s) + 6C(s) ↔ As4(g) + 6CO(g)
– Predict the direction of the shift of the equilibrium position for
this reaction in response to each of the following changes in
conditions.
– a. Addition of CO
– b. Addition or removal of C or As4O6
– c. Removal of As4
Solution
– a. Le Châtelier’s principle predicts that the shift will be away
from the substance whose concentration is increased. The
equilibrium position will shift to the left when CO is added.
– b. Since the amount of a pure solid has no effect on the
equilibrium position, changing the amount of C or As4O6 will
have no effect.
– c. If gaseous As4 is removed, the equilibrium position will shift
to the right to form more products. In industrial processes the
desired product is often continuously remove from the
reaction system to increase the yield.
The Effect of a Change in Pressure
– Basically, there are three ways to change the pressure of a
reaction system involving gaseous components at a given
temperature:
1. Add or remove a gaseous reactant or product at constant
volume.
2. Add an inert gas (one not involved in the reaction) at constant
volume.
3. Change the volume of the container.
– When an inert gas is added at constant volume, there is no
effect on the equilibrium position. The addition of an inert gas
increases the total pressure but has no effect on the
concentrations or partial pressures of the reactants or
products (assuming ideal gas behavior). Thus the system
remains at the original equilibrium position.
– When the volume of the container is changed, the
concentrations (and thus the partial pressures) of both
reactants and products are changed. We could calculate Q and
predict the direction of the shift.
– The central idea is that when the volume of the container
holding a gaseous system is reduced, the system responds by
reducing its own volume. This is done by decreasing the total
number of gaseous molecules in the system.
– To see how this works, we can rearrange the ideal gas law to
give
V = (RT/P)n
– or at constant T and P
V∝n
– That is, at constant temperature and pressure, the volume of
a gas is directly proportional to the number of moles of gas
present.
– Suppose we have a mixture of gaseous nitrogen, hydrogen,
and ammonia at equilibrium (Fig. 6.8). If we suddenly reduce
the volume, what will happen to the equilibrium position? The
reaction system can reduce its volume by reducing the
number of molecules present.
– Consequently, the reaction
N2(g) + 3H2(g) ↔ 2NH3(g)
– will shift to right, since in this direction four molecules (one of
nitrogen and three hydrogen) react to produce two molecules
(of ammonia), thus reducing the total number of molecules
present. The phenomenon is illustrated in Fig. 6.9.
– The opposite also true. When the container volume is
increased, the system will shift in the direction that increases
it volume. An increase in volume in the ammonia synthesis
will produce a shift to the left to increase the total number of
gaseous molecules present.
FIGURE 6.9
(a) A mixture of NH3(g), N2(g), and H2(g) at equilibrium. (b) The
volume is suddenly decreased. (c) The new equilibrium position for
the system containing more NH3, less N2, and less H2. The reaction
N2(g) + 3H2(g) ↔ 2NH3(g) shift to the right (toward the side with
fewer molecules) when the container volume is decreased.
FIGURE 6.10
(a) Brown NO2(g) and colorless N2O4(g) at equilibrium in a syringe. (b)
The volume is suddenly decreased, giving a greater concentration of
both N2O4 and NO2 (indicated by the darker brown color). (c) A few
seconds after sudden volume decrease, the color becomes a much
lighter brown as the equilibrium shifts from brown NO2(g) to
colorless N2O4(g). This is predicted by Le Châtelier’s principle, since in
the equilibrium 2NO2(g) ↔ N2O4(g) the product side has the smaller
number of molecules.
Example 6.5
– Predict the shift in equilibrium position that will occur for each
of the following processes when the volume is reduced.
– a. The preparation of liquid phosphorus trichloride:
P4(s) + 6Cl2(g) ↔ 4PCl3(l)
– b. The preparation of gaseous phosphorus pentachloride:
PCl3(g) + Cl2(g) ↔ PCl5(g)
– c. The reaction of phosphorus trichloride with ammonia:
PCl3(g) + 3NH3(g) ↔ P(NH2)3(g) + 3HCl(g)
Solution
– a. Since P4 and PCl3 are a pure solid and a pure liquid,
respectively, we need to consider only the effect of the
decrease in volume on Cl2. The position of the equilibrium will
shift to the right, since the reactant side contains six gaseous
molecules and product side has none.
– b. Decreasing the volume will shift the given reaction to the
right, since the product side contains only one gaseous
molecule and the reactant side has two.
– c. Both sides of the balanced reaction equation have four
gaseous molecules. A change in volume will have no effect on
the equilibrium position. There is no shift in this case.
The Effect of a Change in Temperature
– It is important to recognize that although the changes we
have just discussed may alter the equilibrium position, they do
not alter the equilibrium constant (assume ideal behavior).
– The effect of temperature on equilibrium is different,
however, because the value of K changes with temperature.
We can use Le Châtelier’s principle to predict the direction of
the change.
– The synthesis of ammonia from nitrogen and hydrogen
releases energy (is exothermic). We can represent this
situation by treating energy as a product:
N2(g) + 3H2(g) ↔ 2NH3(g) + energy
– If energy in the form of heat is added to this system at
equilibrium, Le Châtelier’s principle predicts that the shift will
be in the direction that consumes energy, in this case to the
left.
– Note that this shift decreases the concentration of NH3 and
increases the concentration of N2 and H2, thus decreasing the
value of K. The experimentally observed change in K with
temperature for this reaction is indicated in Table 6.3. The
value of K decreases with increased temperature, as
predicted.
– On the other hand, for a reaction that consume energy (an
endothermic reaction), such as the decomposition of calcium
carbonate,
Energy + CaCO3(s) ↔ CaO(s) + CO2(g)
– an increase in temperature will cause the equilibrium to shift
to the right and the value of K to increase.
– In summary, to use Le Châtelier’s principle to describe the
effect of a temperature change on a system at equilibrium,
treat energy as a reactant (in an endothermic process) or a
product (in an exothermic process), and predict the direction
of the shift as if an actual reactant or product is added or
removed.
– Although Le Châtelier’s principle cannot predict the size of the
change in K, it can correctly predict the direction of the
change.
– Table 6.4 shows how various changes affect the equilibrium
position of the endothermic reaction N2O4(g) ↔ 2NO2(g)
Shifting the N2O4(g) → 2NO2(g) equilibrium by changing the
temperature. (a) At 100°C the flask is definitely reddish brown due to
a large amount of NO2 present. (b) At 0°C the equilibrium is shifted
toward colorless N2O4(g).
(a) 100°C
(b) 0°C
6.9 Equilibria Involving Real Gases
– To gain some appreciation for the effect of nonideal behavior
on the calculation of equilibrium constants, consider the data
in Table 6.5, which show the value of Kp (at 723 K) for the
reaction
N2(g) + 3H2(g) ↔ 2NH3(g)
– calculated from the (uncorrected) observed equilibrium
pressure (Pobs) at various total pressures. Note that Kpobs,
defined as
obs)2/(P obs)(P obs)3
N2
H2
3
Kpobs = (PNH
– increase significantly with total pressure. This result make
sense in view of the fact that, as we discussed in Section 5.10.,
Pobs < Pideal for a real gas at pressure above 1 atm.
– One common method for finding the limiting value (the “true”
value) of Kp is to measure Kp at various value of total pressure
(constant temperature) and then to extrapolate the results to
zero pressure.
– Another way to obtain the true value of Kp is to correct the
observed equilibrium pressures for any nonideal behavior.
– We might represent the activity of the ith gaseous component
as
ai = γiPiobs/Pref
– Where γi is called the activity coefficient for correcting Piobs to
the ideal value. In general, for equilibrium pressures of 1 atm
or less, the value of Kp calculated from the observed
equilibrium pressures is expected to be within about 1% of
the true value. However, at high pressures the deviations can
be quite severe, as illustrated by the data in Table 6.5.
Homework
– .
93