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Unformatted text preview: CHAPTER 4 M 4.1 INTRODUCTION Key idea: You
must be able to ex—
ploit known equilib
ria to calculate species concen
trations efﬁciently Manipulating Equilibrium
Expressions You learned in Chapter 3 that the concept of chemical equilibrium has a
thermodynamic basis. At this point, you can write equilibrium expressions
and calculate equilibrium constants from Gibbs free energy of formation
values. Given enough thermodynamic data, you could caICulate the
equilibrium constant for any equilibrium. However, there are many instances in environmental engineering and
science where you do not have sufﬁcient thermodynamic data to calculate
the equilibrium constant for an equilibrium of interest to you. However,
you may know equilibrium constants for related equilibria. How can
information about similar equilibria be used to determine the concentra—
tions of chemical species at equilibrium? To calculate species concentra
tions efficiently, you must be able to exploit known equilibria. Throughout
this text, you shall be manipulating systems of equilibrium expressions to
calculate species concentrations at equilibrium. In this chapter, some
simple rules for the manipulation of equilibrium expressions will be
presented. Three points will be emphasized throughout this chapter. First, as
discussed in Section 4.2, chemical equilibria can be interpreted as both
chemical expressions and mathematical Statements. Although the chemical
and mathematical forms of equilibria are used for different purposes, they ' should convey consistent information about the way in which species and species concentrations relate to the equilibrium expression. Second,
equilibrium constants can have units associated with them (Section 4.3).
Third, kno ledge of an equilibrium constant for one reaction allows for the
calculation? he equilibrium constants for a number of related reactions. To
calculate additional equilibrium constant, you must master the skills of
reversing (Section 4.4), changing the stoichiometric constants (Section
4.5), and adding equilibria (Section 4.6). With these skills, new equilibria and equilibrium constants can be created, as discussed in Section 4.7. 67 63 A PROBLEMSOLVING APPROACH TO AQUATIC CHEMISTRY 4.2 CHEMICAL AND MATHEMATICAL FORMS OF EQUILIBRIA Key idea: Equilib—
ria can be written as
chemical expressions ‘ and mathematical equations Key idea: Chemical
expressions show the
reactants, products,
and stoichiometric relationship
between chemical species a. Key idea: Chemical
expressions do not
show the relation—
ship between chemical species
concentrations _ Key idea: Never
place brackets ([3) or
braces ({ }) around
chemical species names when
writing chemical expressions 4.2.1 Chemical expressions
Chemical equilibria can be written in two ways: chemical expressions and mathematical expressions. Chemical expressions are idealized ways of
writing the chemical reactions that occur at equilibrium. As an example,
consider the dissociation of ammonium (NHJ) to'arnmonia (NH3) and F.
For this chemical equilibrium, you can write the chemical expression NH4+=Nn3+rr Chemical expressions have several important characteristics. First,
chemical expressions describe the relationships between chemical species.
In particular, chemical expressions show the reactants, products, and
stoichiometry of the chemical reactions. For the ammoniumlammonia
example, the chemical expression shows that each mole of ammonium that
dissociates produces 1 mole of ammonia and 1 mole of H3 By convention,
reaction stoichiometric coefﬁcients equal to 1 are omitted in the chemical
expressions. Second, the = symbol in the chemical expression signiﬁes that the
reaction is at equilibrium. Thus, the chemical expression tells us that the
reaction is reversible and spontaneous. This means that the reactions
proceed in each direction. In this way, the symbol = is not a mathematical
sign of equality, but rather a shorthand notation for .—_~_ In the example, the
chemical expression reveals that ammonium dissociates to form 1 mole of
ammonia and 1 mole of H and also that for each mole of ammonia and Il+
that combine, 1 mole of ammonium is formed. Third, chemical expressions have a serious limitation. They do not tell
you the relationship between chemical species concentrations. In the
ammonium example, the chemical expression does not mean that the
concentration of ammonium equals the sum of the concentrations of
ammonia and Hi (i.e., the chemical express does not imply that [NHJ] = [N113] + {H1} When writing chemical expressions, never place brackets
or braces around the names ofthe chemical species. 4.2.2 Mathematical equations Equilibria also can be written as mathematical equations. To write an
equilibrium as a mathematical equation, set the equilibrium constant equal
to the product of the product species concentrations raised to their
stoichiometric coefﬁcients divided by the product of the reactant species
concentrations raised to their stoichiometric coefﬁcients (as in Section 3‘9_1).i For the example above: T In this chapter, the term stoichiometric coeﬁicient will be used to mean the
stoichiometric coefﬁcients in reactions (i.e., reaction stoichiometric coefﬁcients). #——'_—_—_ Example 4.1 : Interconversion of Chemical Expressions and
Mathematical Statements Write the mathematical form of
the equilibrium given by
FeOH3(s) = Fe“ + SOH’
Write the chemical expression
for the equilibrium given by:
F6390 4): (5)
K :[Fe“13[P043—11 Solution: ‘
For the ﬁrst equilibrium, set K equal to the product of the spe—
cies concentrations raised to
their signed stoichiornetric
coefﬁcients (assuming
concentrations are nearly equal
to activities). Thus K= [Fe3‘][0H‘}3I[Fe0H,(s)] For the second equilibrium, interpret the numerator term as
a product and the denominator
terms as reactants. The chemi— cal expression is
illi‘e2+ + 2130,1 ‘ =
F33(PO4)2(3) , Key idea: Chemical
expressions of
equilibria and math—
ematical equations of equilibria
contain the same information
and must lead to consistent
conclusions Le Chatelier’s principle: a
change to a thermodynamic
function controlling the equilib—
rium will cause the system to
adjust to minimize the change 4. MANIPULATING EQUILIBRIUM EXPRESSIONS 69 K = [NH3]EH+1/[NH4+1 The equationK = [NH3][H+]/[NH4+1 is the proper mathematical relationship
between species concentrations. In Chapter 3, the symbol v, for the stoichiometric coefﬁcient of species 1‘ in an equilibrium was introduced. Recall that stoichiometric coefﬁcients
of reactants are considered negative and stoichiometric coefficients of
reactants are considered positive. Thus, you can write in general: K =r1s} “i mun where {i} and [t] are the activity and concentration of species i, respec
tively. The ammonium example demonstrates the relationship between chemical expressions of equilibrium and mathematical statements of equili
brium. In forming the mathematical equations, you are translating the chemical expressions into mathematical equations. Two other illustrations
are shown in Example 4.1. It is important to be able to translate quickly between chemical expressions and mathematical equations. 4.2.3 An example of chemical expressions and mathematical
equations: Le Chatelier’s principle Chemical expressions and mathematical equations are two ways to write
the information contained in chemical equilibria. As you have seen, each
approach has its uses. However, the two ways of writing equilibria must
lead to consistent conclusions. Throughout this chapter, you will see that
chemical expressions and mathematical equations must be consistent in the
trends they predict in'species concentrations. One example in which predictions from chemical expressions and
mathematical equations can be compared is Le Chatelier’sprinchtle (after
Henry—Louis Le Chatelier, 1850—1936; see the Historical Note at the end
of the chapter). For a system initially at equilibrium, Le Chatelier’s
principle states that a change to a thermodynamic function controlling the
equilibrium (e.g., a change in the system temperature, system pressure, or
species concentration) will cause the system to adjust to minimize the
change. Le Chatelier’s principle usually is applied to chemical expres
sions. For the ammonium/ammom‘a example, assume the system is at
equilibrium. Now, imagine that the Hir concentration is increased. T in this chapter, it will be assumed that the solutions are dilute. Thus, activities and concentrations are nearly equal, and
tions for activities in equilibrium expressions (see Section 3.9.1). you can use concentrations as approxima— l
l
l
E 70 A PROBLEMSOLVlNG APPROACH TO AQUATIC CHEMISTRY _———I—l—'— Example 42: Le Chatelier’s
Principle Using Le Chatelier’s principle,
explain why bleach manufactur
ers bubble chlorine gas into
sodium hydroxide solutions
(rather than into pure water)
when making bleach. The per
tinent equilibria are C12(g) + H20 = HOC1+ H? + C1"
H20 = H+ + OH’ Solution: From the ﬁrst equilibrium, the
bleach (here, HOCl or hypo—
chlorous acid) concentration is
increased when (311(g) is in
creased and] or when [H’] is de—
creased. From the second
equilibrium, [FF] is decreased
when [OI1'] is increased (since
the activity of water, a pure
liquid, is ﬁxed; see Section
4.3.2). Thus, bleach manufac—
turers add NaOH to increase the
HOC1 concentration (thus saving
money by reducing the mass of
water shipped per kg of bleach). __________._._._.———————————~ Thoughtful Pause
What will happen to the NHJHJ“ [NI14+ equilibrium if the
concentration of H+ is increased? _._——————‘—"—‘"—"—_'—‘_— An increase in the El concentration will shift the system away from
equilibrium. According to Le Chatelier’s principle, the system will adjust to
minimize the impact of the increase in the 1L1)r concentration. To accomplish
this, the equilibrium (NFL,+ = NH3 + H”) will shiﬂ to the left, with a
resulting increase in the ammonium concentration. is Le Chatelier’s principle consistent with the mathematical version of
equilibrium information? In the mathematical expression (K =
[NH3][H+]/'[NH4+]), K is constant at constant temperature and pressure.
Think of this equation as purely a mathematical construct divorced from the
chemistry: K = constant = xy/z. If you make y larger but keep K constant,
then one of the following statements must be true: (1) x decreases, (2) z
increases, or (3) both it decreases and 2 increases. Back to the chemistry: to
maintain a constant amount of N in the system, you cannot decrease the
ammonia concentration without increasing the ammonium concentration by
an equal amount. Similarly, you cannot increase [NHJ] without decreasing
[NH3] by an equal amount. Thus, the ammonia concentration must decrease
and the ammonium concentration must increase. In the symbols used
previously, at decreases and 2 increases. The mathematical statement of
equilibrium tells you that the ammonium concentration must increase (and
the ammonia concentration must decrease) as H" is increased. Thus, Le
Chatelier’s principle (usually applied to the chemical expression) is
consistent with the mathematical version of equilibrium information.
Another instance of Le Chatelier’s principle is shown in Example 4.1T 4.3 UNITS OF EQUILIBRIUM CONSTANTS Key idea: Equilib—
rium constants are
dimensionless, but
units are associated with equi
librium constants as a reminder
of the units of the species in the
equilibrium 4.3.1 Introduction What are the units of equilibrium constants? Recall from Section 3.9.2 that
equilibrium constants are dimensionless, because each term in the equilib
rium constant is normalized to a standard state having the same concentration
(or activity) units as the species concentration. You can write mathematical
expressions such as iog(K) and AG” = —RTln(K) without concern about the units of K. T Le Chatelier’s principle is a powerful tool for understanding the qualitative
behavior of simple chemical systems. The great American chemist and double Nobel
laureate Linus Pauling wrote (Pauling, 1964): “When you have obtained a grasp of
Le Chatelier’s principle, you will be able to think about any problem of chemical
equilibrium that arises, and, by use of a simple argument, to make a qualitative
statement about it... Some years after you have ﬁnished your college work, you may
have forgotten all the mathematical equations relating to chemical equilibrium.
I hope, however, that you have not forgotten Le Chatelier’s principle.” Example 4.3: Units Associated
with Equilibrium Constants Using typical concentration u
nits, what units should be asso—
ciated with the ﬁrst equilibrium
in Example 4.2? The equilib—
rium is
C12(g) + H20 = HOC1+
H+ + Ct Solution: Typical concentration units for
gases, pure liquids (such as
water), and dissolved species
are atm, none, and moi/L, re
spectively. Thus, you would
associate the units of m013tL3—
atm (or Matatm) with the ﬁrst
equilibrium in Example 4.2. 4. MANIPULATING EQUILIBRIUM EXPRESSIONS 71 Although equilibrium constants are inherently dimensionless, it is
convenient to associate units with equilibrium constants to reﬂect the
concentration units ofthe species in the equilibrium expression. This is an
accounting trick to aid in manipulating equilibria and will be used
occasionally throughout this text. In addition, writing units with equilib
rium constants allows one another way to check whether the manipulation
of the equilibria and equilibrium constants was performed properly; the
units associated with the final equilibrium constant should match the final
equilibrium expression. 4.3.2 Concentration scales To be able to associate units with equilibrium constants, you must choose
a consistent set of concentration scales. Four types of species are found
commonly in equilibria in aqueous systems: dissolved species, pure solids,
pureliquids, and gases. As discussed in Chapter 2, dissolved species
concentrations in dilute solution usually are exp'resscd using the molarity
concentration scale. Thus, dissolved species concentrations commonly are
written in molfL = M. For example, the equilibrium constant for ammo
nium dissociation (K = [NH3][H+}/[NH4+]) may be written with units of
MZIM = M to indicate the units of [NH3], [E], and [NHK]. Pure solids and pure liquids usually employ the mole fraction
concentration scale. Pure solids and pure liquids have a mole fraction of
unity. As a result, pure solids and pure liquids (e. g., water) do not affect
the units of equilibrium constants Thus, for the reaction CaCO 3(s)= Ca2
+ C032" , you can write K = [Ca2+}[CO32']/[CaC03(s)] = {C321 [C03] The equilibrium constant K usually has associated units of M2. You shall
frequently make use of the following reaction: 1120= H” + OH‘. For this
reaction, you can write = [H+i[0H']/[H20} = [H1 [OH'] where K usually has associated units of M2.
Gases usually employ the partial pressure concentration scale. Thus,
gases commonly have concentrations of aim. For example, for the reactibn
03(aq)= 03(g), you can write K= [03(g)]/[O3(aq)] with typical associated
units of atm/ M atmL/mol Another illustration of associating units with
equilibrium constants may be found in Example 4.3. 4.3.3 Using units with equilibrium constants Recall that a general rule of quantitative analysis is to avoid comparing
values with different units. For example, it is improper to conclude that an
automobile withp: gas mileage of 30 miles per gallon is more fuel efﬁcient E
E
g 72 A 13ROBLEMSOLV1NG APPROACH TO AQUATIC CHEMISTRY Key idea: Take
great care when
comparing equili—
brium constants with different
units than an automobile with a gas mileage of 12.8 km/L. In fact, the gas mileage is about the same for the two vehicles. Similarly, you cannot compare the
values of equilibrium constants with dijjbrentassociated units. For example, K for the equilibrium H31304 = P03— + 3H+ is 3.2><10’22 and K for the equilibrium HCN = CN’ + H+ is 6.3X10'“. Can we use this information to
compare the ability of phosphoric acid (H3PO4) and hydrocyanic acid (HCN)
to produce H”? The answer is no: the equilibrium constant for the ﬁrst
equilibrium has units of M3 (: mol3/L3) associated with it, whereas the
equilibrium constant for the second equilibrium has units of M associated with it. The equilibrium constants have different units and cannot be compared. The units associated with equilibrium constants are a reminder
that you should not compare values with different units. 4.4 REVERSING EQUILIBRIA reversing an equilibrium: an
equilibrium is reversed when its
reactants and products are inter—
changed a e \ Key idea: AGﬂm
vaiues of reverse
reactions are equal in
absolute value but opposite in
sign and K values of reverse '
reactions are reciprocals of one
another One of the takehome lessons from this chapter is that knowledge of one
equilibrium constant allows you to calculate easily the equilibrium
constants of related equilibria. In this Section, you will see that knowing
an equilibrium constant allows for the calculation of the equilibrium
constant for the reaction formed when reactants and products in the original
reaction are interchanged. Interchanging reactants and products is called
reversing an equilibrium. _ For the example equilibrium of this chapter (NH: = NH3 + If), the
equilibrium constant, K, is about SXlO‘w M = 109'3 Mi What is the
equilibrium constant for the reverse reaction (NIl3 ¥ IrI+ = NHf)? You
can answer this question in two ways: by examining the chemical
expression and by examining the mathematical equation for K. From the
chemical expression, it is clear that AG” for the ammonia/H+ association rxn reaction is —l times 13ng for the ammonium dissociation reaction. Thoughtful Pause
Why are the AG?“ of reverse reactions equal in absolute value
but opposite in sign? ' T Equilibrium constants can take values over many orders of magnitude. In
addition, you know that log(K) values are proportional to AG?“ and therefore have
special meaning. Thus, we usually express K values as the antilog base 10 ; in other "
words, as 10‘. M
Example 4.4: Reversing Equilibria Find the equilibrium constant
for HgCl+ = Hg2+ + Cl“ if
AG?” ‘4 —38.54 kJ/mol for the
equilibrium: Hg2+ + Cl“ = HgCF' Solution: If AG?“ for the equilibrium Hg2+
+ or = chr is —33.54
kJ/mol, then AG?“ for there
verse reaction (HgCl+ = Hg2+ +
Cl‘) is +3854 kJ/mol. Thus, K for HgCF dissociation at 25°C is K = exp(—AG‘:m/RD
= 1.76X107 Alternatively, you could calcu—
late the equilibrium constant for
the equilibrium lIg2+ + Cl" =
HgCl+ from AG?“ (K =
5.70X106) and note that the
equilibrium constant for the
equilibrium HgCl+ = Hg2+ +
C1' is 1/’5.70><106 = 1.76X0‘7. ______‘ 4. MANH’ULATING EQUELIBRIUM EXPRESSIONS 73 For any reaction, AG ﬁn, £21453. = 2 lid'5 71, — Z inICTDJ» Thus, products reactants
AG?” of reverse reactions are equal in absolute value but opposite in sign,
since the reactants and products are simply interchanged for reverse
AGI'DJ'” reactions. Now, since K = e R T , K for the reverse reaction must be equal
to the reciprocal of K for the original reaction. For the example, K for the
NH3 + H+ association reaction is l/(K for the ammonium dissociation
reaction) = 1/10‘9‘3 M3 10”"3 M'l. Thus, examination of the chemical
expression tells you that K for one reaction is the reciprocal of K for the
reverse reaction. The mathematical form of the equilibrium also should yield that K for
one reaction is the reciprocal of K for the reverse reaction. This is obvious
by inspection. For example, the ammonium dissociation reaction has an
equilibrium constant equal to [NH3]{H*]/[NH4+], whereas the equilibrium
constant for the ammoniale association reaction is {NHfj/(INHQ [Hi]; K
for the NH3 + H+ association reaction is the reciprocal of K for the amine—
nium dissociation reaction. . You can Show that K for one reaction is the reciprocal of K for the
reverse reaction in the general case by using the product notation of Section
4.2.2. For any equilibrium (where activity and concentration are nearly interchangeable): K = HM” . For the reverse reaction, vim, = —v,. since
1' reactants and products are reversed in the reverse reaction. Thus: . v7 . —y 1
Krev =Hlll =Hll] ‘ =K' See also Example 4.4. 4.5 EFFECTS OF STOICHIOMETRY 4.5.1 Linear independence
Compare the following two reactions: NIL+ = NH3 + H+ with K = K; and
2NH4+ 2 2NH3 + 2H+ with K = K2. The chemical expressions convey the
same information: for each mole of ammonium dissociated, 1 mole of
ammonia and 1 mole of HJr are formed. (In addition, for each mole of
ammonia associating with 1 mole of Hi 1 mole of ammonium is formed.) Since the chemical expressions NH: = NH3 + H“ and ZNIL,+ = ZNH3
+ IZIIJr contain the same information, it would be improper to use both
equilibria in describing a chemical system. Why? You cannot use more than
one expression that contains the same information in any physical system.
For example, if you were solving an algebraic system for the values of x and
y, you could not use both the equation x + 2y 2 5 and the equation 3x + 6y =
15; the two equations convey the same information. When two equilibria differ only by a constant multiple of their stoichio
metric coefﬁcients, we say the equilibria are linearly dependent. (The 74 A PROBLEMSOLVING APPROACH TO AQUATIC CHER/[ISTRY Example 4. 5: Changing
Stoichiometry Find the equilibrium constant
for HCN Z H+ + CN' if AG?“ for the reaction VzHCN = V2H+
+ 1/2 CM" is +27 lemol. Solution: If AG?“ for the reaction 1/2HCN
= 1/2H+ + ‘ACN' is +27 kJ/mol,
then AG:m for the reaction
HCN = m+ CN’ is 2(+27
kJ/mol) = +54 kJJ'rnol. Thus, K
for HCN = H“ + CN‘ at 25°C
is K = exp(AGan/RT)
= exp{(—54 kJ/m01)/
[(8.314X10'3kllmol
°K)(298°K)}}
= 3.42X10_w You also could calculate the
equilibrium constant for 1AzHcN
= ‘/2H+ + V2CN’ from AG?“ (K
= 1.85X10'5) and note that the
equilibrium constant for HCN
= H++ CN' is (1.85X10'5)2 =
3.4ZXEO‘ 1°. —ll————__ 3 Key idea: Multiply
ing stoichiometric
coefﬁcients by a con
stant multiplies AG; by the
constant and raises K to the
constant definition of linear dependence is expanded in Section 4.6.2.) You will need to seek a set of linearly independent equilibria to describe a chemical
system. 4.5.2 Free energy and equilibrium constants
Return to two reactions: NH“+ = NH3 + H+ with K = K1 and ZNHJ =
2NH3 + 2H+ with K = K2. What is AG:m for each equilibrium? Since AG:m =21}: (719,1. , the standard Gibbs free energy of the second equilibrium should be twice the standard Gibbs free energy of the ﬁrst equilibrium: doubling the stoichiometric coefﬁcients values doubles AG?”
What about K? You know that K2 : exp(mAG:m,2/RD
= exp(—2AG‘:M_1/RI)
= [exp(_AGonm,i/RD]2
= Klz Thus, K for 2NH4+ = ZNH3 + 2H” is the square of K for NH4+ = NH3 +
if. Is this result consistent with the mathematical form of the equilibrium
information? For the ﬁrst equilibrium, K] = [NH3] [HTENHJ], whereas for I the second equilibrium K2 2 [NH3]2[H+]2I[NH4+]2 = K12. As expected, the mathematical form of the equilibria also reveals that K for the second
equilibrium is the square of K for the ﬁrst equilibrium. In general, let K be
the equilibrium constant for an equilibrium. If you multiply each of the
stoichiometric coefficients by c, then the equilibrium constant for the new equilibrium isH[i]"“’i = (mt) =Kc.
l l The effects of stoichiornetry can be summarized as followa Multiply
ing each stoichiometric coefﬁcient in an equilibrium expression by a
constant does not convey any new information about the equilibrium.
However, multiplying stoichiometric coefficients by a constant, 0, results
in an equilibrium with AG?” equal to 0 times the AG?“ for the original
equilibrium and K equal to K for the original equilibrium raised to the c
power. You shall ﬁnd that it is sometimes convenient to track log(K)
instead of K. Multiplying stoichiometric coefﬁcients by a constant, c,
multiplies AG?“ by c and multiplies log(K) by 0 [since clog(K) = log(K“")].
The effects of changing stoichiometric constants also are shown in
Example 4.5. It should be noted that reversing equilibria (discussed in Section 4.3)
is really a special case of changing reaction stoichiometry. Recall that we
usually think of stoichiomctric coefﬁcients for reactants as negative and
stoichiometric coefficients for products as positive. Therefore, reversing
an equilibrium (i.e., switching reactants and products) is equivalent to E ' 4. MANIPULATING EQUILIBRIUM EXPRESSIONS 75 multiplying each stoichiometric coefficient by — 1. Thus, in reversing an ,. 3‘
equilibrium, the sign of the Gibbs free energy of reaction is multiplied by z‘? i‘
—l and the new equilibrium constant is the reciprocal of the original
equilibrium constant (i.e., K is raised to the —l power). 4.6 ADDING EQUILIBRIA 4.6.1 Free energy and equilibrium constants ' ‘
You can form new equilibria by adding existing equilibria. For example, 3? ‘
how would you create a new equilibrium to show the relationship between
ammonium, ammonia, hydroxide ion (OH'), and water? You can
accomplish this by adding the following two equilibria: NH; = NH3 + W reaction 1
H“ + OH" = H2O reaction 2 Adding:
NH4++H++OH'=NH3+H++H,O E; l l Eliminating the common species on both sides of the equilibrium (here,
only IF appears on both sides): NH4" + on 2 NH3 + H20 reaction 3 E '; How do AG” and K for the new equilibrium (reaction 3) compare to AG” rm 11m and K for the original cquilibria (reactions 1 and 2)? You know from .
Section 3.2.2 that free energies are additive. Thus, you might expect that i
ACE”,3 = AG“ + AG" ,2. For the equilibrium constant: rm,l rxn i ' K3 =exp(—AG7.,.,3/Rn
‘ 2 exp[—(AG:W],1 + AG?xn,2)/Rn I I i
= 6199(AGlxn,1/RDBXP(_AGZH,2/RD E =K,K, i '
E Therefore, adding equilibria results in a new equilibrium with Gibbs ﬂee
. Key idea: When energy equal to the sum of the Gibbs free energies of the individual
E ’ adding equilibria, equilibria and with an equilibrium constant equal to the product of the
1 add AG"ml values and individual equilibrium constants. Recall that if K3 = KIKZ, then log(K3) =
! multiply K values [or add logUCl) + log(K2). Thus, when adding equilibrium expressions, add AG“
5 rxn 10800 V6131163] values and multiply K values or, equivalently, add A612,, values and add
log(K) values. This analysis points a nice trick involving log(K) values: i
logﬂ?) values are additive when equilibria are added. With this fact, you
often can calculate equilibrium constants for new equilibria in your head. The mathematical form of the equilibria should give the same result.
Indeed: 76 A PROBLEMSOLVING APPROACH TO AQUATIC CHEMISTRY linearly independent equilibria:
a set of equilibria where no
equilibrium expression can be
formed by linearly combining
some or all of the other
equilibria in the set K1 = {NH31[H1/[NH4+]
K2 = [HZO}/{H+][OH'], and: K3 = [NHallH20l/iNH4Wl0H'] It is clear thatK3 =K1K2. In general, ile = [Hi] V‘ and K2 2 H[i] ”I then
r 1‘ K3, the equilibrium constant formed by adding reactions 1 and 2, is given
byniilviw' = K1K2> 4.6.2 Linear independence revisited It is important to note that reaction 3 does not convey any information in
addition to the information conveyed by reactions 1 and 2. In fact, given
any two of reactions 1, 2, or 3, it is possible to create the other reaction by
judiciously adding and/ or reversing the reactions in hand. It is possible to
extend the concept of linearly independent equilibria developed in Section
4.4: a set ofequilibria is linearly independent if no equilibrium expression can be formed by linearly combining some or all of the other equilibria in 7 i the set. Here, linearly combining means reversing equilibria, multiplying
stoichiometric coefficients by a constant, and! or adding equilibria. Another
illustration of linear independence is shown in Example 4.6. 4.7 CREATING EQUILIBRIA Example 4.6: Linear Indepen—
dence Is the following set of equilibria
linearly independent? Al(OH)3(s) = A13++ 30H A13++01r = anon)2+ AKOH)2+ + ZOH’ =
Al(0H)3(S) Solution: The set is not linearly independ—
ent. The third equilibrium
can be formed by reversing
and adding the ﬁrst two equil
ibria: Al3++ 301? = Al(OH)3(s) + Agog)“ : in3+ + OH'
Al(OH)2++20H" = Al(0H)3(s) In performing equilibrium calculations, you often have to manipulate
knoWn equilibria to write a target equilibrium in the form desired. Two
skills are required: balancing chemical reactions and manipulating equilibria. 4.7.1 Balancing chemical reactions
Conservation of mass (Section 3.2.2) requires that chemical reactions are balanced. Balancing reactions is a ﬁve—step process (see also Appendix C,
Section (1.1.2). The process will be illustrated with the equilibrium
between hypochlorous acid (HOCl, a common disinfectant) and chloride (or). Step 1: Write the known reactants on the left and known products
on the right ' In the example, write: HOCl = (31* Step 2: Adjust stoichiometric coefﬁcients to balance all elements
except H and 0 i
E
i
i ”'ﬂI—‘wmmmww'm W , .,_,,__.....,, ._....,.., x Key idea: To bal
ance a chemical
reaction, balance all
elements except H and 0, add
water to balance 0, add 11“ to
balance H, and add e' to bal—
ance the charge Example 4. 7: Balancing
Chemical Reactions Balance the equilibrium
between thiosulfate (82032") and
sulfate (3042‘). Solution: Follow the procedure in the
text: Step 1: Write the known species
3203} = SO42—
Step 2: Balance all but 0, H
32032" = 28042"
Step 3: Add H20 to balance 0
$2032" + 513120 = 28042“
Step 4: Add 1? to balance H
82032" + 5H20 =
28042" + 10H+
Step 5: Add e" to balance the
charge
82032" + SHZO =
2304* + 10H+ + 8c" The balanced reaction is s,0,2 + 511,0 =7
250: + 10H+ + Se‘ 4. MANIPULATING EQUILIBRIUM EXPRESSIONS 77 In the example, all elements except H and 0 already are balanced
and you still have: HOCl = Cl". If you were balancing the
equilibrium between molecular chlorine (C12) and chloride, you
would adjust the stoichiometric coefﬁcients to balance the element
Cl. You could accomplish this in either one of two equivalent
ways: Cl2 = 2Cl' or 1/2Cl2 = Cl". Step 3: Add water (H20) to balance the element 0 In the example, oxygen will be balanced if you add water as a
product: HOCl : Cl" +HZO. Step 4: Add 11* to balance the element H In the example, hydrogen will be balanced if you add one H‘” as a
reactant: HOCl + H+ = C1" + H20. At this point inthe process, all
elements should be balanced. Step 5: Add electrons (e‘) to balance the charge In the example, the leﬁ~hand side has a net charge of +1 and the
righthand side has a net charge of — 1. You must add two e‘ as
reactants: HOCl + H” + 2e" = C1’ + H20. The equilibrium is now
balanced. Although it may seem slow and unwieldy at ﬁrst, it is recommended that
you follow this process step by step when balancing reactions. Another
illustration of balancing reactions is given in Example 4.7. The balancing
procedure is extended in Problem 4.7 at the end of the chapter. 4.7 .2 Manipulating equilibria After balancing the desired reaction, it may be possible to relate it to other
equilibria by reversing equilibria, multiplying stoichiometric coefﬁcients by
a constant, and/or adding equilibria. It is important to master these skills
before you proceed to the mechanics of equilibrium calculations in Part 1] of
this text. Consider the following example. In developing remediation strategies
for a landﬁll leachate, you wish to know the equilibrium constant for the
dissolution of cadmium carbonate solid to form divalent cadmium andcarbon
dioxide gas. You search tables of equilibrium constants and ﬁnd the
following information: CdCO s = Cd2+ + CO 2‘ K = 10l1.66 Mzt l Recall from Section 4.3 that the typical units of solids and liquids are the mole
fraction. The mole fractions of pure solids [e.g., CdCO3(s)] and pure liquids (e.g.,
H20) are unity. Thus, the units of pure solids and pure liquids do not appear in the
equilibrium constants. ‘ *ng? 73 A PROBLEMSOLVlNG APPROACH TO AQUATIC CHEMISTRY HCO3' —— CO32'+PF K2=10“‘°'3M
i H2603 = Hco; +H+ K3=10’5'3M
9 H2CO3 = C02(g)+H20 K4= 1015 atm—L/rnol You can ﬁnd the equilibrium constant for the target equilibrium by (1)
writing the balanced equilibrium, and (2) manipulating the known equili
bria to match the target equilibrium. First, balance the reaction by using
the ﬁve—step process outlined in Section 4.6. (You may Wish to try this
ﬁrst on your own.) Step 1: Write the known reactants on the left and known products a
on the right moors) 2 car + C02(g) E Step 2: Adjust stoichicmetrie coefﬁcients to balance all elements
except H and O ‘ f _ All elements except H and O are balanced in Step 1
‘ Step 3: Add water (1110) to balance the element 0
. . cacors) = Cd2+ + C02(g) +1120
‘ Step 4: Add H” to balance the element H
CdCO3(s) + ZIF = ca2+ ‘l‘ C02(g) + H20
Step 5: Add electrons (e") to balance the charge
The charges are balanced in Step 4. The balanced reaction is: cacogs) + 2m 2 Cd2+ + C02(g) + H20 Now you need to combine the existing equilibria to obtain CdC03(s) + 2H+
= Cd2+ + C02(g) + H20. Start with the K1 equilibrium. It has CdCO3(s)
and Cd2+ in the desired places as a reactant and product, respectively. You
now have: ‘ cursors) = Cdz++C032‘ ‘ K1=10’”'56M2 You do not want carbonate (C032') as a product. Thus, you need to add an
equilibrium with carbonate as a reactant to cancel the carbonate as a
product in the equilibrium you are building. This can be accomplished by
reversing the K2 equilibrium and adding: e“ g Key idea: Equilib—
rium constants can
i" be calculated by
linearly combining other equi—
libria or from AG" ” values H 4. MANIPIEATING EQUILIBRIUM EXPRESSIONS 79 CdCO3(S) = Cd2+ + C0327 Kl : 1041.56 M2
CdC03(S) + PF 3 Cd2+ + HCO3' K6 : KIKS : 104.35 M Notice that you take the reciprocal of K2 when you reverse the K2 equilib
rium. Also, you multiply the equilibrium constants when adding reactions. Again, you do not want bicarbonate (HCO3') as a product. To eliminate
HCO3“, reverse the K3 equilibrium and add: CdCO3(S) + H+ = C(ZIZ‘lw + HCO3_ K6 2107 ['35 M
w
CdCO3(s) + 2Ht = CdB + H2003 K3 = 6K? 2 104.94 Finally, add the K4 equilibrium to eliminate H2CO3 and have C02(g) as a
product: Cdco,(s) +2er+ = Cd2+ + 1er003 K, =10“94
+H co = C0 +H 0 K =1015 atm—L/mol
Cdco,(s) + 2H+ m Cd2+ + CO2(g) + H20 '
m=mm
= 106'44 atmL/mol Thus, the target equilibrium (cadmium carbonate solid in equilibrium with
divalent cadmium and carbon dioxide gas) has an equilibrium constant
equal to 106'44 atmL/mol. 4.7.3 Why learn how to manipulate equilibria? In Section 4.7.2, you calculated an equilibrium constant for a target
equilibrium by linearly combining other equilibria. Thoughtful Pause
From Chapter 3, is there another way to calculate the
equilibrium constant for the target equilibrium? You know from Chapter 3 that you can calculate K from exp(~AGfx,,/RT).
The standard Gibbs free energy of formation values for CdCO3(s), H‘”, Cd“,
C02(g), and H20 are, respectively, #6694, 0, —77 .58, —394.37, and
—237.18 kJ/mol. These values give ASL” = —39.73 kJ/mol orK= 106'96 at
25°C. This is about three times larger thanthe value of K = 106"34 atm
L/rnol determined in Section 4.7.2. Such discrepancies are not unusual,
given the uncertainty in the thermodynamic data and errors in the
equilibrium constants of the constituent equilibria. It is important to
become adept at calculating equilibrium constants both from linearly
combining other equilibria and from thermodynamics. 80 A PROBLEMSOLVING APPROACH TO AQUATIC CHEMISTRY Key idea: You
must iearn to manip—
ulate equilibria since
thermodynamic data (equili—
brium constants and AGf" val—
ues) are not always available Why should you learn how to calculate equilibrium constants by
manipulating known equilibria when you can always calculate equilibrium
constants from thermodynamics? In a reasonably complex aqueous
system, it is unlikely you will be able to look up all required equilibria in
the form you want. In addition, the AGf " values may not be available for
the species of 1nterest. If AG“ values are not available, then AG?“ cannot be calculated. Thus, it is extremely important to be able to manipulate
equilibria to calculate equilibrium constants. 4.7.4 A systematic approach to manipulating equilibria: an
advanced concept
The process described in Section 4.7.2 of linearly combining equilibria to create a target equilibrium may seem somewhat haphazard at ﬁrst. For
simple systems, it is often expedient to play with the equilibria as jigsaw puzzle pieces until the target equilibrium emerges (as was done in the 
~ cadmium carbonate example). For more complex systems, you may wish to take a more systematic
approach as follows. List the n constituent equilibria that are to be linearly
combined to form the target equilibrium. Let m be the total number of
chemical species in the n constituent equilibria and in the target equilib~
riurn. Write the stoichiometric coefﬁcients for each species in each
equilibrium, using the usual notation that stoichiometric coefﬁcients of
reactants are negative and the stoichiometric coefﬁcients of products are
positive. Let vU be the stoichiometric coefﬁcient of specres j in constituent
equilibrium' 1. Now, multiply the stoichiometric coefﬁcients of constituent
equilibrium 1' by a constant 6,. For each species j, sum the values of V951
over all the equilibria and set the sum equal to the stoichiometric coefﬁ cient of spec1es J for the target equilibrium (=v Urge, 1) :21011’ U — —VU . This will yield a linear set of m equations (one for each species) 1n n unknowns
(cl, ..., an) that can be solved to ﬁnd the cl. values. From the c, values, you
can determine the linear operations to perform on the constituent equilibria
to form the target equilibrium and thus the target equilibrium K. As an example, consider the CdCOU(s) problem of Section 4. 7. 2. The
target equilibrium ls: Cdc0,(s) + 2H+— Cd2+ + co,(g) + H 0 K, Inger: The n = 4 constituent equilibria are: Eq. 1: CdCO3(s) : Cd2++ (3032’ K1 2 104355
Eq. 2: HCO; = (3032‘ + H+ K2 = 10103 _
Eq. 3: HUGO;5 = HC03‘ + H+ K3 2 1075.3
002%) + H20 K4 = 101'5 ll Eq. 4: H2C03 . “ii i
t
E
E
g
E
i
.1
2:
§.
E
i
E 4. MANIPULATING EQUTLIBRIUM EXPRESSIONS 31 The system has m = 8 chemical species: CdCO3(s), Cd“, C02(g), H2003,
HCO3', C032", HF, and H20. Writing the matrix of stoichiometric coef—
ﬁcients: Equil. CdC03(s) Cd“ c_02(g) H2C03 Hco; co; H“ H20 1 el 1 0 0 0 l 0 0
2 0 0 0 0 —1 1 1 0
3 O 0 0 l 1 O l 0
4 0 0 1 e1 0 0 0 1
Target —l 1 1 0 0 0 —2 1 Multiplying the stoichiometric coefﬁcients of constituent equilibrium iby
unknowns cf: Equil. CdC03(s) Cd2+ C02(g) H2603 HC03' cof‘ H+ H20 1 ‘5': cl 0 0 0 cl 0 0
2 0 0 0 0 ecz c2 c2 0
3 0 0 0 c3 c3 0 c3 0
4 O 0 C4 —C4 0 0 0 C4
Target —l 1 l 0 0 O *2 l Summing the columns for each chemical species and setting each sum
equal to the stoichiometric coefﬁcient for the target equilibrium yields
eight linear equations: _cl=_1 64:1 _62+C3=0 62+03:_2
C121 _03_C4=0 6140220 C4=1 This system of linear equations can be solved (by inspection in this simple
case) to reveal: c1 = 1, c2 = —1, c3 = — 1, and 04 T— 1. In other words, to
obtain the target reaction, add equilibrium 1, the reverse of equilibrium 2
(Le, multiply the stoichiometric coefficients of the species in equilibrium
2 by — l), the reverse of equilibrium 3, and equilibrium 4. Since free
energies are additive: A(?‘,'Wm.g‘,_,,f = AGroml — AG"m2 — AG?“3 + 13ij“ and
Kluge, = K1K4/(K2K3). In general, the n linear and independent equations'can
be solved by the methods of linear algebra to determine the n values of CI, ..., c". In matrix notation: i
i
i
e
i
g 82 A PROBLEMSOLVING APPROACH TO AQUATIC CHEMISTRY 4.8 SUMMARY Val V12 Vim ‘31 V21 V22 Vim ‘72 _ ' ~ ' ' _(vrarg2!l Vlargell Vurgem)
1"Isl V111 "' vim cm where, again, Va. is the stoichiometric coefﬁcient of species j in constituent
equilibrium i and Va. is the stoichiometric coefﬁcient of species j in the target equilibrium. In general for this approach: K m“, = ﬁKff .
i=1 Chemical equilibria can be interpreted as both chemical expressions and
mathematical statements. Chemical expressions show relationships
betWeen chemical species and mathematical statements Show relationships
between species concentrations. The chemical and mathematical forms of
equilibria convey similar information about the way in which species and
species concentrations relate to the equilibrium expression. Knowing the equilibrium constant of one reaction allows you to
calculate the equilibrium constants for a number of related reactions.
Reversing an equilibrium results in a new equilibrium with K equal to the
reciprocal of the K of the original equilibrium (Knew = llKold). Multiplying
the stoichiomeuic coefﬁcients of an equilibrium by a constant (say, e)
results in a new equilibrium with K equal to the K of the original equilib
rium raised to the constant (Knew = 01;). Finally, adding two equilibria
results in a new equilibrium with K equal to the product of the equilibrium
constants of the original equilibria (K = KMMKDM). new 4.9 PART I CASE STUDY: CAN METHYLMERCURY BE FORMED
CHEMICALLY IN WATER? Recall that the Part 1 case study involves deciding whether methylmercury,
CH3Hg+, can be formed in water from CH4(aq) and Hg”. To answer this
question, you must write a balanced reaction between reactants and
products. Following the procedures introduced in this chapter, you can
Show that: CH4(aq) + Hg2+ = CHJHg‘” + H . From Section 3.11, the A5? values for CH4(aq), Hg”, and CH3Hg+ are, respectively, 734.4, +164.4, and +1129 kI/mol. The A5; value for HP is
0 kJ/Inol. Thus, AG” = —l7.1 kJ/mol, andK = eXp(—AG‘fm/RD = 103"). mm é 4. MANIPULATING EQUILIBRIUM EXPRESSIONS 83 To see if methylmercury formation is signiﬁcant, you will have to‘do
a' small equilibrium calculation. If concentrations can be used for activities, recall that:
K 2 [CH3Hg+l[Fl/([CHdanngzj Assume for the moment that the only methane—containing species are
CH,,(aq) and CH3Hg+ and the only Hgcontaining species are Hg2+ and
CH3Hg‘L. (This is a weak assumption: other Hgcontaining species likely
exist in this system.) If [CH3Hg+] is given the symbol x, then: [CH4(aq)]
: total methane — x and [Hg2+] = total Hg — x. Thus: K = x[H+]/[(total methane — x)(total Hg _ x)] ForK= 103'“, pH 7, total methane = 10‘5 M, and total Hg = 10’7 M, you can
calculate that the CII3l:lg+ concentration is about 10"7 M. Thus, under these
conditions, the (ElI3Elg+ concentration is small but much larger than the
concentration of Hg2+. SUMMARY OF KEY IDEAS 0 You must be able to exploit known equilibria to calculate species
concentrations efﬁciently O Equilibria can be written as chemical expressions and mathematical
equations 0 Chemical expressions show the reactants, products, and stoichiometric
relationship between chemical species but not the relationship between
chemical species concentrations 0 Chemical expressions do not show the relationship between chemical
species concentrations 0 Never place brackets (U) or braces ({ }) around chemical species names
when writing chemical expressions 0 Chemical expressions of equilibria and mathematical equations of
equilibria contain the same information and must lead to consistent conclusions 0 Equilibrium constants are dimensionless, but units are associated with
equilibrium constants as a reminder of the units of the species in the equilibrium PRINCIPLE HenriLouis Le Chatelier 34 A PROBLEMSOLVING APPROACH TO AQUATIC CHEMISTRY 0 Take great care when comparing equilibrium constants with different
units 0 AG” values of reverse reactions are equal in absolute value but rxn opposite in sign, and K values of reverse reactions are reciprocals of
one another 0 Multiplying stoichiometric coefﬁcients by a constant multiplies AG‘fm
by the constant and raises K to the constant 0 When adding equilibria, add AG?“ values and multiply K values [or
add log(K) values] 0 To balance a chemical reaction, balance all elements except H and 0,
add water to balance 0, add W to balance H, then add e" to balance the
charge 0 Equilibrium constants can be calculated by linearly'combining other equilibria or from AGZM values a You must learn to manipulate equilibria since thermodynamic data
(equilibrium constants and AG]? values) are not always available ‘ HISTORICAL NOTE: HENRILOUIS LE CHATELIER AND LE CHATELIER’S Who was the man behind Le Chatelier’ s Principle? Henri Le Chatelier was
a contemporary of J. Williard Gibbs (see the Historical Note in Chapter 3).
Le Chatelier, son of an engineer, was trained as an engineer at Ecole des
Mines in Paris. He worked for the French government as a mining
engineer in the Corps des Mines for two years. Rather unexpectedly, he
was offered a professorship in chemistry at his alma mater. This was the
beginning of a career in academia that would span 30 years at Parisian
universities, including Ecole des Mines, Ecole Polytechnique, College de
France, and the Sorbonne. In contrast with Gibbs’s theoretical brilliance, Le Chatelier was
interested in the applications of chemistry. His training as a problem—
solving engineer was reﬂected in his accomplishments. Le Chatelier’s
work on cements, thermodynamics, combustion, ammonia synthesis, and
metallurgy was ahnost always motivated by practical problems. The principle for which his name is linked was reﬁned by Le Chatelicr
through a series of papers published in the 18805. This version of the
principle, published in 1880, echoes Newton’s Third Law of Motion:
“Every change of one of the 'factors of an equilibrium occasions a
rearrangement of the system in such a direction that the factor in question 4. MANIPULATING EQUILIBRIUM EXPRESSIONS 85 experiences a change in a sense opposite to the original change.” (Le
Chatelier, 1888). Le Chatelier devoted the later years of his career to social issues related
to his science and engineering background, including educational reform
and the scientific management of industry (called Taylorism, aﬁer
Frederick W. Taylor). As a teacher, he changed chemical education by
emphasizing general principles over memorization. It is not difﬁcult to
imagine the hundreds of students in ﬁn de sie‘cle lecture halls learning
about Le Chatelier’s Principle from Henri Le Chatelier himself. PROBLEMS 4.1 If you add two equilibria (with equilibrium constants K1 and K2) to get a third equilibrium (with
equilibrium constant K3), you know from Section 4.6 that K3 = KIKZ. Use this information to “prove” that
Gibbs ﬂee energies are additive. 4.2 The equilibrium constant for the equilibrium H20 = H+ + OH‘ is 10'14 M2. What is the equilibrium
constant for the equilibrium 3H” + 3OH‘ = 3H20? What units are associated with this equilibrium
constant? 4.3 An environmental engineer is seeking to ﬁnd a way to recover silver from a metal ﬁnishing operation.
She ﬁnds in a textbook an equilibrium constant of 106'3 for the reaction 1/2Ag20(s) + H” = Ag” + 1/2H20.
What is the equilibrium constant for the equilibrium containing the same information but with all stoichio—
metric coefficients integers less than 3? 4.4 An equilibrium expression is added to its reverse reaction to form a new equilibrium. What is the
equilibrium constant of the new equilibrium? What is the Gibbs free energy of reaction for the new
equilibrium? Do your answers make sense? 4.5 Balance the equilibrium between chromate ((390%) and trivalent chromium (Cry). 4.6 Write a balanced equilibrium for the reaction with Al(OH)2+ as a reactant and AKOH);+ as a product. 4.7 In balancing reactions, you sometimes may wish to write a final equilibrium in terms of OH' rather than
H+. How would you modify the balancing procedure of Section 4.7.1 to produce an equilibrium in terms of
OH"? Illustrate your procedure with the chlorine example found in Section 4.7.1 or an equilibrium of your
choosing. 86 A PROBLEMSOLVING APPROACH TO AQUATIC CHEMISTRY 48 Using your answer to Problem 4.6 and the constituent equilibria below, what is the equilibrium constant for the balanced equilibrium between Al(OlI)2+ and Al(OH);? A1(OH)3(S) + 3F = A13+ + 31—120 K1 : 1010.3 M‘Z
A1(OH)3(S) = AKOH 2+ + 20H‘ K2 = 10222 M3
Al3+ + 20H“ = AI(OH)2’r K3 = 1018.5 Mz
H20 = H++OH_I K4=10—14M2 4.9 Are the four constituent equilibria in Problem 4.8 linearly independent? Explain your answer. 4.10 Are the ﬁve equilibria in Problem 4.8 [four constituent equilibria plus the target equilibrium between
A1(OI?[)2+ and Al(OH)2+] linearly independent? Explain your answer. 4.11 By trial and error, determine the equilibrium constant for the equilibrium between FeC03(s) as a
reactant and Fe(0H)2(s) and C02(g) as products. Use some or all of the constituent equilibria below. FeC_O3(s) = Fe2+ + (3032' K1: 10*”68 M2
Fe(OH)2(s) = Fe2+ + ZOH‘ K2 = 1045'1 M3
HCO; = (3032— + H+ K3 = 10—10'3 M
H2003 = H00; + I—F K. = 10*63 M H2C03 = C02(g) + H20 K5 = 1015 atm~Ll1nol
H* + OH" = H20 K6=1014 M’2 4.12 Repeat Problem 4.11 using the systematic method of Section 4.7.3. ...
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 Fall '11
 JenniferJay
 Chemistry

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