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CHEMICAL EQUILIBRIUM

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Title: CHEMICAL EQUILIBRIUM


1
CHEMICAL EQUILIBRIUM
2
CHEMICAL EQUILIBRIUM
the dynamic state in which rates of the forward
and reverse reactions are identical.
aA bB cC dD
Kgt1 ? Forward reaction is favoured
Equilibrium constant a,b,c,d stoichiometry
coefficients A, B, C, D concentrations
of A, B, C, D
  • in standard state
  • For solutes 1M
  • For gases 1 atm
  • For solids and pure liquids X 1
  • In these conditions K is dimensionless.

3
Equilibrium const. for a reverse reaction K1
1/K
cC dD aA bB
Equilibrium const. for two reactions added K3
K1 x K2
HA H A-
H C CH
HA C CH A-
4
THERMODYNAMICS
The equilibrium constant is derived from the
thermodynamics of a chemical reaction.
ENTHALPY
?H - enthalpy change is the heat absorbed or
released during reaction
?H lt 0 Heat liberated Exothermic reaction
?H gt 0 Heat absorbed Endothermic reaction
5
ENTROPY
?S a measure of disorder of a substance
Gas ? Liquid ? Solid
Decrease in disorder
?S gt 0 Products more disordered than reactants
?S lt 0 Products less disordered than reactants
6
FREE ENERGY
Gibbs free energy ?G ?H - T?S
?G lt 0 The
reaction is favoured Kgt1
?G gt 0 The
reaction is not favoured Klt1
Free energy and equilibrium
OR
7
LE CHATELIERS PRINCIPLE
When a system at equilibrium is disturbed, the
direction in which the system proceeds back to
equilibrium is such that the change is partially
offset.
If reaction is at equilibrium and reactants are
added (or products removed), the reaction goes
to the right.
If reaction is at equilibrium and products added
( or reactants are removed), the reaction goes to
the left.
8
Recall Q Reaction quotient
has the same form as equilibrium constant (K),
but the solution concentrations do not have to be
equilibrium concentrations.
Thus at equilibrium Q K
According to Le Châtelier If Q gt K ? reaction
will proceed in the reverse direction If Q lt
K ? reaction will proceed in the forward
direction
9
A B C D
At equilibrium
A 0.002M B 0.025M C 5.0M
D 1.0M
1x105 at 250C
Say we add double reactant A, A 0.004M
0.5x105 at 250C
Q lt K
To reach equilibrium Q K and the reaction
must go to the right ? spontaneous
10
THE EFFECT OF TEMPERATURE ON K
?G ?H - T?S
Independent of T
For endothermic reactions (?H gt 0) K increases
if T increases.
For exothermic reactions (?H lt 0) K decreases
if T increases.
11
K AND SOLUBILITY
K Ksp (solubility product) when the equilibrium
reaction involves a solid salt dissolving to give
its constituent ions in solution.
Recall Solid 1 Saturated solution in
equilibrium with undissolved solid
Thus if an aqueous solution is left in contact
with excess solid, the solid will dissolve until
Ksp is satisfied. Thereafter the amount of
undissolved solid remains constant.
12
Example Calculate the mass of PbCl2 that
dissolves in 100 ml water. (Ksp 1.7x10-5 for
PbCl2)
Initial
(solid)
Final
(solid)
m 0.45 g
13
THE COMMON ION EFFECT
  • Now add 0.03M NaCl to the PbCl2 solution
  • We added 0.03M Cl-

Initial
(solid)
Final
(solid)
For this system to be at equilibrium when Cl-
is added, the Pb2 decreases (reverse
reaction). this is an application of the Le
Chateliers principle and is called THE COMMON
ION EFFECT
The salt will be less soluble if one of its
constituent ions is already present in the
solution.
14
The Common Ion Effect - experiment
15
THE NATURE OF WATER AND ITS IONS
H does not exist on its own in H2O ? forms H3O
H3O
16
In aqueous solution, H3O is tightly associated
with 3 molecules of H2O through exceptionally
strong hydrogen bonds.
One H2O is held by weaker ion-dipole attraction
17
Can also form H5O2 cation ? H shared by 2 water
molecules
H3O2- (OH-.H2O) has been observed in solids
18
AUTOPROTOLYSIS
Water undergoes self-ionisation ? autoprotolysis,
since H2O acts as an acid and a base.
H2O H2O H3O OH-
The extent of autoprotolysis is very small. For
H2O Kw H3OOH- 1.0 x 10-14 (at
25oC)
19
pH
pH -logH Approximate definition of pH
pH pOH -log(Kw) 14.00 at 250C
20
It is generally assumed that the pH range is
0-14. But we can get pH values outside this
range. e.g. pH -1 ? H 10 M This is
attainable in a strong concentrated acid.
21
ACTIVITY
Equilibrium constant
A, B, C, D concentrations of A, B, C, D
  • BUT in a real solution all charged ions are
  • surrounded by ions with opposite charge ionic
    atmosphere
  • hydrated - surrounded by tightly held water
    dipoles

22
Adding an inert salt to a sparingly soluble
salt increases the solubility of the sparingly
soluble salt.
inert salt a salt whose ions do not react
with the compound of interest
WHY?
Consider BaSO4 (Ksp 1.1x10-10) as the
sparingly soluble salt and KNO3 as the inert
salt. In solution
The cation (Ba2) is surrounded by anions
(SO42-, NO3-) ?net positive charge is reduced
The anion (SO42-) is surrounded by cations (Ba2,
K) ?net negative charge is reduced
? attraction between oppositely charged ions
is decreased.
23
The net charge in the ionic atmosphere is less
than the charge of the ion at the center.
The ionic atmosphere decrease the attraction
between ions.
The greater the ionic strength of a solution, the
higher the charge in the ionic atmosphere. Each
ion-plus-atmosphere contains less charge and
there is less attraction between any particular
cation and anion.
24
Activity of the ion in a solution depends on its
hydrated radius not the size of the bare ion.
25
IONIC STRENGTH, µ
A measure of the total concentration of ions in
solution. The more highly charged an ion, the
more it is counted.
Where ci concentration of the ith species
zi charge for all ions
in solution
26
Example Find the ionic strength of 0.010 M
Na2SO4 solution.
27
Effect of ionic strength on solubility
Explain all 4 cases
28
ACTIVITY COEFFICIENTS
To account for the effect of ionic strength,
concentrations are replaced by activities.
Activity coefficient
Activity of C
And general form of equilibrium constant is
29
  • Activity coefficient
  • Measure of deviation of behaviour from ideality
    (ideal ? ? 1)
  • Allows for the effect of ionic strength

At low ionic strength activity coefficients ?
1 and K ? concentration equilibrium
Thus for the sparingly soluble salt BaSO4,
dissolving in the presence of the inert salt
KNO3 Ksp aBa aSO4 Ba2?Ba
SO42-?SO4 ?If more BaSO4 dissolves in the
presence of KNO3, Ba2 and SO42- increases
and ?Ba and ?SO4 decreases
30
ACTIVITY COEFFICIENTS OF IONS
Extended Debye-Huckel equation relates activity
coefficients to ionic strength
at 250C
? effective hydrated radius of the ion
31
Effect of Ionic Strength, Ion charge and Ion Size
on the Activity Coefficient
(Over the range of ionic strength from 0 to 0.1M)
  1. As ionic strength increases, the activity
    coefficient decreases. ? ? 1 as ? ? 0
  1. As the charge of the ion increases, the departure
    of its activity coefficient from unity increases.
    Activity corrections are much more
    important for an ion with a charge of ?3 than one
    with the charge ?1.

Activity coefficients for differently charged
ions with a constant hydrated radius of 500pm.
3. The smaller the hydrated radius of the ion,
the more important activity effects become.
32
Obtain values for ? from the table
Use interpolation to find values of ? for ionic
strengths not listed
33
How to interpolate - SELF STUDY!!
Linear interpolation
34
At high ionic strengths activity coefficients of
most ions increase
Concentrated salt solutions are not the same as
dilute aqueous solutions ? different
solvents
H in NaClO4 solution of varying ionic strengths
35
pH AND ACTIVITY COEFFICIENTS
pH -logH Approximate definition of pH
The real definition of pH is
NOTE A pH electrode measures activity of H and
NOT concentration
36
SYSTEMATIC TREATMENT OF EQUILIBRIUM
Chemical equilibrium provides a basis for most
techniques in analytical chemistry and
application of chemistry to other disciplines
such as like biology, geology etc.
The systematic treatment of equilibrium gives us
the tool to deal with all types of complicated
chemical equilibria.
The systematic procedure is to write as many
independent algebraic equations as there are
unknowns (species) in the problem. This includes
all chemical equilibrium conditions two
balances charge and mass balances.
37
CHARGE BALANCE
The sum of positive charges in solution equals
the sum of negative charges.
Charge neutrality
E.g. An aqueous solution of KH2PO4 and KOH
contains the following ionic species
H, OH-, K, H2PO4-, HPO42-, PO43- The charge
balance is
H K OH- H2PO4- 2HPO42-
3PO43-
The coefficient in front of each species
the magnitude of the charge on the ion
38
MASS BALANCE
Conservation of matter.
Quantity of all species in a solution containing
a particular atom must equal the amount of that
atom delivered to the solution.
E.g. Mass balance for 0.02 M phosphoric acid in
water
0.02 M H3PO4 H2PO4- HPO42- PO43-
39
SYSTEMATIC TREATMENT OF EQUILIBRIUM Step 1.
Write the pertinent reactions. Step 2. Write the
charge balance equation. Step 3. Write the mass
balance equations. Step 4. Write the equilibrium
constant for each chemical reaction. Step 5.
Count the equations and unknowns. Step 6. Solve
for all the unknowns.
40
E.g. The ionization of water
H2O H OH- Kw 1.0x10-14 at 250C
Find the concentrations of H and OH- in pure
water
Step 1. Pertinent reaction
only one above.
Step 2. Charge balance
(1)
Step 3. Mass balance
(2)
Step 4. Equilibrium constants
the only one (3)
Step 5. Count equations and unknowns
2 eq. and 2 unknowns
Step 6. Solve.
For pure water the ionic strength approaches 0
and we can write eq.3 as
41
E.g. The solubility of Hg2Cl2
Find the concentration of Hg22 in a saturated
solution of Hg2Cl2
Step 2. Charge balance (1)
Step 3. Mass balance
(2)
Step 4. Equilibrium constants (3)
(4)
Step 5. Count equations and unknowns 4 eqs. and
4 unknowns
Step 6. Solve. Using eqn 2 we can write eqn 3 as
42
THE DEPENDENCE OF SOLUBILITY ON pH
Coupled equilibria the product of one reaction
is reactant in the next reaction
Problem The mineral fluorite, CaF2, has a cubic
crystal structure and often cleaves to form
nearly perfect octahedra. Find the solubility of
CaF2 in water.
43
Step 1. Pertinent reaction
Step 2. Charge balance
(1)
Step 3. Mass balance
Some fluoride ions react to give HF. (2)
Also
44
CaF2(s) Ca2 2F- Ksp F- H2O HF
OH- Kb H2O H OH- Kw
Step 4. Equilibrium constants
(3)
(4)
(5)
Step 5. Count equations and unknowns
5 eqs. and 5 unknowns
Step 6. Solve
45
To simplify the problem let us solve it for a
fixed pH 3
That means H and OH-
Then from eqn 4 Kb HFOH-/F-
HF/F- Kb/OH- 1.5x10-11/1.0x10-11 1.5
Thus HF 1.5F-
Substitute HF in the eqn 2 F- HF
2Ca2
F- 1.5F- 2Ca2
And F- 0.80Ca2
Using this expression in eqn 3 Ksp
Ca2F-2 Ca2(0.80Ca2)2
Now find F- 3.1x10-4 M HF 4.7x10-4 M
Thus Ca2 (Ksp/0.802)1/3 3.9x10-4M
46
NOTE To fix the pH of a solution an ionic
compound is added. Thus the charge balance
equation as written not longer holds.
pH dependence of the conc. of Ca2, F- and HF in
a saturated solution.
Also OH- H HF No longer holds
47
Applications of coupled equilibria in the
modeling of environmental problems
Found Ca in acid rain that has washed off
marble stone (largely CaCO3) increases as the
H of acid rain increases.
CaCO3(s) 2H(aq) ? Ca2(aq) CO2(g)
H2O(l)
48
Al is usually locked into insoluble minerals
e.g. kaolinite and bauxite. But due to acid rain,
soluble forms of Al are introduced into the
environment. (Similarly with other minerals
containing Hg, Pb etc.)
Total Al as a function of pH in 1000 Norwegian
lakes.
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