Hein and Arena

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Chemical Equilibrium Chapter 16
Larry Emme
Chemeketa Community College
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Reversible Reactions
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reversible reaction - a chemical reaction in
which the products formed react to produce the
original reactants.
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The reaction between NO2 and N2O4 is reversible.
N2O4 is formed
N2O4 decomposes when heated forming NO2
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reaction to the right (forward)
reaction to the left (reverse)
Ice water
Hot water
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Rates of Reaction
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• The rate of a reaction is variable. It depends
  on

• concentrations of the reacting species
• reaction temperature
• presence or absence of catalysts
• the nature of the reactants

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Chemical Equilibrium
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equilibrium A dynamic state in which two or more
opposing processes are taking place at the same
time and at the same rate.
chemical equilibrium The state in which the rate
of the forward reaction equals the rate of the
reverse reaction in a chemical change.
At equilibrium the concentrations of the products
and the reactants are not changing.
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A saturated salt solution is in equilibrium with
solid salt.
salt crystalsare dissolving
Na and Cl- are crystallizing
At equilibrium the rate of salt dissolution
equals the rate of salt crystallization.
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Le Chateliers Principle
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Henri LeChatelier
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In 1888, the French chemist Henri LeChatelier set
forth a far-reaching generalization on the
behavior of equilibrium systems.
This generalization, known as LeChateliers
Principle, states
If a stress or strain is applied to a system in
equilibrium, the system will respond in such a
way as to relieve that stress and restore
equilibrium under a new set of conditions.
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Effect of Concentration on Equilibrium
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• For most reactions the rate of reaction increases
  as reactant concentrations increase.

• The manner in which the rate of reaction changes
  with concentration must be determined
  experimentally.

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An equilibrium is disturbed when the
concentration of one or more of its components is
changed. As a result, the concentration of all
species will change and a new equilibrium mixture
will be established.
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The system is at equilibrium
results in C and D being produced faster than
they are used.
results in A and B being used faster than they
are produced.
increases the rate of the forward reaction
Increasing the concentration of B
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The system is again at equilibrium
In the new equilibrium
concentration of A has decreased
concentrations of B, C and D have increased
After enough time has passed, the rates of the
forward and reverse reactions become equal.
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Effect of Concentration Changeson the Chlorine
Water Equilibrium
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Effect of C2H3O2Concentration Changes on pH
Add 0.100 mol NaC2H3O2
Add 0.200 mol NaC2H3O2
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Effect of Pressure on Equilibrium
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• Changes in pressure significantly affect the
  reaction rate only when one or more of the
  reactants or products is a gas and the reaction
  is run in a closed container.

• The effect of increasing the pressure is to
  increase the concentrations of any gaseous
  reactants or products.

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Increase Pressure
increases CO2 concentration
CaCO3(s) CaO(s) CO2(g)
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Decrease Pressure
decreases CO2 concentration
CaCO3(s) CaO(s) CO2(g)
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• In a system composed entirely of gases, a
  increase in the pressure of the container will
  cause the reaction and the equilibrium to shift
  to the side that contains the smallest number of
  molecules.

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Increase Pressure
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Increase Pressure
Equilibrium does not shift. The number of
molecules is the same on both sides of the
equation.
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Effect of Temperature on Equilibrium
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When the temperature of a system is raised, the
rate of reaction increases.
The rate of the reaction that absorbs heat is
increased to a greater extent, and the
equilibrium shifts to favor that reaction. When
the process is endothermic, the forward (left to
right) reaction is increased. When the process is
exothermic, the reverse (right to left) process
is increased.
In a reversible reaction, the rates of both the
forward and the reverse reactions are increased
by an increase in temperature.
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Heat may be treated as a reactant in endothermic
reactions.
At room temperature very little CO forms.
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Effect of Catalystson Equilibrium
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A catalyst is a substance that influences the
rate of a reaction and can be recovered
essentially unchanged at the end of the reaction.
A catalyst does not shift the equilibrium of a
reaction. It affects only the speed at which the
equilibrium is reached.
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Energy Diagram for an Exothermic Reaction
Activation energy the minimum energy required
for a reaction to occur.
A catalyst speeds up a reaction by lowering the
activation energy.
A catalyst does not change the energy of a
reaction.
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AlCl3
PCl3(l) S(s) ? PSCl3(l)
Very little thiophosphoryl chloride is formed in
the absence of a catalyst because the reaction is
so slow.
In the presence of a catalyst the reaction is
complete in a few seconds.
MnO2
?
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Equilibrium Constants
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At equilibrium the rates of the forward and
reverse reactions are equal, and the
concentrations of the reactants and products are
constant.
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The equilibrium constant (Keq) is a value
representing the unchanging concentrations of the
reactants and the products in a chemical reaction
at equilibrium.
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For the general reaction
at a given temperature
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For the reaction
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For the reaction
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The magnitude of an equilibrium constant
indicates the extent to which the forward and
reverse reactions take place.
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When the molar concentrations of all species in
an equilibrium reaction are known, the Keq can be
calculated by substituting the concentrations
into the equilibrium constant expression.
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Calculate the Keq for the following reaction on
concentrations of PCl5 0.030 mol/L, PCl3 0.97
mol/L and Cl2 0.97 mol/L at 300oC.
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Ionization Constants
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In addition to Kw, several other ionization
constants are used.
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Ka
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When acetic acid ionizes in water, the following
equilibrium is established
Ka is the ionization constant for this
equilibrium.
Ka is called the acid ionization constant.
Since the concentration of water is large and
does not change appreciably, it is omitted from
Ka.
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At 25oC, a 0.100 M solution of HC2H3O2 is 1.34
ionized and has an H of 1.34 x 10-3 mol/L.
Calculate Ka for acetic acid.
The moles of unionized acetic acid per liter are
0.100 mol/L 0.00134 mol/L 0.099 mol/L
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Substitute these concentrations into the
equilibrium expression and solve for Ka.
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What is the H in a 0.50 M HC2H3O2 solution?
The ionization constant, Ka, for HC2H3O2 is 1.8 x
10-5.
The equilibrium expression and Ka for HC2H3O2
are
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What is the H in a 0.50 M HC2H3O2 solution?
The ionization constant, Ka, for HC2H3O2 is 1.8 x
10-5.
The equilibrium expression and Ka for HC2H3O2
are
HC2H3O2 at equilibrium is 0.50 x.
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Substitute these values into Ka for HC2H3O2.
HC2H3O2 0.50 - x
Assume x is small compared to 0.50 - x. Then 0.50
x ? 0.50
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Substitute these values into Ka for HC2H3O2.
HC2H3O2 0.50 - x
Solve for x2.
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Take the square root of both sides of the
equation.
Making no approximation and using the quadratic
equation the answer is 2.99 x 10-3 mol/L, showing
that it was justified to assume Y was small
compared to 0.5.
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Calculate the percent ionization in a 0.50 M
HC2H3O2 solution.
The ionization of a weak acid is given by
The percent ionization is given by
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Calculate the percent ionization in a 0.50 M
HC2H3O2 solution.
The ionization of acetic acid is given by
The percent ionization of acetic acid is given by
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Ionization Constants (Ka) of Weak Acids at 25oC
Acid Formula Ka Acid Formula Ka
Acetic HC2H3O2 1.8 x 10-5 Hydrocyanic HCN 4.0 x10-10
Benzoic HC7H5O2 6.3 x 10-5 Hypochlorous HOCl 3.5 x10-8
Carbolic HC6H5O 1.3 x 10-10 Nitrous HNO2 4.5 x10-4
Cyanic HCNO 2.0 x 10-4 Hydrofluoric HF 6.5 x10-4
Formic HCHO2 1.8 x 10-4
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Ionic Equilibria Problems
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• Use Ka as a measure of acid strength. Let 1 Hi
  and 5 Lo

Acid Ka order
HC2H3O2 1.8 ? 105
H2S 6 ? 108
HF 7 ? 10 4
HC2Cl3O2 2 ? 101
HCN 7.2 ? 1010
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• Use Ka as a measure of acid strength. Let 1 Hi
  and 5 Lo

Acid Ka order
HC2H3O2 1.8 ? 105 3
H2S 6 ? 108 4
HF 7 ? 10 4 2
HC2Cl3O2 2 ? 101 1
HCN 7.2 ? 1010 5
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• Find the pH in 0.1 M HCN if
• Ka 7.2 ? 1010

Step 1 ionize the acid
Step 2 write the Ka expression
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Step 3 let x H CN HCN 0.1 M
x 0.1 M
Step 4 substitute into Ka expression
Step 5 solve for x
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Step 6 solve for pH
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• Find H in 0.1 M H2CO3 if
• Ka 4.0 ? 1017

Step 1 ionize the acid
Step 2 write the Ka expression
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Step 3 let x H , CO32 ½ x
H2CO3 0.1 M ½ x 0.1 M
Step 4 substitute into Ka expression
Step 5 solve for x
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• Find the pH of a 0.001 M solution of bombastic
  hydroxide (BmOH) if
• Kb 1.6 ? 1010

Step 1 ionize the base
Step 2 write the Kb expression
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Step 3 let x Bm OH BmOH
0.001 M x 0.001 M
Step 4 substitute into Kb expression
Step 5 solve for x
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Step 6 solve for pOH
Step 7 solve for pH
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Find the Ka for a 0.01 M HCN solution if the pH
6.3
Step 1 ionize the acid
Step 2 write the Ka expression
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Step 3 use pH to find H
H 10pH 106.3 5.0 ? 107
H CN, HCN 0.01- 5.0 ? 107 0.01
Step 4 substitute into Ka expression
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Solubility Product Constant
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The solubility product constant, Ksp, is the
equilibrium constant of a slightly (sparingly)
soluble salt.
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Silver chloride is in equilibrium with its ions
in aqueous solution.
The equilibrium constant is
Rearrange
The amount of solid AgCl does not affect the
equilibrium.
The concentration of solid AgCl is a constant.
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Ksp for Sparingly Soluble Solids

• To find Ksp
• 1. ionize the salt
• 2. write the Ksp expression
• 3. determine the Ms of the ions
• 4. substitute into the Ksp expression

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The solubility of AgCl in water is 1.3 x 10-5
mol/L.
Because each formula unit of AgCl that
dissolves yields one Ag and one Cl- , the
concentrations of the two ions are equal.
Ag Cl- 1.3 x 10-5 mol/L
Ksp AgCl-
The Ksp has no denominator.
(1.3 x 10-5)(1.3 x 10-5)
1.7 x 10-10
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Molar Solubility for Sparingly Soluble Solids

• 1. ionize the salt
• 2. write the Ksp expression
• 3. let x equal the amount of salt that
• dissolves
• 4. find concentrations of ions in terms of x
• 5. substitute in the Ksp expression
• 6. solve for x

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The Ksp value for lead sulfate is 1.3 x 10-8.
Calculate the solubility of PbSO4 in moles per
liter.
The equilibrium equation of PbSO4 is
The Ksp of PbSO4 is
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The Ksp value for lead sulfate is 1.3 x 10-8.
Calculate the solubility of PbSO4 in moles per
liter.
Substitute x into the Ksp equation.
The solubility of PbSO4 is 1.1 x 10-4 mol/L.
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Solubility Product Constants (Ksp) at 25oC
Compound Ksp Compound Ksp
AgCl 1.7 x 10-10 CaF2 3.9 x 10-11
AgBr 5 x 10-13 CuS 9 x 10-45
AgI 8.5 x 10-17 Fe(OH)3 6 x 10-38
AgC2H3O2 2 x 10-3 PbS 7x 10-29
Ag2CrO4 1.9 x 10-12 PbSO4 1.3 x 10-8
BaCrO4 8.5 x 10-11 Mn(OH)2 2.0 x 10-13
BaSO4 1.5 x 10-9
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Practice
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• The Ksp value for silver carbonate is 6.2 x
  10-12. Calculate the solubility of Ag2CO3 in
  moles per liter.

If x amount of salt that dissolves, then Ag
2x, CO32 x
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4x3 6.2 x 10-12
x3 1.6 x 10-12
x 1.2 x 10-4 mol/L
The solubility of silver carbonate is 1.2 x 10-4
mol/L
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• The Ksp value for calcium phosphide is 1.7
  x 10-40. Calculate the solubility of Ca3P2 in
  moles per liter.

Solution in lecture!
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Acid-Base Properties of Salts
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hydrolysis is the term used for the general
reaction in which a water molecule is split.
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Salts that contain the anion of a weak acid
undergo hydrolysis.
The net ionic equation for the hydrolysis of
sodium acetate is
The water molecule splits.
The solutionis basic.
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Salts that contain the cation of a weak base
undergo hydrolysis.
The net ionic equation for the hydrolysis of
ammonium chloride is
The water molecule splits.
The solutionis acidic.
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Salts derived from a strong acid and a strong
base do not undergo hydrolysis.
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Ionic Composition of Salts and the Nature of the
Aqueous Solutions They Form
Type of salt Nature ofAqueous Solution Examples
Weak base-strong acid Acid NH4Cl, NH4NO3
Strong base-weak acid Basic NaC2H3O2
Weak base-weak acid Depends on the salt NH4C2H3O2, NH4NO2
Strong base-strong acid Neutral NaCl, KBr
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Buffer Solutions The Control of pH
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A buffer solution resists changes in pH when
diluted or when small amounts of acid or base
are added.
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A weak acid mixed with its conjugate base form a
buffer solution.
Sodium acetate when mixed with acetic acid forms
a buffer solution.
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A weak acid mixed with its conjugate base form a
buffer solution.
If a small amount of HCl is added, the acetate
ions of the buffer will react with the H of the
HCl to form unionized acetic acid.
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A weak acid mixed with its conjugate base form a
buffer solution.
If a small amount of NaOH is added, the acetic
acid molecules of the buffer will react with the
OH of the NaOH to form water.
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A liter of a buffered solution contains 0.1 moles
of H2CO3 and 0.05 moles of NaHCO3. Find the pH if
Ka for the acid is 1.0 ? 107
Solution in lecture!
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The End