Chemical Equilibrium
Chapter 15
Homogeneous equilibrium:
When all reacting species are in the
same phase.
Heterogeneous equilibrium:
When all reacting species are
NOT in the same phase.
Heterogeneous Equilibrium:
When reacting species are in
DIFFERENT phases.
SOLID and LIQUID phases are
EXCLUDED from the equilibrium
expression because they do not have
“concentrations”
Chemical Equilibrium
A state achieved when
the RATES of the forward and
reverse reactions are EQUAL
&
the concentrations of the reactants and
products remain constant.
Equilibrium Constant
The equilibrium constant (Keq) is
independent of concentration changes,
but
dependent on the temperature.
Writing Equilibrium Constant Expressions
For the general reaction
αA+βB ↔cC+dD
Forward Rate = kF [A]α [B]β
Reverse Rate = kR [C]c [D]d
At Equilibrium
Forward Rate = Reverse Rate
Forward Rate = Reverse Rate
Writing Equilibrium Constant Expressions
For the general reaction
αA+βB ↔cC+dD
At Equilibrium
kF [A]α [B]β = kR [C]c [D]d
Put CONSTANTS on one side of equation
And
Concentrations on other side of equation
Equilibrium Constant
• The equilibrium expression compares
reactant & product concentrations.
[
Products]
K=
r
[Reactants]
p
Size of Equilibrium Constant
If Kc > 103, products predominate
If Kc < 10–3, reactants predominate
If Kc is in the range 10–3 to 103,
appreciable concentrations of both
reactants and products are present.
Writing Equilibrium Constant Expressions
αA+βB ↔cC+dD
At Equilibrium
kF [A]α [B]β = kR [C]c [D]d
K eq
kF
=
=
kR
C
c
α
A
[D ]
β
[B ]
d
The Concept of Equilibrium
The point at which the rate of decomposition
N2O4(g) → 2NO2(g)
equals the rate of dimerization
2NO2(g) → N2O4(g).
is a dynamic equilibrium.
The equilibrium is dynamic because the reaction
has not stopped: the opposing rates are equal.
N 2O 4(g )
2 N O 2(g )
At equilibrium, as much N2O4 reacts to form
NO2 as NO2 reacts to re-form N2O4
The double arrow implies the process is dynamic
Forward reaction: Rate = kf[N2O4]
Reverse reaction: B → A Rate = kr[NO2]
At equilibrium kf[N2O4] = kr[NO2]
The Direction of the Chemical
Equation and Keq
An equilibrium can be approached from any direction.
For Example:
N2O4(g)
2NO2(g)
has
K eq =
2
NO 2
P
PN 2O 4
= 6.46
In the reverse direction:
2NO2(g)
K eq =
PN 2O 4
2
PNO
2
N2O4(g)
1
= 0.155 =
6.46
Other Ways to Manipulate Chemical
Equations and Keq Values
The reaction
Has
2N2O4(g)
K eq =
4NO2(g)
4
NO 2
2
N 2O 4
P
P
• which is the square of the equilibrium
constant for N O (g)
2NO (g)
2 4
2
Equilibrium Concentration
Amounts of components are given as
molarity { moles solute / liters of solution
or
partial pressure of a gas
]2
[NO2
Kc = [
N2O4 ]
Kp =
2
PNO2
PN2O4
Equilibrium Constant
Write the Kp expressions for:
2 N2O5(g) ↔ 4 NO2(g) + O2(g)
Kp = (P NO2 )4 (P O2) / (P N2O5)2
Equilibrium Constant
Write the Kc expressions for:
2 N2O5(g) ↔ 4 NO2(g) + O2(g)
KC = [NO2 ]4 [ O2] / [N2O5]2
2 N2O5(g) ↔ 4 NO2(g) + O2(g)
The Kp and Kc expressions :
Kp = (P NO2 )4 (P O2) / (P N2O5)2
KC = [NO2 ]4 [ O2] / [N2O5]2
Is Kp = KC ?
No !
Relation between Gas Pressure
and Concentration
•
•
•
•
P V = n R T Ideal Gas Equation
Concentration (M) = moles / Liter
M=n/V
Rearranging P V = n R T
n
P
M= =
V RT
When are KP and KC EQUAL ?
For which of the following is KP = KC
(a) CO2(g) + C(s) ↔ 2 CO(g)
(b) Hg(l) + Hg2+(aq) ↔ Hg22+(aq)
(c)2Fe(s) + 3H2O(g) ↔ Fe2O3(s) + 3H2(g)
(d) 2 H2O(l) ↔ 2 H2(g) + O2(g)
Homogeneous And Heterogeneous
Equilibria
• Homogeneous Equilibrium When all
reacting species are in the same phase.
• Heterogeneous Equilibrium: When all
reacting species are NOT in the same phase.
Heterogeneous Equilibrium:
When reacting species are in
DIFFERENT phases.
SOLID and LIQUID phases are
EXCLUDED from the equilibrium
expression because they do not have
“concentrations”
Equilibrium Constant
The equilibrium concentrations at 1000 K
for the reaction
CO(g) + Cl2(g) ↔ COCl2(g)
[CO] = 1.2x10–2 M,
[Cl2]=0.054 M,
[COCl2] = 0.14 M
Calculate Kc and Kp.
are
CO(g) + Cl2(g) ↔ COCl2(g)
[CO] = 1.2x10–2 M,
[Cl2]=0.054 M,
[COCl2] = 0.14 M
Kc = [COCl2] / [CO][Cl2]
Kc = 0.14 /(1.2x10–2 )(0.054 ) = 21578.2984
and
KC = KP (RT)
Why ?
Kp = KC / RT = KC / (0.0821)(1000) =
Sulfur dioxide + Oxygen = Sulfur trioxide
SO2 (g) + O2 (g) = SO3 (g)
2 SO2 (g) + O2 (g) = 2 SO3 (g)
[
SO3 ]
KC =
2
[SO2 ] [O2 ]
2
The Initial Change Equilibrium method
Is used to calculate equilibrium constant
2 SO2 (g) + O2 (g) = 2 SO3 (g)
Initial Moles
#
Change in Moles
–2x
Equilibrium
# - 2x
#
-x
# - x
Finally Concentrations at Equilibrium
#
+ 2x
# + 2x
The Initial Change Equilibrium method
Is used to calculate equilibrium constant
1.00 mole of SO2 and 1.00 mole of O2 are put
into a 1.00 liter flask. At equilibrium, 0.0925
mole of SO3 is formed. Calculate Kc at
1000 K for the reaction
2 SO2 (g) + O2 (g) = 2 SO3 (g)
The Initial Change Equilibrium method
Is used to calculate equilibrium constant
2 SO2 (g) + O2 (g) = 2 SO3 (g)
Initial Moles
1.00
Change in Moles -0.925
Equilibrium
0.075
Equil. Conc 0.075 M
1.00
0
-0.925/2
+0.925
0.537
0.925
0.537 M
0.925 M
The Initial Change Equilibrium method
Is used to calculate equilibrium constant
[
SO3 ]
KC =
2
[SO2 ] [O2 ]
2
2
0.925
2
K =
= 2.8 x10
C
2
0.075 0.537
Methane (CH4) reacts with hydrogen
sulfide to yield H2 and carbon disulfide
What is the value of Kp if the partial
pressures in an equilibrium mixture
are 0.20 atm of CH4, 0.25 atm of H2S,
0.52 atm of CS2, and 0.10 atm of H2?
How do you solve the problem?
•
•
•
•
•
1st Write Reaction
2nd Balance Reaction
3rd Write Equilibrium Constant Equation
4th Put Equilibrium Pressures in Equation
Do the Arithmetic
Using Equilibrium Constants
The equilibrium constant (K c ) for
the formation of nitrosyl chloride,
2 NO(g) + Cl2(g) ↔ 2 NOCl(g)
from nitric oxide & chlorine gas is
6.5 x 10 4 at 35 °C.
2NO(g) + Cl2(g) ↔ 2NOCl(g) Kc= 6.5x104
2.0 x 10–2 moles of NO,
8.3 x 10–3 moles of Cl2, &
6.8 moles of NOCl
are mixed in a 2.0–L flask.
Is the system at equilibrium?
If not, what will happen?
2NO(g) + Cl2(g) ↔ 2NOCl(g) Kc= 6.5x104
K = 6 . 5 x 10
4
[ NOCl ] 2
=
[ NO ] 2 [ Cl 2 ]
2.0 x 10–2 moles of NO, 8.3 x 10–3 moles of
Cl2, 6.8 moles of NOCl in a 2.0–L flask.
2
[3.4]
7
= 2.8 x10
−2 2
−3
[1.0 x10 ] [4.15 x10 ]
The reaction quotient (Qc)
Predicts reaction direction.
Qc > Kc
System proceeds to form reactants.
Qc = Kc
System is at equilibrium.
Qc < Kc
System proceeds to form products.
Using Equilibrium Constants
• A mixture of 0.500mol H2 and 0.500mol I2
was placed in a 1.00–L stainless steel flask at
700°C. The equilibrium constant Kc is 57
H2(g) + I2(g) ↔ 2HI(g)
• Calculate the equilibrium concentrations.
Le Chatelier’s principle:
If an external stress is applied to a system at
equilibrium, the system adjusts in such a
way that the stress is partially offset.
Le Chatelier’s principle:
Stress may be changes in
• concentration,
• pressure, volume, or
• temperature
that removes the system from equilibrium.
Concentration Changes:
The concentration stress of an added
reactant or product is relieved by
reaction in the direction that consumes
the added substance.
Concentration Changes:
The concentration stress of a removed
reactant or product is relieved by
reaction in the direction that
replenishes the removed substance.
Use the synthesis of ammonia as an example
N2(g) + 3H2(g) ↔ 2NH3(g)
At equilibrium [N2 ] = 0.50M [H2 ] = 3.00M
and [NH3 ] = 1.98 M at 700K
N2(g) + 3H2(g) ↔ 2NH3(g)
At equilibrium [N2 ] = 0.50M [H2 ] = 3.00M
and [NH3 ] = 1.98 M at 700K
what happens when the concentration
of N2 is increased to 1.50M?
N2(g) + 3H2(g) ↔ 2NH3(g)
when the concentration of N2 is increased
Le Châtelier’s principle says the
reaction will relieve the stress by
converting the N2 to NH3
N2(g) + 3H2(g) ↔ 2NH3(g) Kc = 0.291 at 700K
If H2 is increased
Le Châtelier’s principle tells us the
reaction will relieve the stress by
converting the extra H2 to NH3
N2(g) + 3H2(g) ↔ 2NH3(g) Kc = 0.291 at 700K
If NH3 is removed
Le Châtelier’s principle tells us the
reaction will relieve the stress by
producing more NH3
Volume and Pressure Changes
PV=nRT
Pressure is inversely proportional to volume
Increasing pressure = Decreasing volume
From P V = n R T
P = M (RT)
increasing pressure or decreasing volume
increases concentration.
N2(g) + 3H2(g) ↔ 2NH3(g)
Use the synthesis of ammonia as an example
There are Four (4) units of gas reactants but
only Two (2) units of gas products
Therefore, an increase in pressure would
shift the equilibrium to the right
Le Châtelier’s Principle
Volume and Pressure Changes
Only reactions containing gases are
affected by changes in volume and
pressure.
The reaction of iron (III) oxide
with carbon monoxide occurs in a
blast furnace when iron ore is
reduced to iron metal:
Fe2O3(s) + 3 CO(g) ↔ 2 Fe(l) + 3 CO2(g)
Fe2O3(s) + 3 CO(g) ↔ 2 Fe(l) + 3 CO2(g)
•
Use Le Châtelier’s principle to predict the
direction of reaction when an equilibrium
mixture is disturbed by:
(a) Adding CO
SHIFT TO RIGHT
Fe2O3(s) + 3 CO(g) ↔ 2 Fe(l) + 3 CO2(g)
• Use Le Châtelier’s principle to predict the
direction of reaction when an equilibrium
mixture is disturbed by:
(b) Removing CO2
SHIFT TO RIGHT
Fe2O3(s) + 3 CO(g) ↔ 2 Fe(l) + 3 CO2(g)
• Use Le Châtelier’s principle to predict the
direction of reaction when an equilibrium
mixture is disturbed by:
(c) Increasing Pressure
NO EFFECT
Why ????
Fe2O3(s) + 3 CO(g) ↔ 2 Fe(l) + 3 CO2(g)
• Use Le Châtelier’s principle to predict the
direction of reaction when an equilibrium
mixture is disturbed by:
(d) Adding Fe2O3
NO EFFECT
Why ????
Temperature Changes:
1. Changes the equilibrium constant
2. Endothermic processes are favored when
temperature increases.
3. Exothermic processes are favored when
temperature decreases.
Le Châtelier’s Principle & Temperature
In the first step of the Ostwald process
for synthesis of nitric acid, ammonia
is oxidized to nitrogen monoxide
4NH3(g) + 5O2(g) ↔ 4NO(g) + 6H2O(g)
H° = –905.6 kJ
4NH3(g) + 5O2(g) ↔ 4NO(g) + 6H2O(g)
H° = –905.6 kJ
What happens as temperature is increased ?
Reaction is EXO thermic therefore an
increase in temperature will shift the
equilibrium toward the reactants
Le Châtelier’s Principle
Catalysis: No effect.