Chapter 18: Chemical
Equilibrium
1. The Concept of Equilibrium
 A. Equilibrium exists when two opposing
processes occur at the same rate.
 B. Reversible reactions 2NO2 (g) N2O4(g)
 1. Products take part in a separate reaction to reform the
reactants
 2. A chemical reaction in which the products can
regenerate the original reactants is a reversible
reaction
 3. Use two half arrows to indicate a reversible reaction
 Forward reaction (above arrow) reverse reaction (below
arrow)
1. The Concept of Equilibrium
 C. Chemical Equilibrium
 1. The reaction rate depends on different
factors
 -The concentrations of the substances in
the reaction: reaction rate is proportional
to reactant concentration.
 2. When substance enter into the reaction,
the concentration of the reactants
decreases as the reactant are converted
into products.
 -Therefore the concentration of the
products increase.
The Concept of Equilibrium
 3. Chemical equilibrium is the
point at which the forward
reaction is equal to the rate of the
reverse reaction.
 - The concentrations of the
reactants and products become
constant
The Concept of Equilibrium
 4. Chemical equilibrium is the state in
which the concentration of reactants and
products remain constant with time
because the rate at which they are being
formed is equal to the rate at which they
are consumed in the opposite reaction.
 5. Square brackets [ ] are used to denote
concentration of a substance.
II. The Law of Chemical Equilibrium
 A. The Equilibrium Constant
 1. The Law of Mass Action
 A. Formulated by Guldbert and Waage (1864)
 B. Expresses the relative concentrations of
reactants and products at equilibrium in
terms of quantity called the equilibrium
constant (Keq)
II. The Law of Chemical Equilibrium
 C. In the reaction:
 aA+ bB  cC+ dD
 Keq= [C]c [D]d
[A]a [B]b
a,b,c,d = coefficients for the substances
A,B,C,D = reactants and products
II. The Law of Chemical Equilibrium
 D. This ratio is always a constant value for
a given reaction regardless of initial
concentrations at a given temperature.
 1. This is called The
Law of Chemical
Equilibrium.
 2. The equilibrium can be reached from either
direction.
 E. Equilibrium position: Set of equilibrium
concentrations in an experiment
II. The Law of Chemical Equilibrium
 F. The equilibrium constant
is a measure of the extent
to which a reaction
proceeds to completion.
II. The Law of Chemical Equilibrium
1. Keq >> 1:
The numerator must be larger
than the denominator
The equilibrium concentrations
of the products must be larger
than the reactants
The equilibrium “lies to the
right”
 Keq <<1 :
The numerator must be much
smaller than the denominator
The equilibrium concentrations
of the products must be
smaller than the reactants
The equilibrium “lies to the
left”
II. The Law of Chemical Equilibrium
 B. Homogeneous and Heterogeneous
Equilibria
 1. Homogeneous Equilibria: Equilibrium
conditions for reaction in which all the
reactants and products are in the same state.
 2. Heterogeneous Equilibria: Equilibrium
conditions for reactions in which the
substances involved are in more than one
state.
II. The Law of Chemical Equilibrium
 3. Example of heterogeneous equilibria
 NH4Cl(s)
NH HCl
3 (g) +
(g)
 The concentration of a pure liquid or
solid is basically constant and
unaffected by the temperature. The
concentration is density/molar mass.
 Therefore the concentration of a pure
liquid or solid does NOT change during
a reaction
 The concentrations of liquids and
solids are left out of the equilibrium
expression.
 So, Keq for our example =
[HCl][NH3]
II. The Law of Chemical Equilibrium
 C. The Reaction Quotient
 1. Q: Used to determine if a
reaction is at equilibrium.
 2. Calculated like Keq except that
it uses the concentrations that exist
at the time the measurement is
taken not the equilibrium
concentrations.
II. The Law of Chemical Equilibrium
 3. Example
At 472° C, N2 (g) + 3 H2 (g)
2NH3 (g)
Keq= .105
You measure the concentrations to be
[NH3]=.15M [N2]=.0020M [H2]= .10 M
Find Q and decide if you are at
equilibrium
2
[NH3]
3
[N2] [H2]
Q=
/
2
3
Q= [.15M] / [.0020][.10] =
1.1 x 104
Not at equilibrium because Q does not equal
K
eq
II. The Law of Chemical Equilibrium
 3. Which direction will the
reaction proceed?
Q<Keq : There is too much
of the reactants and too little
of the products  the
reaction will shift to the right
to make more products.
3. Which Direction will a reaction
proceed?
Q>Keq : There is too
much of the products and
too little of the reactants 
the reaction will shift to the
left.
Q=Keq  No Shift.
II. The Law of Chemical Equilibrium
 4. Try: At 448° C, Keq=50.5 for
H2(g)+I2(g)
2HI(g)
 Find Q and predict how the
reaction will shift
 [H2] = .150 M [I2 ]=.175M
[HI]=.950 M
 Q=
/ [H2] [I2 ] =
 [.950 M]2 / [.150 M][.175M]=34.4
 Shifts to the right because
2
[HI]
Q < Keq
III. LeChatelier’s Principle
 A. If a change in conditions is
imposed on a system at equilibrium,
the equilibrium position will shift in the
direction that tends to reduce the
change in conditions
 B. Changes in Concentration
 1. If more of a substance is
added to a reaction at
equilibrium, the concentration of
that substances increases.
A. Reaction will return to
equilibrium by consuming
some of the added substance.
 2. If a substance is removed, its
concentration decreases
 A. The reaction will return to
equilibrium by producing more of
the substance that was removed.
 3. Only the equilibrium shifts, NOT
the equilibrium constant.
III. LeChatelier’s Principle
 C. Changes in Pressure
 1. If the total pressure of a system
increases, the system will shift to reduce
the pressure by proceeding in the
direction that produces fewer molecules
 A. The reaction changes the
equilibrium position but NOT the
equilibrium constant.
D. Effect of Changing Temperature
 1. Example
 H2(g)+I2(g)
2HI(g)+ heat
 Keq=54.5 @ 400° C
 Keq=45.9 @ 490° C
 So raising the temperature
causes the reaction to proceed
less completely to the products
III. LeChatelier’s Principle
 2. If a forward reaction is exothermic,
the reverse reaction is endothermic in
a reversible reaction.
 3. Changing temperature DOES
change the value of the
Equilibrium Constant.
III. LeChatelier’s Principle
 E. The Haber Process
1. Haber examined the
reaction :
N2(g)+3H2(g)
2NH3(g)+ heat
III. LeChatelier’s Principle
 A. This reaction reached equilibrium
before producing much ammonia.
 B. Haber developed a process and the
equipment necessary to reach the
pressures and temperatures needed to
produce large amounts of ammonia
III. LeChatelier’s Principle
 i. Ammonia is continuously removed
 ii. Pressure is increased to force a reaction to
the right.
 iii. The forward reaction is exothermic but he
increased the temperature to speed up the
reaction even though it drove the reverse
reaction.
 1. Compensated for this by increasing the pressure.
 2.If he had decreased the temperature, the reaction
moved too slowly.
III. LeChatelier’s Principle
 C. Haber also developed the use
of chlorine as poison gas weapon.
 D. Reactions that go to
completion
 1. Formation of a gas
 H2CO3(aq)
H2O(g) + CO2(g)
III. LeChatelier’s Principle
 2. Formation of a precipitate
 NaCl(aq) +AgNO3(aq) NaNO3 (aq) + AgCl(s)
 3. Formation of a slightly ionized product
 HCl(aq) + NaOH(aq) NaCl(aq) + H2O
III. LeChatelier’s Principle
 G. Common-ion Effect
 1. An equilibrium reaction may be driven in
the desired direction by applying LeChatelier’s
principle
 2. Example: HCl bubbled into NaCl solution
 NaCl (s)
Na+ (aq) + Cl- (aq)
 Keq = [Na+] [Cl-]
 HCl H3O+ +Cl – (aq)
III. LeChatelier’s Principle
 3. Common Ion effect:
 -The addition of an ion common
to two solutes brings about
precipitation or reduced
ionization.
IV. Equilibria of Acids, Bases and
Salts
 A. The acid dissociation constant
 1. When the products and
reactants of a reaction reach
equilibrium, a certain ratio of
their concentrations always has
the same value
IV. Equilibria of Acids, Bases and
Salts
 2. For the reaction:
 HA (acid) + H2O H3O+ (hydronium
ion) + A- (anion)
 Keq= [H3O+][A-] / [HA][H2O]
 Ka = [H3O+][A-] / [HA]
 Ka= acid dissociation constant
IV. Equilibria of Acids, Bases and
Salts
 3. The greater the Ka, the further the
action runs to completion. So Ka is the
measure of the strength of an acid
 4. For diprotic and triprotic acids, each
dissociation takes place in a separate
step.
IV. Equilibria of Acids, Bases and
Salts
 A. Step 1:H2CO3+ H2O H3O+ +CO3  Ka1= [H3O+][HCO3-] / [H2CO3]=4.5x10-7
 B. Step 2: HCO3-+H2O
 Ka = 5.6 x 10-11
2
H3O+ + CO32-
IV. Equilibria of Acids, Bases and
Salts
 B. The Base Dissociation Constant
 1. B(base) + H2O OH- + HB+
(cation)
 2. Kb: the measure of the strength
of the base
 Kb= [HB+][OH-] / [B]
IV. Equilibria of Acids, Bases and
Salts
 C. Calculating Dissociation Constants
 1. Example: acetic acid is a weak
monoprotic acid that dissociates into an
acetate ion and a hydronium ion in
aqueous solution. Calculate Ka for
acetic acid if a 1.0 M solution results in
an equilibrium hydronium concentration
of 0.0042 M
IV. Equilibria of Acids, Bases and
Salts
 2. You try : Ammonia is a weak
base. If the initial concentration of
ammonia is 0.150 M and the
equilibrium concentration of
hydroxide is 0.0016M, calculate
Kb for ammonia.
IV. Equilibria of Acids, Bases and
Salts
 D. Buffers

1. A solution that can resist changes in pH

2. Weak acid and salt of a weak acid

3. Weak base and salt of a weak base
 E. Ionization Constant of Water

Kw=concentration of
[OH-][H30+] = 1.0 x10-14
IV. Equilibria of Acids, Bases and
Salts
 F. Hydrolysis of Salts
 1. Hydrolysis:
 A. reaction between H20 molecules
and ions of a dissolved salt.
 2. Anion Hydrolysis:
if the acid is a weak acid, its
conjugate base (anion) will be
strong enough to remove H+
from H20 molecules to form OHions.
IV. Equilibria of Acids, Bases and
Salts
3. Cation
Hydrolysis:
 if the base is weak its
conjugate acid which is now
the cation will be strong
enough to donate a H+ ion to a
H2O molecule to form H30+
ions
IV. Equilibria of Acids, Bases and
Salts
 G. Hydrolysis in Acid-Base Reactions
 1. Strong acid-Strong base  neutral
solution
 2. Strong acid-Weak base  acidic
solution
 3. Strong base -Weak acid basic
solution
 4. Weak acid-Weak base acidic,
basic, or neutral solution
V. Solubility equilibrium
 A. Solubility product
 1. Soluble: A substance whose
solubility is greater than 1 gram
per 100 grams of H20
 2. Insoluble: A substance
whose solubility is less than 0.1
grams per 100 grams of H2O.
 3. Slightly Soluble: Solubility
between 0.1 and 1.0 gram per 100
grams of H20.
 4. Saturated Solution: contains the
maximum amount of solute possible
at a given temperature in
equilibrium with an undissolved
excess of substances.
V. Solubility equilibrium
 5. Example: Ksp= solubility
product constant: the product of
the molar concentration of its ions
in a saturated solution, each
raised to the power that is the
coefficient of that ion in the
balanced chemical equation
V. Solubility equilibrium
 6. Calculate the solubility
product constant of Lead (II)
Chloride which has a solubility
of 1.0 g/100 g water at 20
degrees Celsius.
V. Solubility equilibrium
 B. Precipitation Calculations
 1. If the ion product is greater than
the Ksp, the salt precipitates.
 2. Example: Will a precipitate form if
20.0mL of 0.010M Barium Chloride
solution is mixed with 20.0mL of
0.0050 M Sodium Sulfate solution?
 C. Read about limitations on
the use of Ksp on page 620.