Transcript Introductory Chemistry: Concepts & Connections 4th Edition

Introductory Chemistry:
Concepts & Connections
4th Edition by Charles H. Corwin
Chapter 16
Chemical
Equilibrium
Christopher G. Hamaker, Illinois State University, Normal IL
© 2005, Prentice Hall
Equilibrium Concept
• Most chemical reactions do not continue until all
of the reactants are used up.
• Most reactions are ongoing, reversible processes;
preceding in both the forward direction to give
products and in the reverse direction to give the
original reactants
• We indicate an equilibrium reaction with a double
arrow:
reactants ⇆ products
Chapter 16
2
Equilibrium Concept Continued
• In an equilibrium reaction, initially the rate of the
forward reaction is very fast.
• As more products are formed, the rate of the
reverse reaction speed up.
• When the rates of the forward and reverse
reactions are the same, the system is at
equilibrium.
forward reaction
reactants ⇆ products
reverse reaction
Chapter 16
3
Collision Theory
• Molecules must collide in order to react.
• In a successful collision, existing bonds are
broken as new bonds are formed and the reactants
are transformed into products.
• This is the collision theory of reactions.
Chapter 16
4
Factors in Successful Collisions
• There are three factors that affect the rate of a
chemical reaction.
1. Collision Frequency:
• When we increase the frequency at which molecules
collide, we increase the rate of reaction. The more
collisions you have, the greater the odds that a
collision will be successful.
2. Collision Energy:
• For a reaction to occur, the molecules must collide
with enough energy to form the new bonds.
Chapter 16
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Factors in Successful Collisions
3. Collision Geometry:
• For a reaction to occur, the molecules must be
oriented in the proper geometry for the reaction to
occur.
• In (a), the reactants have the correct geometry and
products are formed after the collision. In (b), the
reactants do not have the correct geometry and they
do not react.
Chapter 16
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Factors that Effect Reaction Rates
There are three factors that effect the reaction rate:
1. Reactant Concentration:
– As we increase the concentration of the reactant(s),
the molecules are closer together and collide more
frequently. The more collisions, the faster the
reaction.
2. Reaction Temperature:
– As we increase the temperature, we increase the
energy of the reactants. As we increase the energy of
the reactants, the rate of the reaction increases
because of the increased collision frequency and the
collision energy.
Chapter 16
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Factors that Effect Reaction Rates
3. Addition of a Catalyst:
– A catalyst increases the rate of a reaction. A catalyst
increases the number of effective collisions by
creating a more favorable collision geometry.
– A catalyst is not consumed in a reaction.
Chapter 16
8
Energy Barriers in Reactions
• For a chemical reaction to occur, the reactants
must collide with sufficient energy to react.
• This energy is required to achieve the transition
state required to form the products (a).
• Without sufficient energy, the reaction does not
occur (b).
Chapter 16
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Endothermic Reactions
• An endothermic reaction absorbs heat as the
reaction proceeds.
N2(g) + O2(g) + heat ⇆ 2 NO(g)
• A reaction profile shows the energy of reactants
and products during a reaction.
• The highest point on a
reaction profile is the
transition state.
Chapter 16
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Reaction Profiles
• The energy required for reactants to achieve the
transition state is the activation energy, Eact.
• The energy difference between reactants and
products is the heat of reaction, DH.
• The DH for an
endothermic
reaction is
positive.
Chapter 16
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Exothermic Reactions
• An exothermic reaction releases heat as the
reaction proceeds.
NO(g) + O3(g) ⇆ 2 NO2(g) + O2(g) + heat
• The DH for an exothermic reaction is negative.
Chapter 16
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Effect of a Catalyst
• A catalyst is a substance that allows a reaction to
proceed faster by lowering the activation energy.
• The reaction profile shows the effect of a catalyst
on the reaction 2 H2(g) + O2(g) ⇆ 2 H2O + heat
• A catalyst does not
change DH for a
reaction.
• A catalyst speeds up
both the forward and
reverse reactions.
Chapter 16
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Chemical Equilibrium Concept
• A chemical change is a reversible process that can
proceed simultaneously in both the forward and
reverse directions.
• When the rate of the forward and reverse reactions
are proceeding at the same rate, the reaction is in a
state of chemical equilibrium.
3 O2(g) ⇆ 2 O3(g)
• At equilibrium,
ratef (O2 reaction) = rater (O3 reaction)
Chapter 16
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Rates and Equilibrium
• The rate of reaction is the rate at which the
concentrations of reactants decrease per unit time.
3 O2(g) ⇆ 2 O3(g)
• Starting with only O2, as O2 is consumed, the rate
of the forward reaction decreases.
• As O3 is produced, the rate of the reverse reaction
increases. When the rates are equal, equilibrium
is achieved.
Chapter 16
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Law of Chemical Equilibrium
• Consider the following general reaction:
aA+bB ⇆ c C+dD
• The law of chemical equilibrium states that the
molar concentrations of the products (raised to the
powers c and d), divided by the molar
concentrations of the reactants (raised to the
powers a and b), equals a constant.
Chapter 16
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Equilibrium Constant, Keq
• Mathematically, we express the law of chemical
equilibrium as follows:
[C]c[D]d
Keq =
[A]a[B]b
• The constant, Keq, is the general equilibrium
constant.
• The value of Keq varies with temperature. So a
given value of Keq is valid only for a specific
temperature.
Chapter 16
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Writing Equilibrium Constants
• Lets write the equilibrium constant expression for
the reaction 2A ⇆ B.
• Recall, Keq is product(s) over reactant(s), each
raised to its coefficient in the balanced reaction.
Recall, that square brackets represent the molar
concentration of a species.
• The equilibrium constant expression is:
[B]
Keq =
[A]2
Chapter 16
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Homogeneous Equilibria
• A homogeneous equilibrium is a reaction where
all of the products and reactants are in the
physical same state.
• What is the equilibrium constant expression for
the homogeneous equilibrium:
2 SO2(g) + O2(g) ⇆ 2 SO3(g)
[SO3]2
Keq =
[SO2]2[O2]
Chapter 16
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Heterogeneous Equilibria
• A heterogeneous equilibrium is a reaction where
one of the substances is in a different physical
state.
C(s) + H2O(g) ⇆ CO(g) + H2(g)
• The concentrations of liquids and solids do not
change, and they are therefore omitted from
equilibrium constant expressions:
[CO][H2]
Keq =
[H2O]
Chapter 16
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Equilibrium Constant Expressions
• So, when we write equilibrium constant
expressions, we only include the concentrations of
substances in the gas or aqueous state.
• What is the equilibrium constant expression for
the following reaction?
NH4NO3(s) ⇆ N2O(g) + 2 H2O(g)
Keq = [N2O][H2O]2
Chapter 16
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Experimental Determination of Keq
• We can calculate the numerical value of Keq if we
know the concentrations of all of the species in
the reaction.
H2(g) + I2(g) ⇆ 2 HI(g)
• If the concentrations at equilibrium are [H2] =
0.212 M, [I2] = 0.212 M, and [HI] = 1.576 M,
what is Keq?
[HI]2
(1.576)2
Keq =
= 55.3
=
[H2][I2]
(0.212)(0.212)
Chapter 16
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Values of Keq
• It doesn’t matter how much of each substance we
start with, the value of Keq is always 55.3.
Chapter 16
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Shifts in Gaseous Equilibria
• Le Chatleir’s principle states that when a
reversible reaction at equilibrium is stressed by a
change in concentration, temperature, or pressure,
the equilibrium shifts to relieve the stress.
• Lets look at the equilibrium between colorless
N2O4 and brown NO2:
N2O4(g) ⇆ 2 NO2(g)
• If we increase the amount of N2O4, the reaction
shifts to the right to produce more NO2.
• If we increase the amount of NO2, the reaction
shifts to the left to produce
more
N
O
.
2
4
Chapter 16
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Effect of Temperature
• The reaction is endothermic,
N2O4(g) + heat ⇆ 2 NO2(g)
• If we lower the temperature, the reaction shifts to
produce more N2O4.
• If we raise the temperature, the reaction shifts to
produce more NO2.
Chapter 16
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Effect of Pressure
• In a gaseous equilibrium, increasing the pressure
will shift the reaction to the side with fewer gas
molecules.
• In the reaction N2O4(g) ⇆ 2 NO2(g), increasing
the pressure will shift the reaction to the left
producing more N2O4.
Chapter 16
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Effect of an Inert Gas
• If we add an inert gas to a gaseous reaction at
equilibrium, what will happen?
• The volume of the container does not change,
therefore the concentration (and the partial
pressure) of the substances do not change.
• Adding an inert gas has no effect on a system at
equilibrium.
Chapter 16
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Ionization Equilibrium Constant
• The equilibrium constant for the ionization of a
weak acid or base is the ionization equilibrium
constant, Ki.
• What is Ki constant for the ionization of
hydrofluoric acid?
HF(aq) ⇆ H+(aq) + F– (aq)
[H+][F-]
Ki =
[HF]
Chapter 16
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Ionization of a Weak Base
• We can also write a Ki expression for the
ionization of ammonium hydroxide:
NH4OH(aq) ⇆ NH4+(aq) + OH– (aq)
[NH4+][OH-]
Ki =
[NH4OH]
Chapter 16
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Experimental Determination of Ki
• We can calculate the numerical value of Ki if we
know the concentrations of all of the species in
the reaction.
HC2H3O2(aq) ⇆ H+(aq) + C2H3O2– (aq)
• If the concentrations at equilibrium are
[HC2H3O2] = 0.100 M, [H+] = 0.00134 M, and
[C2H3O2–] = 0.00134 M, what is Ki?
[H+][C2H3O2–]
(0.00134)(0.00134)
Ki =
= 1.80 × 10-5
=
[HC2H3O2]
(0.100)
Chapter 16
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Weak Acid-Base Equilibria Shifts
• Lets look at the following weak acid equilibrium:
HF(aq) ⇆ H+(aq) + F– (aq)
• If we increase the amount of HF, the reaction
shifts to the right to produce more H+ and F-.
• If we raise the pH (by adding base, for example),
we decrease [H+], and the reaction shifts to the
right.
• If we add some soluble NaF, we increase the [F-],
and the reaction shifts to the left.
• If we add some soluble NaCl, nothing happens.
Chapter 16
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Solubility Product Equilibria
• Insoluble salts are really very slightly soluble.
• If we add Ag2SO4 to water, some of the slightly
soluble Ag2SO4 dissolves:
Ag2SO4(s) ⇆ 2 Ag+(aq) + SO42-(aq)
• We can write the solubility product equilibrium
constant, Ksp, for the reaction:
Ksp = [Ag+]2[SO42-]
• Recall, we don’t include pure solids or liquids in
equilibrium constant expressions.
Chapter 16
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Experimental Determination of Ksp
• We can calculate the numerical value of Ksp if we
know the concentrations of all of the species in the
reaction.
Mg(OH)2(s) ⇆ Mg2+(aq) + 2 OH-(aq)
• If the concentrations at equilibrium are [Mg2+] =
0.00016 M, and [OH-] = 0.00033 M, what is Ksp?
Ksp = [Mg2+][OH-]2 = (0.00016)(0.00032)2
Ksp = 1.6 × 10-11
Chapter 16
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Solubility Equilibria Shifts
• Lets look at the following solubility equilibrium:
AgCl(s) ⇆ Ag+(aq) + Cl-(aq)
Ksp = [Ag+][Cl-]
• What happens if we add more AgCl?
– Nothing since AgCl does not appear in Ksp
• What happens if we add some soluble NaCl?
– We increase [Cl-], and the equilibrium shifts to the left
producing more solid AgCl.
Chapter 16
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Conclusions
• According to collision theory we can speed up a
reaction in three ways:
– Increasing the concentration of reactants
– Raising the temperature of the reaction
– Adding a catalyst
• An endothermic reaction absorbs heat energy and
an exothermic reaction release heat energy.
Chapter 16
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Conclusions Continued
• The amount of energy necessary to achieve the
transition state is the activation energy, Eact.
• The difference in the energy of the reactants and
products is the heat of reaction, DH.
• A catalyst speeds up a reaction by lowering the
activation energy.
• A catalyst speeds up both the forward and reverse
reactions.
Chapter 16
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Conclusions Continued
• We can write an equilibrium expression for
reactions at equilibrium.
Chapter 16
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Conclusions Continued
• According to Le Chatlier’s principle, a reaction at
equilibrium shifts in order to relieve a stress.
Chapter 16
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