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Equilibrium
Chemical Equilibrium
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
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Equilibrium occurs
when opposing
reactions are
proceeding at equal
rates
Equilibrium can be
reached from either
direction
Ratio of
concentrations will
become constant
Equilibrium Constant

Haber Process
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N2(g) + 3H2(g) ↔
2NH3(g)
aA + bB ↔ cC + dD
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Ke q = (PC)c(PD)d
(PA)a(PB)b
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Ke q = [C]c[D]d
[A]a[B]b
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Equilibrium Expression
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Ke q = equilibrium
constant
Equilibrium Constant
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N2O4(g) ↔ 2NO2(g)
What is the
equilibrium
expression?
What do you notice
about the equilibrium
constant?
Practice
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Write the equilibrium expression for Ke q for the
following reactions:
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2O3(g) ↔ 3O2(g)
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2NO(g) + Cl2(g) ↔ 2NOCl(g)
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Ag+(aq) + 2NH3(aq) ↔ Ag(NH3)2+(aq)
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H2(g) + I2(g) ↔ 2HI(g)
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Cd+2(aq) + 4Br-(aq) ↔ CdBr42-(aq)
Equilibrium Constant
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The magnitude of the
constant can tell us
what the make up
equilibrium mixture is
If Ke q > 1 products
predominate
If Ke q < 1 reactants
predominate
Practice
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The reaction of N2 with O2 to form NO might be
considered a means of “fixing” nitrogen.
N2(g) + O2(g) ↔ 2NO(g)
The value for the equilibrium constant for this
reaction at 25°C is Ke q = 1x10- 3 0. Describe the
feasibility of the reaction for nitrogen fixation.
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The equilibrium constant for the reaction of
H2(g) + I2(g) ↔ 2HI(g) varies with temperature
as follows: Ke q = 794 at 298K; Ke q = 54 at
700K. Is the formation of HI favored more at
Equilibrium Constant
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Because equilibrium can be reached from either
direction how we write equilibrium expressions
is arbitrary.
N2O4(g) ↔ 2NO2(g)
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In terms of the forward reaction
Keq = (PN O2)2 = 6.46 (@ 100°C)
PN 2 O 4
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In terms of the reverse reaction
Keq = PN 2 O 4 = 0.155 (@ 100°C)
(PN O2)2
Equilibrium Constant
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N2O4(g) ↔ 2NO2(g)
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Keq = (PN O2)2 = 6.46 (@ 100°C)
PN 2 O 4
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2N2O4(g) ↔ 4NO2(g)
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Keq = (PN O 2)4
(PN 2 O 4)2
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The equilibrium constant of a reaction that has been
multiplied by a number is the equilibrium constant
raised to that power
Equilibrium Constant
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2NOBr(g) ↔ 2NO(g) + Br2(g)
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Ke q= 0.42
What is the equilibrium expression?
Br2(g) + Cl2(g) ↔ 2BrCl(g)
Ke q= 7.2
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What is the equilibrium expression?
The sum of these two equations is
2NOBr(g) + Cl2(g) ↔ 2NO(g) + 2BrCl(g)
Ke q= 3.0
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What is the equilibrium expression?
Practice
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Given the following information,
HF(aq) ↔ H+(aq) + F-(aq)
Ke q= 6.8x10- 4
H2C2O4(aq) ↔ 2H+(aq) + C2O42 -(aq)
3.8x10- 6
Ke q=
determine the value of the rate constant for the
following reaction:
2HF(aq) + C2O42 -(aq) ↔ 2F-(aq) + H2C2O4(aq)
Units of Keq
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Equilibrium constants have no units (are
dimensionless) even though we put in
concentrations and pressures.
What we actually put in are ratios of these
quantities to a certain reference point.
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For concentrations the reference is 1M
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For pressures the reference is 1atm
Keq = (PN O 2/Pr e f)2
(PN 2 O 4/Pr e f)
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This allows us to put both pressures and
Heterogeneous Equilibria
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Homogeneous equilibria – all substances are in
the same state
Heterogeneous equilibria – substances are in
different states
PbCl2(s) ↔ Pb+ 2(aq) + 2Cl-(aq)
Keq = [Pb2 +][Cl-]2
[PbCl2]
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If a pure solid or liquid is involved in a
heterogeneous equilibrium its concentration is
not included in the expression
Rules for Equilibrium Expressions
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Partial pressures of gases are
substituted into the equilibrium
constant expression.
Molar concentrations of
dissolved species are
substituted into the equilibrium
constant expression.
Pure solids, pure liquids, and
solvents are not included in the
equilibrium constant
expression.
Practice
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Write the equilibrium constant expressions for
each of the following reactions:
CO2(g) + H2(g) ↔ CO(g) + H2O(l)
SnO2(s) + 2CO(g) ↔ Sn(s) + 2CO2(g)
Sn(s) + 2H+(aq) ↔ Sn2 +(aq) + H2(g)
Practice
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Each of the following mixtures was placed in a
closed container and allowed to stand. Which of
the mixtures is capable of attaining the equilibrium
expressed by this reaction - CaCO3(s) ↔ CaO(s)
+ CO2(g).
a) pure CaCO3
b) CaO and a pressure of CO2 greater than the
value of Ke q
c) some CaCO3 and a pressure of CO2 greater than
the value of Ke q
d) CaCO3 and CaO
Calculating Equilibrium Constants
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To calculate the equilibrium constants for a
reaction we need to know the equilibrium
concentration for at least one of the substances.
We will do this using the following procedure
1. Tabulate the known initial and equilibrium
concentrations of all species in the equilibrium
constant expression.
2.For all species for which both the initial and equilibrium
concentrations are known, calculate the change in
concentration that occurs as the system reaches
equilibrium.
3.Use the stoichiometry of the reaction (that is, use the
Calculating Equilibrium Constants
Enough ammonia is dissolved in 5.00 liters of
water at 25°C to produce a solution that is
0.00124M in ammonia. The solution is then
allowed to come to equilibrium. Analysis of the
equilibrium mixture shows that the
- 4 M. Calculate
+ - is 4.64x10
concentration
of
OH
NH3(aq) + H2O(l) ↔ NH4 (aq) + OH (aq)
at 25°C for the0Mreaction:
InitialKe q 0.0124M
0M
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ChangeNH3(aq) + H2O(l) ↔ NH4+(aq) + OH-(aq)
Equilibrium
4.64x10- 4 M
Practice
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Sulfur trioxide decomposes at high
temperatures in a sealed container: 2SO3(g) ↔
2SO2(g) + O2(g). Initially the vessel is charged
at 1000K with SO3(g) at a partial pressure of
0.500 atm. At equilibrium the SO3 partial
pressure is 0.200 atm. Calculate the value of Ke
q at 1000K.
Answer: 0.338
Predicting the Direction of Reaction
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If you are given a set
of concentrations or
pressures and the
equilibrium constant
you can determine in
which direction a
reaction will proceed.
Substitute the given
[]'s or pressures in to
the equilibrium
expression.
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The result is called
Practice
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At 448°C the equilibrium constant, Keq, for the
reaction H2(g) + I2(g) ↔ 2HI(g) is 51. Predict
how the reaction will proceed to reach
equilibrium at this temperature if we start with
2.0x10- 2 mol of HI, 1.0x10- 2 of H2, and 3.0x10- 2
mol of I2 in a 2.00L container.
Practice
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At 1000K the value of Keq for the reaction
2SO3(g) ↔ 2SO2(g) + O2(g) is 0.338. Calculate
the value of Q, and predict the direction in
which the reaction will proceed toward to attain
equilibrium if the initial partial pressures of
reactants are PSO3 = 0.16 atm; PSO2 = 0.41atm;
PO2 = 2.5atm.
Calculating Equilibrium
Concentrations
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In a system where we know a few of the
equilibrium pressures or concentrations and the
equilibrium constant, we can solve for the ones
we don't
In the Haber Process, N2(g) + 3H2(g) ↔
2NH3(g), Ke q = 1.45x10- 5 at 500°C. In an
equilibrium mixture of the three gases, the
partial pressure of H2 is 0.928atm and that of N2
is 0.432atm. What is the partial pressure of NH3
in this equilibrium mixture?
Practice
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At 500K the reaction PCl5(g) ↔ PCl3(g) + Cl2(g)
has a Ke q = 0.497. In an equilibrium mixture at
500K, the partial pressure of PCl5 is 0.860atm
and that of PCl3 is 0.350atm. What is the partial
pressure of chlorine in the equilibrium mixture?
Answer: 1.22atm
Calculating Equilibrium
Concentrations
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In systems where all we know is the equilibrium
constant and the initial concentrations, finding
the equilibrium concentration will require a bit
more algebra.
A 1.000L
flask+ isI (g)
filled
with
1.000mol of H2 and
H2(g)
↔
2HI(g)
2
2.000mol
Initial
59.19 of
atm I2 at
118.4448°C.
atm
0The
atm value of the
equilibrium constant, Ke q, for the reaction at
Change
Equilibrium
448°C is 50.5. What are the partial pressures of
H2, I2 and HI in the flask at equilibrium?
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Practice
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For the equilibrium, PCl5(g) ↔ PCl3(g) + Cl2(g),
the equilibrium constant, Ke q, has the value
0.497 at 500K. A gas cylinder at 500K is
charged with PCl5(g) at an initial pressure of
1.66 atm. What are the equilibrium pressures of
PCl5, PCl3, and Cl2 at this temperature?
Le Châtelier's Principle
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If a system at equilibrium is disturbed by an
change in temperature, pressure, or the
concentration of one of the components, the
system will shift its equilibrium position so as to
counteract the effect of the disturbance.
There are 3 main ways to change equilibrium
1) Adding or removing a reactant or product
2) Changing the pressure
3) Changing the temperature
Changes in Reactant or Product [ ]'s
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If a chemical system
is at equilibrium and
we add a substance
(reactant or product)
the reaction will shift
so as to reestablish
equilibrium by
consuming part of the
added substance.
Removing a
substance will cause
the reaction to move
Changes in Pressure and Volume
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At a constant temperature, reducing the volume
of a gaseous equilibrium mixture causes the
system to shift in the direction that reduces the
number of moles of gas.
Increasing the volume causes a shift in the
direction that produces more gas molecules.
Video
What would happen to equilibrium if a gas that
was not involved in the reaction such as Argon
was added to the following system?
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N2(g) + 3H2(g) ↔ 2NH3(g)
Changes in Temperature
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Changes in [ ] and volume shift equilibrium
without changing the equilibrium constant.
Changes in temperature do cause the
equilibrium constant to change.
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Endothermic:
reactants + heat ↔ products
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Exothermic:
reactants ↔ products + heat
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When the temperature is increased, it is as if we
have added a reactant, or a product, to the
system at equilibrium. The equilibrium shifts in
the direction that consumes the excess reactant
or product, namely heat.
Changes in Temperature
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Endothermic
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Exothermic
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↑ T causes ↑ in Ke q
↓ T causes ↓ in Ke q
Co(H2O)62 +(aq) + 4Cl(aq) ↔ CoCl42 -(aq) +
6H2O(l) ∆H > 0
Practice
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Consider the following equilibrium:
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N2O4(g) ↔ 2NO2(g) ∆H° = 58.0 kJ
In what direction will the equilibrium shift when
each of the following changes is made to a
system at equilibrium:
a) Add N2O4
b) Remove NO2
c) Increase the total pressure by adding N2(g)
d) Increase the volume
e) Decrease the temperature
Effect of Catalysts
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A catalyst increases
the rate at which an
equilibrium is
achieved, but does
not change the
composition of the
equilibrium mixture.
Integrative Practice
At temperatures near 800°C, steam passed over
hot coke (a form of carbon obtained from coal)
reacts to form CO and H2:
C(s) + H2O(g) ↔ CO(g) + H2(g)
The mixture of gases that results is an important
industrial fuel called water gas.
a) At 800°C the equilibrium constant for this
reaction is 14.1. What are the equilibrium partial
pressures of H2O, CO, and H2 in the equilibrium
mixture at this temperature if we start with solid
carbon and 0.100 mol of H2O in a 1.00L vessel?