Download Chapter 15 Chemical Equilibrium

Chemistry, The Central Science, 11th edition
Theodore L. Brown, H. Eugene LeMay, Jr., and
Bruce E. Bursten
Chapter 15
Chemical Equilibrium
Equilibrium
© 2009, Prentice-Hall, Inc.
Overview of Equilibrium – Chapter 15
• write Keq expression, in units of pressure or
concentration
• assess the Keq with regard to relative concentrations
of reactants and products
• manipulate chemical equations and Keq – reciprocal,
multiplication, application of Hess’s Law
• heterogeneous equilibria – what is included in Keq?
• calculate Keq, including use of RICE tables
Equilibrium
© 2009, Prentice-Hall, Inc.
Overview of Equilibrium – Chapter 15
• calculate Q and predict the direction of a reaction by
relating Q to Keq
• calculate equilibrium concentrations of reactants
and/or products when given Keq
• apply Le Châtelier’s Principle to predict responses of
the equilibrium system to changes in reactant or
product concentrations, changes in pressure or
volume, changes in temperature
• describe and explain the effects of catalysts on
equilibrium
Equilibrium
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The Concept of Equilibrium
Chemical equilibrium occurs when a
reaction and its reverse reaction proceed at
the same rate.
Equilibrium
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The Concept of Equilibrium
• As a system
approaches equilibrium,
both the forward and
reverse reactions are
occurring.
• At equilibrium, the
forward and reverse
reactions are
proceeding at the same
rate.
Equilibrium
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A System at Equilibrium
Once equilibrium is
achieved, the
amount of each
reactant and product
remains constant.
Equilibrium
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Depicting Equilibrium
Since, in a system at equilibrium, both
the forward and reverse reactions are
being carried out, we write its equation
with a double arrow.
N2O4 (g)
2 NO2 (g)
Equilibrium
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Dynamic Equilibrium
• A dynamic equilibrium exists when the rates
of the forward and reverse reactions are
equal.
• There is no further change in [reactant] or
[product].
• How fast you get to equilibrium depends on
kinetics.
Equilibrium
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The
Equilibrium
Constant
Equilibrium
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The Equilibrium Constant
• Forward reaction:
N2O4 (g)  2 NO2 (g)
• Rate Law:
Rate = kf [N2O4]
Equilibrium
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The Equilibrium Constant
• Reverse reaction:
2 NO2 (g)  N2O4 (g)
• Rate Law:
Rate = kr [NO2]2
Equilibrium
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The Equilibrium Constant
• Therefore, at equilibrium
Ratef = Rater
kf [N2O4] = kr [NO2]2
• Rewriting this, it becomes
kf
[NO2]2
=
kr
[N2O4]
Equilibrium
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The Equilibrium Constant
The ratio of the rate constants is a
constant at that temperature, and the
expression becomes
kf
Keq =
kr
[NO2]2
=
[N2O4]
Equilibrium
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Summary
1. At equilibrium, the concentrations of
reactants and products no longer change
with time.
2. For equilibrium to occur, neither reactants
nor products can escape from the system.
3. At equilibrium a particular ratio of
concentration terms equals a constant.
Equilibrium
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The Equilibrium Constant
• Consider the generalized reaction
aA + bB
cC + dD
• The equilibrium expression for this
reaction would be
[C]c[D]d
Kc =
[A]a[B]b
Equilibrium
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The Equilibrium Constant
Since pressure is proportional to
concentration for gases in a closed
system, the equilibrium expression can
also be written
(PCc) (PDd)
Kp =
(PAa) (PBb)
Equilibrium
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Keq vs Kc vs Kp
• Keq = the general expression for equilibrium constant
expressions.
• Kc = Keq for which molar concentrations were used to evaluate
the constant (i.e. subscript “c” = concentration).
• Kc includes
– Ka (weak acids) and Kb (weak bases) (Chapter 16)
– Ksp (solubility product) (Chapter 17)
– Kw (water)
• Kp = Keq where “p” stands for pressure
Equilibrium
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Relationship Between Kc and Kp
• From the Ideal Gas Law we know that
PV = nRT
• Rearranging it, we get
n
P=
RT
V
If we express volume in liters the quantity (n/V) is equivalent to molarity.
Equilibrium
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Relationship Between Kc and Kp
Plugging this into the expression for Kp
for each substance, the relationship
between Kc and Kp becomes
Kp = Kc (RT)n
where
n = (moles of gaseous product) - (moles of gaseous reactant)
Equilibrium
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Sample Exercise 15.1 (p. 632)
Write the equilibrium expression for Keq
for these three reactions:
a) 2 O3(g) D 3 O2(g)
b) 2 NO(g) + Cl2(g) D 2 NOCl(g)
c) Ag+(aq) + 2 NH3(g) D Ag(NH3)2+(aq)
Equilibrium
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Practice Exercise 15.1
Write the equilibrium expression for Keq
for these three reactions:
a) H2(g) + I2(g) D 2 HI(g)
b) Cd2+(aq) + 4 Br-(aq) D CdBr42-(aq)
Equilibrium
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Equilibrium Can Be Reached from
Either Direction
As you can see, the ratio of [NO2]2 to [N2O4] remains
constant at this temperature no matter what the initial
concentrations of NO2 and N2O4 are.
Equilibrium
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Equilibrium Can Be Reached from
Either Direction
This is the data from
the last two trials from
the table on the
previous slide.
Equilibrium
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Equilibrium Can Be Reached from
Either Direction
It doesn’t matter whether we start with N2 and
H2 or whether we start with NH3: we will have
the same proportions of all three substances
at equilibrium.
Equilibrium
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Sample Exercise 15.2 (p. 634)
In the synthesis of ammonia from nitrogen
and hydrogen,
N2(g) + 3 H2(g) D 2 NH3(g)
Kc = 9.60 at 300oC,
R = 0.0821 Latm/molK
Calculate Kp for this reaction at this
temperature.
(4.34 x 10-3)
Equilibrium
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Practice Exercise 15.2
For the equilibrium
2 SO3(g) D 2 SO2(g) + O2(g),
Kc is 4.08 x 10-3 at 1000 K.
Calculate the value for Kp.
(0.335)
Equilibrium
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Equilibrium Constants and Units
• Keq – described using activities rather
than [ ] or P  no units
• Activity in an ideal mixture
= [ ] relative to 1 M ([reference])
or
P relative to 1 atm (reference P)
Equilibrium
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Equilibrium Constants and Units
e.g. 0.010 M  activity = 0.010 M = 0.010
1M
same number, no units
e.g. Keq = (PNO2/Pref)2
(PN2O4/Pref)
• Results = units cancel
Molarity and pressure can be used in the same Keq expression
• For pure substances (ie. solids or liquids),
the [reference] = itself,  1
Equilibrium
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What Does the Value of K Mean?
• If K>>1, the reaction is
product-favored;
product predominates
at equilibrium.
Equilibrium
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What Does the Value of K Mean?
• If K>>1, the reaction is
product-favored;
product predominates
at equilibrium.
• If K<<1, the reaction is
reactant-favored;
reactant predominates
at equilibrium.
Equilibrium
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Sample Exercise 15.3
The following diagrams represent three different systems at
equilibrium, all in the same size containers.
a)
Without doing any calculations, rank the three
systems in order of increasing equilibrium
constant, Kc.
b)
If the volume of the containers is 1.0 L and each sphere
represents 0.10 mol, calculate Kc for each system.
Equilibrium
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Practice Exercise 15.3
The equilibrium constant for the reaction
H2(g) + I2(g) D 2 HI(g) varies with
temperature as follows:
Kp = 792 at 298 K; Kp = 54 at 700 K.
Is the formation of HI favored more at
the higher or lower temperature?
Equilibrium
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Equilibrium
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Sample Exercise 15.4 (p. 637)
The equilibrium constant for the reaction of N2
with O2 to form NO equals Kc = 1 x 10-30 at
25oC.
N2(g) + O2(g) D 2 NO(g)
Using this information, write the equilibrium
constant expression and calculate the
equilibrium constant for the following
reaction:
2 NO(g) D N2(g) + O2(g)
Equilibrium
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Practice Exercise 15.4
For the formation of NH3 from N2 and
H2, N2(g) + H2(g) D 2 NH3(g),
Kp = 4.34 x 10-3 at 300oC.
What is the value of Kp for the reverse
reaction?
(2.30 x 102)
Equilibrium
© 2009, Prentice-Hall, Inc.
Overview of Equilibrium – Chapter 15
Where we have been:
• write Keq expression, in units of
pressure or concentration
• assess the Keq with regard to relative
concentrations of reactants and
products
Equilibrium
© 2009, Prentice-Hall, Inc.
Overview of Equilibrium – Chapter 15
Where we are going:
• manipulate chemical equations and Keq – reciprocal,
multiplication, application of Hess’s Law
• heterogeneous equilibria – what is included in Keq?
• calculate Keq, including use of RICE tables
• calculate Q and predict the direction of a reaction by
relating Q to Keq
• calculate equilibrium concentrations of reactants
and/or products when given Keq
Equilibrium
© 2009, Prentice-Hall, Inc.
Overview of Equilibrium – Chapter 15
• apply Le Châtelier’s Principle to predict
responses of the equilibrium system to
changes in reactant or product
concentrations, changes in pressure or
volume, changes in temperature
• describe and explain the effects of
catalysts on equilibrium
Equilibrium
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Manipulating Equilibrium Constants
1.
The equilibrium constant of a reaction in the
reverse reaction is the reciprocal of the equilibrium
constant of the forward reaction.
2.
The equilibrium constant of a reaction that has
been multiplied by a number is the equilibrium
constant raised to a power that is equal to that
number.
3.
The equilibrium constant for a net reaction made up
of two or more steps is the product of the
equilibrium constants for the individual steps.
Equilibrium
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Manipulating Equilibrium Constants
1. The equilibrium constant of a
reaction in the reverse reaction is the
reciprocal of the equilibrium constant of
the forward reaction.
In the opposite direction:
N2O4 (g) D 2 NO2 (g)
2 NO2 (g) D N2O4 (g)
Keq = [NO2]2 = 0.212
[N2O4]
Keq = [N2O4] = 4.72
[NO2]2
Keq for these reactions are reciprocals of each other.
Equilibrium
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Manipulating Equilibrium Constants
2. The equilibrium constant of a reaction that has
been multiplied by a number is the equilibrium
constant raised to a power that is equal to that
number.
N2O4(g) D 2 NO2(g)
Kc = [NO2]2 = 0.212
[N2O4]
2 N2O4(g) D 4 NO2(g)
Kc = [NO2]4 = (0.212)2
[N2O4]2
Equilibrium
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Manipulating Equilibrium Constants
3. The equilibrium constant for a net
reaction made up of two or more steps
is the product of the equilibrium
constants for the individual steps.
Equilibrium
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Sample Exercise 15.5 (p. 638)
Given the following information,
HF(aq) D H+(aq) + F-(aq)
H2C2O4(aq) D 2 H+(aq) + C2O42-(aq)
Kc = 6.8 x 10-4
Kc = 3.8 x 10-6
determine the value of Kc for the following reaction:
2 HF(aq) + C2O42-(aq) D 2 F-(aq) + H2C2O4(aq)
(0.12)
Equilibrium
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Practice Exercise 15.5
Given the following information at 700 K,
H2(g) + I2(g) D 2HI(g)
Kp = 54.0
N2(g) + 3 H2(g) D 2 NH3(g) Kp = 1.04 x 10-4
determine the value of Kp (at 700 K)
2 NH3(g) + 3 I2(g) D 6 HI(g) + N2(g)
(1.51 x 109)
Equilibrium
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Heterogeneous
Equilibrium
Equilibrium
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Heterogeneous Equilibria
• Equilibria in which all reactants and
products are present in the same phase
are called homogeneous equilibria.
• Equilibria in which one or more
reactants or products are present in a
different phase are called
heterogeneous equilibria.
Equilibrium
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The Concentrations of Solids and
Liquids Are Essentially Constant
Both can be obtained by multiplying the
density of the substance by its molar
mass — and both of these are constants
at constant temperature.
Equilibrium
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The Concentrations of Solids and
Liquids Are Essentially Constant
Therefore, the concentrations of solids
and liquids do not appear in the
equilibrium expression.
PbCl2(s) D Pb2+(aq) + 2 Cl-(aq)
Kc = [Pb2+] [Cl-]2
Equilibrium
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CaCO3(s) D CaO(s) + CO2(g)
As long as some CaCO3 or CaO remain in
the system, the amount of CO2 above the
solid will remain the same.
Equilibrium
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Heterogeneous Equilibria
Systems where the solvent is involved
as a reactant or product and the solutes
are present at low concentrations.
e.g. dissociation of a weak acid
H2O(l) + CO32–(aq) D OH–(aq) + HCO3–(aq)
Kc = [OH–][HCO3–] / [CO32–]
Equilibrium
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Sample Exercise 15.6 (p. 640)
Write the equilibrium-constant Kc for
each of the following reactions:
a) CO2(g) + H2(g) D CO(g) + H2O(l)
Kc =
b) SnO2(s) + 2 CO(g) D Sn(s) + 2 CO2(g)
Kc =
Equilibrium
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Practice Exercise 15.6
Write the equilibrium-constant expressions for
each of the following reactions:
a) Cr(s) + 3 Ag+(aq) D Cr3+(aq) + 3 Ag(s)
Kc =
b) 3 Fe(s) + 4 H2O(g) D Fe3O4(s) + 4 H2(g)
Kp =
Equilibrium
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Sample Exercise 15.7 (p. 641)
Each of the following mixtures was placed in a closed
container and allowed to stand. Which of these
mixtures is capable of attaining the equilibrium
CaCO3(s) D CaO(s) + CO2(g)
a) pure CaCO3
b) CaO and a pressure of CO2 > Kp
c) some CaCO3 and a pressure of CO2 > Kp
d) CaCO3 and CaO
Equilibrium
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Practice Exercise 15.7
When added to Fe3O4(s) in a closed
container, which one of the following
substances – H2(g), H2O(g), O2(g) - will
allow equilibrium to be established in
the reaction
3 Fe(s) + 4 H2O(g) D Fe3O4(s) + 4 H2(g)
Equilibrium
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Equilibrium
Calculations
Equilibrium
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Sample Exercise 15.8 (p. 642)
A mixture of hydrogen and nitrogen in a
reaction vessel is allowed to attain equilibrium
at 472oC. The equilibrium mixture of gases
was analyzed and found to contain 7.38 atm
H2, 2.46 atm N2, and 0.166 atm NH3. From
these data calculate the equilibrium constant,
Kp, for
N2(g) + 3 H2(g) D 2 NH3(g)
(2.79 x 10-5)
Equilibrium
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Practice Exercise 15.8
An aqueous solution of acetic acid is found to
have the following equilibrium concentrations at
25oC: [HC2H3O2] = 1.65 x 10-2 M; [H+] = 5.44 x
10-4 M; and [C2H3O2-] = 5.44 x 10-4 M.
Calculate the equilibrium constant, Kc, for the
ionization of acetic acid at 25oC. The reaction is
HC2H3O2(aq) D H+(aq) + C2H3O2-(aq)
(1.79 x 10-5)
Equilibrium
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Equilibrium Calculations
1. Tabulate all the known initial and equilibrium concentrations of the
species that appear in the equilibrium-constant expression.
2. For those 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 coefficients in the
balanced chemical equation) to calculate the changes in concentration
of all the other species in the equilibrium.
4. From the initial concentrations and the changes in concentration,
calculate the equilibrium concentrations. These are then used to
evaluate the equilibrium constant.
Equilibrium
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An Equilibrium Problem (Sample
Exercise 15.9, p. 643)
A closed system initially containing 1.000 x
10-3 M H2 and 2.000 x 10-3 M I2 at 448 C is
allowed to reach equilibrium.
Analysis of the equilibrium mixture shows
that the concentration of HI is 1.87 x 10-3 M.
Calculate Kc at 448 C for the reaction taking
place, which is
H2(g) + I2(g) D 2 HI(g)
Equilibrium
© 2009, Prentice-Hall, Inc.
What Do We Know?
Reaction:
Initially
H2(g)
+
I2(g)
D
2 HI(g)
[H2], M
[I2], M
[HI], M
1.000 x 10-3
2.000 x 10-3
0
Change
At equilibrium
1.87 x 10-3
Equilibrium
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[HI] Increases by 1.87 x 10-3 M
Reaction:
Initially
H2(g)
+
I2(g)
D
2 HI(g)
[H2], M
[I2], M
[HI], M
1.000 x 10-3
2.000 x 10-3
0
Change
+1.87 x 10-3
At equilibrium
1.87 x 10-3
Equilibrium
© 2009, Prentice-Hall, Inc.
Stoichiometry tells us [H2] and [I2] decrease
by half as much.
Reaction:
H2(g)
+
I2(g)
D
2 HI(g)
[H2], M
[I2], M
[HI], M
Initially
1.000 x 10-3
2.000 x 10-3
0
Change
-9.35 x 10-4
-9.35 x 10-4
+1.87 x 10-3
At equilibrium
1.87 x 10-3
Equilibrium
© 2009, Prentice-Hall, Inc.
We can now calculate the equilibrium
concentrations of all three compounds…
Reaction:
H2(g)
+
I2(g)
D
2 HI(g)
[H2], M
[I2], M
[HI], M
Initially
1.000 x 10-3
2.000 x 10-3
0
Change
-9.35 x 10-4
-9.35 x 10-4
+1.87 x 10-3
6.5 x 10-5
1.065 x 10-3
1.87 x 10-3
At equilibrium
Equilibrium
© 2009, Prentice-Hall, Inc.
…and, therefore, the equilibrium constant.
[HI]2
Kc =
[H2] [I2]
=
(1.87 x 10-3)2
(6.5 x 10-5)(1.065 x 10-3)
= 51
Equilibrium
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Practice Exercise 15.9
Sulfur trioxide decomposes at high
temperature in a sealed container:
2 SO3(g) D 2 SO2(g) + O2(g)
Initially the vessel is charged at 1000 K 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 Kp at 1000 K.
(0.338)
Equilibrium
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The Reaction Quotient (Q)
• Q gives the same ratio
the equilibrium
expression gives, but
for a system that is not
at equilibrium.
• To calculate Q, one
substitutes the initial
concentrations of
reactants and products
into the equilibrium
expression.
C  D 

Q
a
b
 A  B
c
d
Equilibrium
© 2009, Prentice-Hall, Inc.
The Reaction Quotient
• Note: Q = Keq only at equilibrium.
• If Q < Keq then the forward reaction must
occur to reach equilibrium.
• If Q > Keq then the reverse reaction must
occur to reach equilibrium.
Products are consumed, reactants are
formed.
Q decreases until it equals Keq.
Equilibrium
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If Q = K,
the system is at equilibrium.
Equilibrium
© 2009, Prentice-Hall, Inc.
If Q > K,
there is too much product, and the
equilibrium shifts to the left.
Equilibrium
© 2009, Prentice-Hall, Inc.
If Q < K,
there is too much reactant, and the
equilibrium shifts to the right.
Equilibrium
© 2009, Prentice-Hall, Inc.
Sample Exercise 15.10 (p. 645)
At 448oC the equilibrium constant, Kc, for the
reaction H2(g) + I2(g)  2 HI(g) is 51.
Predict how the reaction will proceed to reach
equilibrium at 448oC if we start with
2.0 x 10-2 mol of HI, 1.0 x 10-2 mole of H2,
and 3.0 x 10-2 mol of I2 in a 2.00-L container.
(Q = 1.3, so reaction must proceed from left
to right)
Equilibrium
© 2009, Prentice-Hall, Inc.
Practice Exercise 15.10
At 1000 K the value of Kp for the reaction
2 SO3(g) D 2 SO2(g) + O2(g) is 0.338.
Calculate the value for Qp, and predict the
direction in which the reaction will proceed
toward equilibrium if the initial partial
pressures of reactants are PSO3 = 0.16 atm;
PSO2 = 0.41 atm; PO2 = 2.5 atm.
(Qp = 16; Qp > Kp, so reaction will proceed
from right to left)
Equilibrium
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Sample Exercise 15.11 (p. 646)
For the Haber process,
N2(g) + 3 H2(g) D 2 NH3(g), Kp = 1.45 x 10-5 at
500oC.
In an equilibrium mixture of the three gases at
500oC, the partial pressure of H2 is 0.928 atm
and that of N2 is 0.432 atm. What is the
partial pressure of NH3 in this equilibrium
mixture?
(2.24 x 10-3 atm)
Equilibrium
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Practice Exercise 15.11
At 500 K the reaction
PCl5(g) D PCl3(g) + Cl2(g) has Kp = 0.497.
In an equilibrium mixture at 500 K, the
partial pressure of PCl5 is 0.860 atm
and that of PCl3 is 0.350 atm. What is
the partial pressure of Cl2 in the
equilibrium mixture?
(1.22 atm)
Equilibrium
© 2009, Prentice-Hall, Inc.
Warmup – Equilibrium
p.661, 15.31
A mixture of 0.100 mol of NO, 0.050 mol of
H2, and 0.100 mol of H2O is placed in a 1.0
L vessel at 300 K. The following equilibrium
is established:
2 NO(g) + 2 H2(g) D N2(g) + 2 H2O(g)
At equilibrium, there are 0.062 mol of NO.
a) Calculate the equilibrium concentrations of
H2, N2, and H2O.
b) Calculate Keq .
Equilibrium
© 2009, Prentice-Hall, Inc.
Warmup – Equilibrium
p.661, 15.31
A mixture of 0.100 mol of NO, 0.050 mol of
H2, and 0.100 mol of H2O is placed in a 1.0
L vessel at 300 K. The following equilibrium
is established:
2 NO(g) + 2 H2(g) N2(g) + 2 H2O(g)
At equilibrium, there are 0.062 mol of NO.
a) Calculate the equilibrium concentrations of
H2, N2, and H2O.
(0.012 mol, 0.019 mol, 0.138 mol)
b) Calculate Keq . (650)
Equilibrium
© 2009, Prentice-Hall, Inc.
Sample Exercise 15.12 (p. 646)
A 1.000-L flask is filled with 1.000 mol of H2
and 2.000 mol of I2 at 448oC. The value of
the equilibrium constant, Kc, for the reaction
H2(g) + I2(g) D 2 HI(g)
at 448oC is 50.5.
What are the partial pressures of H2, I2, and
HI in the flask at equilibrium?
([H2] = 0.065 M, [I2] = 1.065 M, [HI] = 1.87 M)
Equilibrium
© 2009, Prentice-Hall, Inc.
Practice Exercise 15.12
For the equilibrium, PCl5(g) D PCl3(g) + Cl2(g),
the equilibrium constant, Kp, has the value
0.497 at 500 K.
A gas cylinder at 500 K 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?
(PPCl5 = 0.967 atm, PPCl3 = PCl2 = 0.693 atm)
Equilibrium
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Le Châtelier’s
Principle
Equilibrium
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Le Châtelier’s Principle
“If a system at equilibrium is disturbed
by a 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.”
Equilibrium
© 2009, Prentice-Hall, Inc.
The Haber Process
As P increases, the amount of ammonia present at equilibrium increases.
As T increases, the amount of ammonia at equilibrium decreases.
Equilibrium
Can this be predicted?
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The Haber Process
The transformation of nitrogen and hydrogen into
ammonia (NH3) is of tremendous significance in
agriculture, where ammonia-based fertilizers are of
utmost importance.
Equilibrium
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The Haber Process
If H2 is added to the
system, N2 will be
consumed and the
two reagents will
form more NH3.
Equilibrium
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The Haber Process
This apparatus
helps push the
equilibrium to the
right by removing
the ammonia (NH3)
from the system as
a liquid.
Equilibrium
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Effects of Volume and Pressure
Changes
Le Châtelier’s principle predicts that if
pressure is increased, the system will
shift to counteract the increase.
• That is, the system shifts to remove
gases and decrease pressure.
• An increase in pressure favors the
direction that has fewer moles of gas.
Equilibrium
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Effects of Volume and Pressure
Changes
• In a reaction with the same number of
moles of gas in the products and
reactants, changing the pressure has no
effect on the equilibrium.
• No change will occur if we increase the
total gas pressure by the addition of a
gas that is not involved in the reaction.
Equilibrium
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The Effect of Changes in
Temperature
Co(H2O)62+(aq) + 4Cl–(aq) D CoCl42–(aq) + 6H2O(l) H > 0
Equilibrium
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Sample Exercise 15.13 (p. 653)
Consider the following equilibrium:
N2O4(g) D 2 NO2(g)
Ho = 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?
Equilibrium
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Practice Exercise 15.13
For the reaction
PCl5(g) D PCl3(g) + Cl2(g) Ho = 87.9 kJ
in what direction will the equilibrium shift
when
a) Cl2(g) is removed;
b) the temperature is decreased;
c) the volume of the reaction system is
increased;
d) PCl3(g) is added?
Equilibrium
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Sample Exercise 15.14 (p. 653)
a)Using the standard heat of formation data in
Appendix C, determine the standard enthalpy
change for the reaction
N2(g) + 3 H2(g) D 2 NH3(g)
0
0
2 mol x -46.19 kJ/mol
= -92.38 kJ/mol
b)Determine how the equilibrium constant for
this reaction should change with temperature.
Equilibrium
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Equilibrium
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Practice Exercise 15.14
The enthalpy change for the reaction
2 POCl3(g) D 2 PCl3(g) + O2(g)
-542.2 kJ/mol
-288.07 kJ/mol
= (2 mol x -288.07 kJ/mol) – (2 mol x -542.2 kJ/mol
= + 508.3 kJ
Use this result to determine how the
equilibrium constant for the reaction should
change with temperature.
Equilibrium
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Catalysts
Equilibrium
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Catalysts
Catalysts increase
the rate of both the
forward and reverse
reactions.
Equilibrium
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Catalysts
When one uses a
catalyst, equilibrium
is achieved faster,
but the equilibrium
composition remains
unaltered.
Equilibrium
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Sample Integrative Exercise 15
At temperatures near 800oC, steam passed over hot
coke (a form of carbon obtained from coal) reacts to
form CO and H2:
C(s) + H2O(g) D CO(g) + H2(g)
The mixture of gases that results is an important
industrial fuel called water gas.
a) At 800oC the equilibrium constant for this reaction is
Kp = 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 1.00-L vessel?
b) What is the minimum amount of carbon required to
achieve equilibrium under these conditions?
Equilibrium
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Sample Integrative Exercise 15
c) What is the total pressure in the vessel at
equilibrium?
d) At 25oC the value of Keq for this reaction is
1.7 x 10-21. Is the reaction exothermic or
endothermic?
e) To produce the maximum amount of CO and
H2 at equilibrium, should the pressure of the
system be increased or decreased?
Equilibrium
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