Chemical Equilibrium Chapter 15
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
α A + β B ↔ c C + d D
Forward Rate = kF [A]α [B]ββββ
Reverse Rate = kR [C]c [D]d
At Equilibrium
Forward Rate = Reverse Rate
Forward Rate = Reverse Rate
Writing Equilibrium Constant Expressions
For the general reactionα A + β B ↔ c C + d D
At Equilibrium
kF [A]α [B]ββββ = 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.
[ ][ ]r
p
KReactantsProducts=
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
α A + β B ↔ c C + d DAt Equilibrium
kF [A]α [B]ββββ = kR [C]c [D]d
[ ][ ]βα BD
A
cC
kk
Kd
R
Feq
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The Concept of EquilibriumThe 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.
At equilibrium, as much N2O4 reacts to form NO2 as NO2 reacts to re-form N2O4
The double arrow implies the process is dynamicForward reaction: Rate = kf[N2O4]
Reverse reaction: B → A Rate = kr[NO2]
At equilibrium kf[N2O4] = kr[NO2]
N 2 O 4 ( g ) 2 N O 2 ( g )
The Direction of the Chemical Equation and Keq
An equilibrium can be approached from any direction.For Example:
has
N2O4(g) 2NO2(g)
46.642
2
ON
2NO ==
P
PKeq
In the reverse direction:
2NO2(g) N2O4(g)
46.61
155.02NO
ON
2
42 ===P
PKeq
Other Ways to Manipulate Chemical Equations and Keq Values
The reaction
Has
• which is the square of the equilibrium constant for
2N2O4(g) 4NO2(g)
2ON
4NO
42
2
P
PKeq =
N2O4(g) 2NO2(g)
Equilibrium Concentration
Amounts of components are given as molarity { moles solute / liters of solution
orpartial pressure of a gas
Kc =NO2[ ] 2
N2O4[ ] Kp =PNO2
2
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
RTP
Vn
M ==
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) are
[CO] = 1.2x10–2 M, [Cl2]=0.054 M,
[COCl2] = 0.14 M
Calculate Kc and Kp.
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)
[ ][ ] [ ]2
22
23
OSOSO
KC =
2 SO2 (g) + O2 (g) = 2 SO3 (g)
Initial Moles # # #
Change in Moles –2x -x + 2x
Equilibrium # - 2x # - x # + 2x
Finally Concentrations at Equilibrium
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 1.00 0
Change in Moles -0.925 -0.925/2 +0.925
Equilibrium 0.075 0.537 0.925
Equil. Conc 0.075 M 0.537 M 0.925 M
The Initial Change Equilibrium method Is used to calculate equilibrium constant
The Initial Change Equilibrium method Is used to calculate equilibrium constant
[ ][ ] [ ]2
22
23
OSOSO
KC =
2108.2537.02075.0
2925.0xCK ==
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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
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.
][][][
105.62
2
24
ClNONOCl
xK ==
7322
2
108.2]1015.4[]100.1[
]4.3[x
xx=−−
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 ChangesP V = n R T
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.