Equilibrium
AP Chem
Mr. Nelson
The Concept of Equilibrium
 The
term reversible reaction is
used to describe reactions that
can go in either the forward or
the reverse direction
 Equilibrium is used to specify that
a reversible reaction has reached
an equal rate for both directions
The Concept of Equilibrium
 Chemical
equilibrium occurs
when opposing reactions are
proceeding at equal rates
◦ Concentrations of the
reactants/products remains
constant
The Concept of Equilibrium
N 2 ( g )  3H 2 ( g )  2 NH 3 ( g )

Its important to
note that forward
and reverse
reactions still
occur, although
outwardly no
apparent change
occurs
The Concept of Equilibrium
 Note
that the equilibrium
concentrations are independent of
initial concentrations
 A mathematical relationship exists
between concentration of the
reactants and products once
equilibrium has been reached
The Concept of Equilibrium

Fish Tank Example
Connection between Equilibrium &
Kinetics

For generalized reaction:
A⇋B

Forward reaction:
AB
◦ Forward Rate Law: Ratef = kf[A]

Reverse reaction:
BA
◦ Reverse Rate Law: Rater = kr[B]
Connection between Equilibrium &
Kinetics
By definition, at equilibrium forward and
reverse rates must be equal, so:
Ratef = Rater
kf[A] = kr[B]
 Also, concentrations remain constant,so:

B  k f
A kr
 constant  K eq
The Equilibrium Constant

The equilibrium constant (Keq) can be
determined at constant temperature for
any reaction at equilibrium:
aA+ bB  cC + dD
[C]c [D]d
Keq = [A]a [B]b
Keq is unitless
The Equilibrium Constant
Note that only equilibrium
concentrations can be placed into an
equilibrium constant expression!
 Also, no solids or liquids are included in
an equilibrium constant expression

 The reason for this is their concentrations remain
fairly constant in equilibrium
 Note that they are still required for the reaction to
proceed however
The Equilibrium Constant
Kc
Kw
The Many
Types of
Equilibrium
Constants,
Keq
Ka/Kb
Kp
Ksp
Concentration vs. Pressure
Equilibria

Equilibrium expressions can be written in
terms of concentration (Kc) measured in
molarity or pressure (Kp) measured in
atm
◦ Example:
N2O4 ( g )  2NO2 ( g )
2
2
[ NO2 ]
Kc 
[ N 2O4 ]
or
P NO2
Kp 
PN 2O4
Concentration vs. Pressure
Equilibria
In general, Kc is not equal to Kp
 The following equation converts between
the two:

K p  K c RT 
n
n = moles of gas products – moles of gas
reactants
◦ Note: When no change in moles, Kp=Kc
Homogeneous vs. Heterogeneous
Equilibria

The term homogeneous equilibrium is used
to describe reactions where all the species
are in the same phase
◦ Example:

N2O4 ( g )  2NO2 ( g )
The term heterogeneous equilibrium is used
to describe reactions where species are in
different phases
◦ Example:
H2CO3 (aq) + H2O (l)  HCO3- (aq) + H3O+ (aq)
Multiple Equilibria
When the products of one reaction are
the reactants of another
 Example:

A + B  C + D K1
C + D  E + F K2
A + B  E + F
Kc
Multiple Equilibria

Using their equilibrium expressions, we
can derive a mathematical relationship:
[ E ][ F ] [C ][ D] [ E ][ F ]
Kc 

x
[ A][ B] [ A][ B] [C ][ D]
K c  K1  K 2
Manipulating Equilibria Expressions

If an equilibrium is written in the reverse, the
following relationship exists:
PCl3 + Cl2  PCl5
[ PCl5 ]
Kc 
[ PCl3 ][Cl2 ]
PCl5  PCl3 + Cl2
[ PCl3 ][Cl2 ]
K c 
[ PCl5 ]
K c  1 / K c
Manipulating Equilibria Expressions

If an equilibrium is multiplied by a factor, the
following relationship exists:
PCl3 + Cl2  PCl5
[ PCl5 ]
Kc 
[ PCl3 ][Cl2 ]
2
2PCl3 + 2Cl2  2PCl5
[ PCl5 ]
K c 
2
2
[ PCl3 ] [Cl2 ]
Kc  Kc
2
The Magnititude of Equilibrium
Constants

What does Keq tell us?
◦ If Keq is very large, products will be
favored
 This implies the numerator is large in the Keq
expression compared to denominator
◦ If Keq is very small, reactants will be
favored
 This implies the denominator is large in the Keq
expression compared to numerator
◦ If Keq is close to 1, then roughly equal
amounts of reactants and products are
present at equilibrium
The Reaction Quotient (Q)
 Recall: In
electrochemistry, we
utilized “Q” which is similar in all
respects to “Keq” except in
definition
◦ Q is utilized when given initial
concentrations, not
concentrations at equilibrium
Predicting the Direction of
Reversible Reactions
 If given an initial set of conditions
for a reversible reaction, utilizing
Q and Keq, a prediction of the
direction of the reaction
(towards reactants/towards
products) can be made
Comparing Kc and Qc
Qc<Kc
Qc=Kc
Qc>Kc
• Ratio of prds to rcts is too small
• To reach equilibrium, rcts must go towards prds
• Shifts to the right
• Initial concentrations are at equilibrium
concentrations
• System is at equilibrium
• Ratio of prds to rcts is too large
• To reach equilibrium, prds must go towards rcts
• Shifts to the left
Comparison of Kc and Qc
Qc
Kc
Qc
Kc
Kc
Qc
Reactants → Products
Equilibrium: no net
change
Reactants ← Products
Calculating Equilibrium
Concentrations

Tabulate all known initial and equilibrium
concentrations
◦ When both are known, calculate the change
Using coefficients, calculate changes in
concentration for all participants in a rxn
 Calculate equilibrium concentrations

The ICE Method

This is the process used for determining
equilibrium concentrations
Example #1: Calculating Equilibrium
Concentrations
A mixture of 0.500 mol H2 and 0.500 mol
I2 was placed in a 1.00 L stainless-steel
flask at 430 °C. The equilibrium constant
Kc for the reaction:
H2 (g) + I2 (g)  2 HI (g)
is 54.3 at this temp. Calculate the
concentrations of H2, I2, and HI at
equilibrium.
H2 (g) + I2 (g)  2 HI (g)
Initial (M)
Change (M)
Equilibrium
(M)
0.500
0.500
0
-x
-x
+2x
0.500 – x
0.500 – x
2x
2
[ HI ]
K c  54.3 
[ H 2][ I 2 ]

Substituting, we get:
2
( 2 x)
54.3 
(0.500  x)(0.500  x)

Taking the square root of both sides:
2x
7.37 
(0.500  x)
x = 0.393 M
 So, the
concentrations at equilibrium
would be the following:
◦ [H2] = (0.500 – 0.393) M = 0.107 M
◦ [I2] = (0.500 – 0.393) M = 0.107 M
◦ [HI] = 2 x 0.393 M = 0.786 M
Example #2: Calculating Equilibrium
Concentrations

When 3.0 mol of I2 and 4.0 mol of Br2 are
placed in a 2.0 L reactor at 150oC, the
following reaction occurs until equilibrium
is reached:
I2(g) + Br2(g)  2IBr(g)
Chemical analysis then shows that the
reactor contains 3.2 mol of IBr. What is
the value of the equilibrium constant Kc
for the reaction?
I2 (g) + Br2 (g)  2 IBr (g)
Initial (M)
Change (M)
Equilibrium
(M)
1.5
2
0
-x
-x
+2x
1.5 – x
2–x
1.6
2
[ IBr ]
Kc 
[ I 2][ Br2 ]
 When
initial and final concentrations
are known, calculate the change:
◦ 0 + 2x = 1.6
◦ x = 0.8
 Then
determine all equilibrium
concentrations:
◦ [I2] = 1.5 – 0.8 = 0.7 M
◦ [Br2] = 2 – 0.8 = 1.2 M
◦ [IBr] = 1.6 M

Plug equilibrium concentrations into
equation and solve:
2
[ IBr ]
Kc 
[ I 2][ Br2 ]
2
[1.6]
Kc 
 3.05
[0.7][1.2]
Calculating Equilibrium
Concentrations when Keq is small
When
solving equilibrium
concentrations and Keq is
small, assume the effect of
the change (“x”) is
negligible (for rcts only)
Example: Calculating Equilibrium
Concentrations when Keq is small
 At
100 °C the equilibrium constant
(Kc) for the reaction:
COCl2 (g)  2 CO (g) + Cl2 (g)
is 2.19 x 10-10. If the initial
concentration of [COCl2] = 1.5 M,
calculate the concentrations of all
species at equilibrium.
COCl2 (g)  2 CO (g) + Cl2 (g)
Initial (M)
Change (M)
Equilibrium
(M)
1.5
0
0
-x
+2x
+x
1.5 – x
+2x
+x
2
[CO] [Cl2 ]
Kc 
[COCl2 ]
2
2
( 2 x) ( x)
Kc 
(1.5  x)
(2 x) ( x)

(1.5)
10
4x

1.5
3
4x

4
11
 x
2.19 x10
3.285 x10
4
3
3
8.2125x10
10
3
3
Le Chatelier’s Principle
 If
a stress is applied to a system
in equilibrium, the equilibrium
shifts to relieve that stress
 Forms of “stress”:
◦ Concentration
◦ Pressure
◦ Temperature
Change in Reactant or Product
Concentration
 Adding
a substance shifts a system at
equilibrium away from the added
substance
◦ System must consume added item
 Removing
a substance from a system
at equilibrium shifts the reaction
toward the removed substance
◦ System must replace removed item
Changes in Pressure
 Increase
in pressure forces an
equilibrium to shift in the
direction that reduces the
number of moles of gas in the
system
◦ Example: N2O4 (g)  2 NO2 (g)
◦ An increase in pressure causes a
shift to the left
Changes in Pressure
 Recall: Changes
in volume can also affect
changes in pressure
◦ Increase volume = decrease pressure
◦ Decrease volume = increase pressure
 Also
the addition of an inert gas (one
not in the reaction) can cause an
increase in pressure as well
Changes in Temperature
 Enthalpy
of the reaction plays a
major roll in how heat effects the
equilibrium
◦ Recall: Endothermic (+∆H) means heat
is a reactant, while exothermic (- ∆H)
means it is a product
 Treat
changes in heat in the same
manner that you treat changes in
concentration
Effects of Catalysts on Equilibrium

Recall: Catalysts lower energy barrier
between reactants and products
◦ Ea for both forward and reverse are lowered
to the same extent
Catalysts increase the rate at which
equilibrium is reached
 However, it does NOT change the
composition of the equilibrium mixture.

Solubility Equilibria
 Saturated
solutions of salts are
another type of chemical
equilibria
 Slightly soluble salts establish a
dynamic equilibrium with the
hydrated cations and anions in
solution
The Solubility Product, Ksp
The equilibrium constant, the Ksp, is no
more than the product of the ions in
solution (Recall: solids do not appear)
 For a saturated solution of AgCl, the
equation would be:



AgCl ( s)  Ag (aq)  Cl (aq)

The solubility product expression would
be:


K sp  [ Ag ][Cl ]
Sequence of Steps for Solubility
Equilibria
Solubility of
compound
Molar solubility
of compound
Concentrations
of cations and
anions
Ksp of
compound
Ksp of
compound
Concentrations
of cations and
anions
Molar solubility
of compound
Solubility of
compound
Solubility
 Solubility
can be expressed two
ways:
◦ Molar solubility is the moles of solute
per 1L of saturated solution (mol/L)
◦ Solubility is the number of grams of
solute per 1 L of saturated solution
(g/L)
 Ksp
expressions require molar
solubility
Determining Ksp by Experimental
Measurements
 Use
ICE table,
but with “s” for
change (instead
of “x”)
 “s” is not only
the change, but
the molar
solubility
Determining Ksp: Example #1
The solubility of CaSO4 is found
to be 0.67 g/L. Calculate the
value of Ksp for calcium sulfate.
Solubility of
CaSO4 in
g/L
Molar
solubility of
CaSO4
[Ca2+] and
[SO42-]
Ksp of
CaSO4
Determining Ksp: Example #1
2
2
CaSO4 ( s )  Ca (aq)  SO4 (aq)
Initial (M)
Change (M)
-s
Equilibrium
(M)
2
0
0
+s
+s
s
s
2
K sp  [Ca ][ SO4 ]  s
2
Determining Ksp: Example #1
0.67 g CaSO4 1 mol CaSO4
3

 4.9 x10 M  s
1 L solution 136.2 g CaSO4

As CaSO4 dissolves, we get 1 mole Ca2+, and 1
mole SO42-, so at equilibrium:
2
2
3
[Ca ]  [ SO4 ]  4.9 x10 M
3
3
K sp  (4.9 x10 )(4.9 x10 )  2.4 x10
5
Determining Ksp: Example #2
Lead (II) chloride dissolves to a slight
extent in water according to the
equation:
2

PbCl2 ( s)  Pb (aq)  2Cl (aq)
Calculate the Ksp if the solubility of
lead (II) chloride has been found to be
1.62 x 10-2 M
Determining Ksp: Example #2
Molar
solubility of
PbCl2
[Pb2+] and [Cl]
Ksp of PbCl2
2

PbCl2 ( s)  Pb (aq)  2Cl (aq)
Initial (M)
Change (M)
Equilibrium
(M)
-s
0
0
+s
+2s
s
2s
Determining Ksp: Example #2
2
 2
K sp  [ Pb ][Cl ]  (s)(2s)  4s
2
5
s  1.8 x10 M
5 3
K sp  4(1.8x10 )
K sp  2.33x10
14
3
Calculating Solubility from Ksp:
Example #1
The Ksp for CaCO3 is 3.8 x 10-9 at 25
°C. Calculate the solubility (in g/L) of
calcium carbonate in pure water.
Ksp of CaCO3
Concentrations
of [Ca2+] and
[CO32-]
Molar solubility
of CaCO3
Solubility of
CaCO3
Calculating Solubility from Ksp:
Example #1
2
2
CaCO3 (s)  Ca (aq)  CO3 (aq)
Initial (M)
Change (M)
-s
Equilibrium
(M)
2
0
0
+s
+s
s
s
2
K sp  [Ca ][CO3 ]  s
2
Calculating Solubility from Ksp:
Example #1
9
3.8 x10  s
5
s  6.16 x10 M
2
5
6.16 x10 mol CaCO3 100 g CaCO3

1 L solution
1 mol CaCO3
3
 6.16 x10 g / L
Predicting Precipitation Reactions
using Q

Using the reaction quotient, Q, we can
◦ Decide whether a ppt will form
◦ Calculate ion concentration required to begin
the ppt. of an insoluble salt
Q < Ksp
Q = Ksp
Q > Ksp
• System has
too much
reactant
• No ppt
• System at
equilibrium
• System has
too much
product
• Forms ppt!
• No ppt
Predicting Precipitation Reactions
 Steps
1. Determine insoluble product using
solubility rules
2. Calculate cation and anion
concentration of insoluble product (if
mixed)
3. Calculate Q
4. Compare to Ksp
Example: Predicting Precipitation
Exactly 200 mL of 0.0040 M BaCl2
are mixed with exactly 600 mL of
0.0080 M K2SO4. Will a precipitate
form? (Ksp of insoluble salt is 1.1 x
10-10)
The Common Ion Effect
 Le
Chatelier’s Principle can describe
the effect of an added common ion
◦ For example, if NaCl were added to a
saturated solution of PbCl2, what effect
would occur?
◦ Increase product concentration shifts
equilibrium towards reactants, forming
more ppt.
Solubility and the Common Ion
Effect

The presence of a
common ion will
cause the
equilibrium to shift
so that even less
of the substance
(with smaller Ksp
value) will dissolve
The Common Ion Effect

Be aware: pH can affect solubility
◦ The addition of an acid or base could have a
side reaction, decreasing concentration of an
ion
◦ If an acid were added to the following, what
effect would be observed?
Mg (OH ) 2 ( s)  Mg 2 (aq)  2OH  (aq)
 H+ combines with OH-, so [OH-] is
decreased, forcing equilibrium to the right
Example: Common Ion Effect
Calculate the solubility of silver chloride
(in g/L) in a 6.5 x 10-3 M silver nitrate
solution. (Ksp = 1.6 x 10-10)
◦ Since silver nitrate is a soluble, strong
electrolyte:


AgNO3 (s) 
 Ag (aq)  NO3 (aq)
H 2O
6.5 x 10-3 M
6.5 x 10-3 M
Example: Common Ion Effect


AgCl( s)  Ag (aq)  Cl (aq)
Initial (M)
Change (M)
-s
Equilibrium
(M)
6.5 x 10-3
0.00
+s
+s
(6.5 x 10-3 + s)
s
Because Ksp is so small, s must be very small in comparison to 6.5 x
10-3, so we approximate that 6.5 x 10-3 + s  6.5 x 10-3
K sp  1.6 x10
10
3
 (6.5x10 )s
8
s  2.5 x10 M
Selective Precipitation

In certain problems, you will have to
determine which of two possible
precipitates will form
◦ The salt with the lower Ksp will precipitate
first

To determine the amount of an ion
needed to precipitate, recall that Ksp
concentrations are saturated, so anything
greater than this will form a ppt.
Example: Selective Precipitation
A solution contains 1.0 x 10-4 M Cu+ and
2.0 x 10-3 M Pb2+. If a source of I- is added
gradually to this solution, will PbI2 (Ksp =
1.4 x 10-8) or CuI (Ksp = 5.3 x 10-12)
precipitate first? Specify the concentration
of I- necessary to begin precipitation of
each salt