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Transcript
```General Concepts of
Chemical Equilibrium
1
In this chapter you will be introduced to
basic equilibrium concepts and related
calculations. The type of calculations
you will learn to perform in this chapter
will be essential for solving equilibrium
problems in later chapters. The ease of
solving equilibrium problems will
depend on the skills you will master in
this chapter.
2
The Equilibrium Constant
For chemical reactions that do not proceed to
completion, an equilibrium constant can be written
as the quotient of multiplication of molar
concentrations of products divided by that of
reactants, each raised to a power equal to its number
of moles.
aA + bB D cC + dD
Where, the small letters represent the number of
moles of substances A, B, C, and D. The equilibrium
constant is written as:
Keq = ([C]c [D]d)/([A]a [B]b)
3
In a chemical reaction, as reactants start to react
products start to form. Therefore, reactants
continuously decrease and products
continuously increase till a point is reached
where eventually no change in concentrations
can be detected. This is the point of equilibrium
which is a point where the rate of the forward
reaction (product formation) equals the rate of
backward reaction (reactants formation). In fact
equilibrium implies continuous transformation
between infinitesimally small amounts of
products and reactants.
4
The rate of the forward reaction diminishes with time, while that of
the backward reaction increases, until they are equal.
A large K means the reaction lies far to the right at equilibrium.
Progress of a chemical reaction
5
6
Calculation of Equilibrium Constants
We know from thermodynamics that a reaction
occurs spontaneously if it has a negative DG
(Gibbs free energy) where:
DG = DH - TDS
Where, DH is the change in enthalpy of the
reaction and DS is the change in entropy. At
standard conditions of temperature and
pressure (standard state) we have the
standard free energy DGo where:
DGo = DHo - TDSo
7
The standard free energy is related to
equilibrium constant by the relation:
DGo = - RT ln K
K = e-DGo/RT
R is the gas constant (8.314 deg K-1mol-1)
It should be clear that DGo gives us good
information about the spontaneity of the
reaction but it offers no clue on the rate at
which the reaction may occur.
8
Lecture 14
Equilibrium Reactions, cont…
Equilibrium Calculations
Le Chatelier's Principle
The equilibrium concentrations of
reactants and products can be changed
by applying an external stress to the
system, e.g. increasing or decreasing
the concentration of a reactant or
product, changing temperature or
pressure. The change occurs in a
direction which tends to counterpart
the applied stress. For example:
10
a.
Increasing the temperature of an
exothermic reaction will shift the reaction to
left (more reactants and less products). The
opposite is observed if the reaction is
exothermic or if heat is removed from an
exothermic reaction.
b.
In a reaction where the number of
gaseous molecules produced is more than
the number of reacting gaseous molecules,
increasing the pressure of the system will
shift the reaction toward the reactants in an
attempt to decrease the number of moles and
vice versa. Reactions in solutions are usually
insensitive to changes in pressure.
11
c.
Increasing the concentration of a reactant or
removing a product will result in a shift of reaction
towards more products and vice versa.
d.
Some reactions can be facilitated by addition of a
catalyst (a substance that is not a part of reactants
or products but its presence makes the reaction
faster). The catalyst does not change the position of
equilibrium but makes the time required to reach this
equilibrium point shorter.
Le Chatelier's principle can be advantageously used to
force reactions that are close to completion to
proceed to completion. This is usually done by
addition of more reagent in a gravimetric procedure,
allowing a gas to escape if one of the products is a
gas, etc..
12
Stepwise Equilibrium Constants
Sometimes a chemical equilibrium can be
written as a multi step equilibrium. The
overall equilibrium constant for the reaction
is the multiple of stepwise equilibrium
constants. Look at the stepwise equilibrium
below:
A2B D A + AB
K1 = [A][AB]/[A2B]
AB D A + B
K2 = [A][B]/[AB]
K1K2 = [A]2[B]/[A2B]
13
Now look at the equilibrium
A 2B D 2 A + B
K = [A]2[B]/[A2B]
From K and K1K2 we see that they are
equal. This is valid for any multi step
equilibrium reaction as will be seen
later.
14
Calculations Using Equilibrium
Constants
Here, we are faced with three situations
which should be considered
separately:
1.
When the equilibrium constant is
very small. Most reactants do not
undergo a reaction and very little
products are produced. Usually the
concentration of products can be
neglected as compared to reactants
concentrations.
15
2.
When the equilibrium constant is very
high. In this case, most reactants disappear
and the reaction container contains mainly
the products. In calculations, one can
neglect the concentration of remaining
reactants as compared to concentrations of
products.
3.
A situation where the equilibrium
constant is moderate. Appreciable amounts
of reactants are left and appreciable amounts
of products are also formed in the reaction
mixture. One can not neglect the reactants or
the products concentration. Following are
examples on each case.
16
Example
Calculate the equilibrium concentration of A and B in a
0.10 M solution of weak electrolyte AB if the
equilibrium constant of the reaction is 3.0x10-6.
Solution
AB D A + B
First we should look at the value of the equilibrium
constant to have an appreciation of what is going on.
It is clear that we have a small equilibrium constant
which suggests that very little products may have
been formed and thus we build our solution on the
assumption that AB will mainly remain unreacted
except for a very little concentration x.
17
Therefore, AB concentration will decrease by x
and A and B will be formed in a
concentration equals x for each. The
following
table
represents
what
is
happening.
18
K = [A][B]/[AB]
Substitution of equilibrium concentration gives
K = (x)(x)/(0.1-x) = 3.0 x 10-6
In order to solve this equation, we should make
a justified assumption which we have
discussed above; that is x is very small as
compared to 0.10 M and later we will check
whether our assumption is valid or not.
19
assume 0.10 >>x. We have
3.0 x 10-6 = x2/0.10, thus we have
x = 5.5 x 10-4
Now we should check the validity of our
assumption that 0.10 >>x by calculating
the relative error
Relative error = (5.5x10-4/0.10) x 100 =
0.55%
20
In this course, we will consider our
assumption valid if the relative error is
below 5%. Therefore, our assumption is
valid and we have:
[A] =5.5x10-4 M, [B] = 5.5x10-4 M,
and
[AB] = (0.10 – 5.5x10-4) ~ 0.10 M
21
In the reaction
A+BDC+D
If 0.2 mol of A is mixed with 0.5 mol of B in 1.0
L, find the equilibrium concentrations of A,
B, C, and D. The equilibrium constant of the
reaction is 4.0x10-9.
It is clear that we have a small equilibrium
constant which suggests that very little
products may have been formed and thus we
build our solution on the assumption that A
and B will mainly remain unreacted except
for a very little concentration x.
22
Therefore, A and B concentration will
decrease by x and C and D will be
formed in a concentration equals x for
each. The following table represents
what is happening.
23
K = [C][D]/[A][B]
Substitution using equilibrium
concentrations results in
K = (x)(x)/(0.20 – x)(0.50 – x)
Now from the value of the equilibrium
constant we can assume that 0.20 >> x.
Then we have
4.0x10-9 = x2/(0.20x0.50)
x = 2.0x10-5
24
Now check the validity of the assumption
by calculating the relative error
Relative Error = (2.0x10-5/0.20) x 100 = 0.01%
The assumption is valid and the
concentrations of the different species
are
[A] = (0.20 – 2.0x10-5) ~ 0.20 M
[B] = 0.50 M
[C] = [D] = 2.0x10-5M
25
In the reaction
A+B DC+D
If 0.2 mol of A is mixed with 0.5 mol of B in 1.0 L,
find the equilibrium concentrations of A, B, C,
and D. The equilibrium constant of the reaction
is 4.0x108.
It is clear that we have a high equilibrium constant
which suggests that very little reactants may
have been left and thus we build our solution on
the assumption that A will mainly be of a very
little concentration x as a result of the chemical
equilibrium. B is already present in excess and
an equivalent amount to A reacts while the rest
of B remains in unreacted.
26
K = [C][D]/[A][B]
Substitution using equilibrium concentrations results in
K = (0.20 – x) (0.20 – x)/(x(0.30 + x))
Now from the value of the equilibrium constant we can
assume that 0.20 >> x. Then we have
4.0x108 = 0.040/0.3x
x = 3.3x10-10
27
Now check the validity of the assumption
by calculating the relative error
Relative Error = (3.3x10-10 /0.20) x 100 =
0.005%
The assumption is valid and the
concentrations of the different species
are
[A] = 3.3x10-10 M
[B] = (0.30 +3.3x10-10) = 0.30 M
[C] = [D] = (0.20 - 3.3x10-10) ~ 0.20 M
28
A and B react as follows
A+2BD2C
Assume 0.10 mol of A reacts with 0.20 mol of B in
1 L. If K = 1.0x1010 , find the equilibrium
concentrations of A, B, and C.
Solution
We have a high equilibrium constant which
suggests that most reactants are changed into
products except for a tiny equilibrium
concentration. Therefore, at equilibrium we
mainly have C.
29
K = [C]2/[A][B]2
K = (0.20 – 2x)2/x(2x)2
Again, we are justified to assume that 0.20 >> 2x since
the equilibrium constant is high. We then have:
1.0x1010 = (0.2)2/4x3
x = 1.0x10-4
30
Now let us check for our assumption by
determining the relative error
Relative error = (2 x 1.0x10-4 /0.20) x 100 = 0.10%
The assumption is valid and the
equilibrium concentrations are
[A] = x =1.0x10-4 M
[B] = 2 x = 2.0x10-4 M
[C] = (0.20 – 2.0x10-4) ~ 0.20 M
31
In the reaction
A+BDC+D
If 0.2 mol of A is mixed with 0.5 mol of B in 1.0
L, find the concentration of A, B, C, and D.
The equilibrium constant of the reaction is
0.30.
It is clear that we do not have a clearly small or
high equilibrium constant which suggests
that both products and reactants will coexist
appreciably in the reaction mixture at
equilibrium. Therefore, A and B
concentration will decrease by x and C and D
will be formed in a concentration equals x for
each.
32
K = [C][D]/[A][B]
K = (x)(x)/(0.20 – x)(0.50 – x)
Now from the value of the equilibrium constant we can
not assume that 0.20 >> x. However, let us do that
and find out what we get
0.30 = x2/(0.20x0.50)
x = 0.17
33
Now check the validity of the assumption by calculating the
relative error
Relative Error = (0.173/0.20) x 100 = 87%
The assumption is invalid and we must not neglect x in
comparison with 0.20, therefore the equilibrium should be
as follows
K = (x)(x)/(0.20 – x)(0.50 – x)
0.30 = x2 /(0.10 – 0.70x +x2)
Rearrangement of the relation above results in
0.70x2 + 0.21x -0.030 = 0
34
The solution of the quadratic equation
ax2 + bx + c = 0 is:
x = {(- b + (b2 – 4 ac)1/2)/2a}
Substitution gives
x = {(- 0.21 + (0.212 – 4 *0.7*(-0.03))1/2)/2*0.7}
x = 0.11 M
Therefore, we have
[A] = 0.20 – 0.11 = 0.09 M
[B] = 0.50 – 0.11 = 0.39 M
[C] = [D] = 0.11 M
35
Lecture 15
Equilibrium Calculations, Cont….
Systematic Approach
The Common Ion Effect
We have discussed earlier that changing the
concentration of a product or reactant has a
significant effect on equilibrium. Addition of
a reactant forces the reaction to proceed
forward. Addition of a product forces the
reaction in the backward direction, a
consequence of Le Chatelier's principle. A
common ion is usually a product ion which
means that reactants concentration will
increase on the expense of products. Let us
look at the following example:
37
Calculate the equilibrium concentrations of A,
B, and AB in a 0.10 M AB (weak electrolyte)
in presence of 0.20 M B. The equilibrium
constant is 3.0x10-6.
AB D A + B
Solution
Be aware that we have previously solved this
problem but without addition of extra amount
of B. Therefore, compare the answers.
38
The equilibrium constant is very small therefore
only tiny amounts of AB will dissociate to yield A
and B. The initial concentration of B is 0.20,
therefore the equilibrium concentration of B will
be 0.20 + x. The following table summarizes the
situation before and after equilibrium:
39
Writing the equilibrium constant expression we
get:
K = x (0.20 + x)/(0.10 – x)
Since K is very small we can assume that 0.1
>> x. This gives:
3.0x10-6 = 0.20x/0.1
x = 1.5x10-6
Let us calculate the relative error to check for
the validity of our assumption
40
Relative error = (1.5x10-6/0.10) x 100 = 1.5x10-3 %
The error is less than 5% therefore our
assumption is OK.
The concentrations of the different
species in present of 0.20 M common
ion will be:
[A] = 1.5x10-6 M
[B] = 0.20 + 1.5x10-6 ~ 0.20 M
[AB] = 0.10 – 1.5x10-6 ~ 0.10
41
Systematic Approach to
Equilibrium Calculations
This Type of approach to equilibrium calculations uses
what is called mass balance and charge balance
concepts. These will be discussed below shortly. By
mass balance we mean that when we dissolve an
amount of dissociating substance, the analytical
concentration of that substance will be equal to the
sum of the concentrations of the different species
derived from it ( mass conservation). Also, solutions
are neutral and thus the concentration of the
negative charges is equal to the concentration of the
positive ones (charge balance, from
electroneutrality).
42
Write mass and charge balance equations for 0.100 M HOAc
in water.
Solution
First, we write the equilibrium:
HOAc D H+ + OAcH2O D H+ + OHNow we deal with the problem and find the charge balance
by placing the positive ions on one side and the negative
ions on the other side.
[H+] = [OH-] + [OAc-]
43
The mass balance equation represents the
concentration of added HOAc where the
analytical concentration added to water was
divided into two parts one represents
undissociated HOAc and the other represents
dissociated HOAc (OAc-). Therefore:
0.100 = CHOAc , the mass balance equation is:
0.100 = [HOAc] + [OAc-]
44
Write a mass and charge balance equations for
0.100 M [Ag(NH3)2]Cl.
Solution
First, write all possible equilibria which might take
place in solution. These would be:
[Ag(NH3)2]Cl  Ag(NH3)2+ + Cl-
45
Ag(NH3)2+ D Ag(NH3)+ +NH3
Ag(NH3)+ D Ag+ + NH3
NH3 + H2O D NH4+ + OHH2O D H+ + OHNow we can write the charge balance equation:
[H+] + [Ag+] + [NH4+] + [Ag(NH3)2+] + [Ag(NH3)+] = [OH-] +
[Cl-]
The mass balance equation can then be written where
all species containing ammonia came from the
original compound [Ag(NH3)2+]Cl.
46
Also, the concentration of ammonia in
Ag(NH3)2+ is twice its concentration since
each mol of Ag(NH3)2+ contains two moles of
ammonia. Therefore, the mass balance
equation for ammonia will be:
0.200 M = 2[Ag(NH3)2+] + [Ag(NH3)+] + [NH3] +
[NH4+]
Another mass balance equation for silver can be
written as
[Ag+] + [Ag(NH3)2+] + [Ag(NH3)+] = [Cl-] = 0.1 M
47
Write mass and charge balance equations for a
solution of H2S.
Solution
First let us set up the equations representing
equilibrium of H2S in water
H2S D HS- + H+
HS- D H+ + S2H2O D H+ + OH-
48
The charge balance equation can be written as
before but considering that the charge
concentration of S2- is twice as the
concentration of S2-. This means that each
mole of S2- contains two moles of charges. The
charge balance equation is:
[H+] = [OH-] + [HS-] + 2[S2-]
Mass balance equation will be:
CH2S = [H2S] + [HS-] + [S2-]
49
Write a mass and charge balance equations for
CdS in solution.
CdS D Cd2+ + S2S2- + H2O D HS- + OHHS- + H2O D H2S + OHH2O D H+ + OHThe charge balance equation can be simply
derived from the equilibria above but taking
in consideration that charge concentration
on cadmium is twice the concentration of
50
The charge concentration on sulfide is
also twice the sulfide concentration.
The equation will be:
[H+] + 2[Cd2+] = [OH-] + [HS-] + 2[S2-]
Mass balance equation:
[Cd2+] = [H2S] + [HS-] + [S2-]
51
Write a charge balance equation for a
solution containing KI and AlI3.
KI g K+ + IAlI3 = Al3+ + 3 IH2O D H+ + OHThe equation can be written directly
considering that the concentration of the
charge on Al3+ is three times aluminum
concentration, therefore we have:
[I-] + [OH-] = [H+] + [K+] + 3[Al3+]
52
Equilibrium Calculations Using
the Systematic Approach
This section will only be briefly presented since it will
not be our approach in solving equilibrium
problems. T depends on building up relations as the
equilibrium constant, solubility product,
autoprotolysis, mass balance, charge balance, etc..
so as to produce equations at least equaling the
number of unknowns. The systematic approach is
not easily presented in a part of a lecture but needs
enough time for mastering in order to be able to
solve equilibrium problems effectively. Therefore, we
are justified to keep it as simple as possible in our
discussion.
53
Example
Use the systematic approach to calculate the
equilibrium concentrations of A, B, and AB in a 0.10
M AB solution. The equilibrium constant is 3.0x10-6.
AB D A + B
Solution
We do not have charges in the chemical equation;
therefore do not attempt to write a charge balance
equation.
54
The equilibrium constant is
K = [A][B]/[AB]
The mass balance equation can be formulated as:
CAB = [AB] + [A]
0.10 = [AB] + [A]
From the value of the equilibrium constant we
know that [AB] >> [A]
Therefore, [AB] = 0.10
55
Also, AB dissociates into A and B only,
therefore [A] = [B]
Substitution in the equilibrium constant expression
gives
3.0x10-6 = [A]2/0.10
[A] = 5.5x10-4 M
The relative error = (5.5x10-4 / 0.10) x 100 =
0.55% which means that A is really much
smaller than AB
[A] = [B] = 5.5x10-4 M
[AB] = 0.10 - 5.5x10-4 ~ 0.10 M
56
Example
Using the systematic approach Calculate the
equilibrium concentrations of A, B, and AB in
a 0.10 M AB (equilibrium constant is 3.0x10-6)
in presence of 0.20 M MB (strong
electrolyte)where the equilibrium below takes
place.
AB D A+ + BMB g M+ + B57
Mass balance for BCB- = [B-] from strong electrolyte + [B-]from dissociation of AB
[B-]from dissociation of AB = [A+]
CB- = [B-] from strong electrolyte + [A+]
A+ concentration is negligible as compared to
amount of B- from strong electrolyte due to very
small K and common ion effect seen earlier.
Therefore we have
[B-] = 0.20 M
We also have from mass balance that
CAB = [AB] + [A+]
58
The same argument can also be used where the
very low K suggests that [AB] >> [A+], therefore:
0.10 = [AB]
Now, substitution in the equilibrium constant
expression ew get:
K = [A+][B-]/[AB]
3.0x10-6 = [A+] * 0.20/0.10
[A+] = 1.5x10-6 M
59
The relative error = (1.5x10-6/0.10) x 100 =1.5x10-3 %
Therefore, the concentrations of the
different species are:
[A+] = 1.5x10-6 M
[B] = 0.20 + 1.5x10-6 ~ 0.20 M
[AB] = 0.10 – 1.5x10-6 ~ 0.10 M
60
Name:
Student Number:
What is the titer of a 1.58 g/L KMnO4
(FW = 158 mg/mmol) in terms of
mg Na2O2 (FW = 34 mg/mmol). The
equation is:
5Na2O2 + 2KMnO4 + 16H+ g
2Mn2+ + 5O2 + 8H2O + 10Na+ + 2K+
61
Lecture 16
Equilibrium Calculations, Cont….
Activity and Activity Coefficients
Activity and Activity Coefficients
In presence of strong electrolytes of diverse ions, weak
electrolytes tend to dissociate more. This behavior is
attributed to the shielding effects exerted by the ions of
the strong electrolyte on the ionic species of opposite
signs from the weak electrolyte. The concentration of an
ion, from a weak electrolyte, in presence of strong
electrolyte is called the activity of the ion and is defined
by the relation:
ai = Ci fi
Where ai is the activity of the ion i, Ci is the concentration of
ion i, and fi is the activity coefficient of ion i.
63
Properties of the activity Coefficient
1.
2.
3.
4.
65
The activity coefficient of uncharged species is
unity.
The activity coefficient of ions in dilute
solutions (less than 10-4 M) is very close to
unity and in fact can be approximated to unity.
As the concentration of electrolyte (diverse
ions) increases, the activity coefficient
decreases and activity becomes less than
concentration
The activity coefficient is highly dependent on
the charge of diverse ions where as the charge
increases, the activity coefficient decreases
significantly
Ionic Strength
A property which represents the strength of ions in
solution can be seen using the term ionic strength
(m) which can be defined as:
m = 1/2 SCiZi2
Zi is the charge on the ion i. If we look carefully to
properties of activity coefficients listed above (3 and
4) it is clear that the term ionic strength combines
the effects of concentration and charge of ions and
thus will be advantageously used as an important
factor affecting fi.
66
Example
Calculate the ionic strength of
KNO3 solution.
Solution
m = 1/2 SCiZi2
m = 1/2 (CK+ Z2 K+ + CNO3- Z2NO3-)
m = 1/2 (0.2 x 12 + 0.2 x 12) = 0.2
67
a 0.2 M
Calculate the ionic strength of a 0.2 M K2SO4
solution.
Solution
m = 1/2 SCiZi2
m = 1/2 (CK+ Z2 K+ + CSO42- Z2SO42-)
m = 1/2 (2 x 0.2 x 12 + 0.2 x 22) = 0.6
Compare between the value of the ionic
strength in this and previous example and
see the effect of charge on the answer.
68
Calculate the ionic strength of the
solution containing 0.3 M KCl and 0.2 M
K2SO4.
m = 1/2 SCiZi2
m = 1/2 (CK+ Z2 K+ + CCl- Z2Cl- + CSO42- Z2SO42-)
m = 1/2 (0.7 x 12 + 0.3 x 12 + 0.2 x 22) = 0.9
We have substituted 0.7 for CK+ since this is the overall
potassium ion concentration but we could make it the
sum of two terms:
{ CK+ Z2 K+}from KCl + { CK+ Z2 K+}from K2SO4 which will also give
the same result.
69
Calculation of Activity coefficients
The activity coefficients of an ion i can be calculated
provided that three pieces of information are known.
These are:
1. The ionic strength of the solution
2. The charge of the ion
3. The effective diameter of the hydrated ion (ai) in
angstrom (Ao)
The Debye-Huckel equation can be used for such a
calculation where:
- log fi = (0.51 Zi2 m1/2)/(1 + 0.33 ai m1/2)
70
In situations where the effective diameter of the ion is 3
Ao (you can assume this situation to solve assigned
problems) we can write
- log fi = (0.51 Zi2 m1/2)/(1 + m1/2)
However the Debye-Huckel equation gives an
approximate value for the activity coefficient. A
better relation is the Davies modification of the
Debye-Huckel equation. The equation is:
- log fi = (0.51 Zi2 m1/2)/(1 + 0.33 ai m1/2) – 0.10 Zi2 m
In our calculations we will only use the Debye-Huckel
equation for estimation of activity coefficients.
72
Example
Calculate the activity coefficients of K+ and
SO42- in a 0.002 M K2SO4 assuming an
effective diameter of 3 Ao for both ions.
Solution
To apply Debye-Huckel equation we need to
determine the ionic strength of the solution
m = 1/2 SCiZi2
m = 1/2 (0.002 x 2 x 12 + 0.002 x 22) = 0.006
73
- log fi = (0.51 Zi2 m1/2)/(1 + m1/2)
- log fK+ = (0.51 x 12 0.0061/2)/(1 + 0.0061/2) = 0.037
fK+ = 10-0.037 = 0.918
For sulfate we have
- log fSO42- = (0.51 x 22 x 0.0061/2)/(1 + 0.0061/2) = 0.147
fSO42- = 10-0.147 = 0.71
If we repeat the calculation for a more concentrated
solution (0.02 M K2SO4) the ionic strength will be 0.06
and the activity coefficients for K+ and SO42- will be
0.79 and 0.38, respectively. Therefore, you should
appreciate the effect of concentration on both ionic
strength and activity coefficients as well.
74
Thermodynamic Equilibrium Constant
The equilibrium constant expressions which use
concentrations are called concentration equilibrium
constants. For example, in the reaction:
AB D A + B
We can write
K = [A][B]/[AB]
This is the so called concentration equilibrium
constant in order to differentiate it from the
thermodynamic equilibrium constant which uses
activities rather than concentrations.
75
KTh = aA aB/aAB
Since a = C f, we can write
KTh = [A] fA [B] fB /[AB] fAB
KTh = ([A][B]/[AB]) (fA fB/fAB)
Substituting K for [A][B]/[AB] we get
KTh = K (fA fB/fAB)
76
Since fi = 1 for very dilute solutions (< 10-4 M)
then we can say that K approximates KTh as
the ionic strength of the solution approaches
zero. Therefore, we should calculate K for
any equilibrium especially when the ionic
strength has a value away from zero. You
should also know that equilibrium constants
listed in appendices are all thermodynamic
equilibrium constants, i.e. measured at very
low concentrations where the ionic strength
approaches zero.
Calculate the concentration equilibrium
constant for the dissociation of AB if fA+, fBare 0.6 and 0.7, respectively. The
thermodynamic equilibrium constant is 2.0 x
10-8.
Solution
KTh = K (fA+ fB-/fAB)
Substitution for fA+, fB- and remembering that
fAB = 1 since AB is not charged, we get
2 x 10-8 = K (0.6 * 0.7)/1
K = 5 x 10-8
78
Calculate the percent dissociation of a 1.0x10-4 M AB in
water. The thermodynamic equilibrium constant is
2.0 x 10-8.
AB D A+ + B-
79
The equilibrium constant is very small, therefore we
can assume that 1.0x10-4 >> x. Substitution in the
equilibrium constant expression gives:
2.0x10-8 = x2/1.0*10-4
x = 1.4x10-6 M
The relative error = (1.4x10-6/1.0x10-4) x 100 = 1.4%
Therefore the assumption is valid
% dissociation = (amount dissociated/initial amount) x
100 = (1.4x10-6/1.0x10-4) x 100 = 1.4%
80
Example
Calculate the percent dissociation of a 1.0x10-4 M AB in
presence of diverse ions where the ionic strength is
0.1 and fA+, fB- are 0.6 and 0.7, respectively. The
thermodynamic equilibrium constant is 2.0 x 10-8.
AB D A+ + B-
81
We first calculate the equilibrium constant since we
can not use the thermodynamic equilibrium constant
in presence of diverse ions:
KTh = K (fA+ fB-/fAB)
Substitution for fA+, fB- and remembering that fAB = 1
since AB is not charged, we get
2 x 10-8 = K (0.6 * 0.7)/1
K = 5 x 10-8
Now we can substitute in the equilibrium constant
expression where we have
82
5x10-8 = x2/1.0*10-4
x = 2.2x10-6 M
The relative error = (2.2 x10-6/1.0x10-4) x 100 = 2.2 %
Therefore the assumption is valid
% dissociation = (amount dissociated/initial amount) x
100 = (2.2 x 10-6/1.0x10-4) x 100 = 2.2 %
If you compare the results in this and previous
examples you will see that the percent dissociation
increased from 1.4 to 2.2% which represents an
increase of:
{(2.2 – 1.4)/1.4}x 100 = 57%
83
Therefore, it is clear that dissociation can
significantly increase in presence of
diverse ions. However, I would like to
emphasize that we will use
concentrations rather than activities
throughout this course and will neglect
the effects of diverse ions except in
situations where I ask you to work
differently.
84
Name:
Student Number:
In the equilibrium:
A+BDC+D
If 0.50 mol of A is mixed with 0.20 mol
of B in a 1 L solution, find the
concentrations of all species at
equilibrium. K=4*10-5
85
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