Spencer L. Seager
Michael R. Slabaugh
www.cengage.com/chemistry/seager
Chapter 8
Reaction Rates and Equilibrium
Jennifer P. Harris
LEARNING OBJECTIVES/ASSESSMENT
When you have completed your study of this chapter, you should be able to:
1. Explain ENTROPY and how this works with the classical energy changes of
exothermic/endothermic to allow spontaneous reactions.
2. Explain reaction rates in general (non-quantitative) terms.
3. Use the concept of molecular collisions to explain reaction characteristics. (Section 8.3;
Exercise 8.20)
4. Represent and interpret the energy relationships for reactions by using energy diagrams.
(Section 8.4; Exercise 8.26)
5. Explain how factors such as reactant concentrations, temperature, and catalysts influence
reaction rates. (Section 8.5; Exercise 8.30)
6. Relate experimental observations to the establishment of equilibrium. (Section 8.6; Exercise
8.38)
7. Write equilibrium expressions based on reaction equations, and use equilibrium constants in
a qualitative way to understand equilibria. (Section 8.7)
8. Use Le Châtelier’s principle to predict the influence of changes in concentration and reaction
temperature on the position of equilibrium for a reaction. (Section 8.8; Exercise 8.52)
ENTROPY
• Entropy is a measurement or indication of the disorder (or
randomness) of a system. The more disorderly or mixed up a
system is, the higher its entropy and the more favorable to
develop. You can always replace the word “randomness” for
entropy to make a statement easier to understand.
• By contrast heat energy is referred to as enthalpy.
• If enthalpy increases, heat is released (exothermic) which
also is favorable to a process occurring.
Examples of changes in Entropy
Predict which of the following has an increase in entropy and
where the change in entropy contributes to favorability.
• The simple diffusion of a gas in air or a soluble substance in
solution.
• H2O(s)  H2O(l)
• H2O(l)  H2O(g)
• C6H12O6 (s) + 6 O2(g)  6 CO2(g) + 6 H2O(g)
• NaCl(aq) + AgNO3(aq)  AgCl(s) + NaNO3(aq)
FACTORS THAT INFLUENCE FAVORABILITY
• Overall favorability are dependent on the sum of both entropy
and enthalpy. In practical terms, this means exothermic and
becoming more random.
• If both are favorable, then the overall process is favorable.
• If both are unfavorable, then the overall process is
unfavorable.
• Where entropy and enthalpy are competing in favorability,
enthalpy generally wins out since enthalpy values are usually
larger than enthalpy values.
ENERGY CONSIDERATIONS
• Although Entropy considerations are important, we will for the
most part limit our overall energy considerations to heat
energy (enthalpy).
• Exothermic reactions tend to be favorable and are written as
Reactants = Products + heat
• Endothermic reactions tend to be unfavorable and are written
as Reactants + heat = Products
• Exothermic reactions can be spontaneous (occur without any
obvious input of energy or stimulus) but usually require a kick
start (energy of activation). E.g. H2 and O2 don’t react without
a spark.
• Even though endothermic reactions are unfavorable, they
can still occur but energy must be put into the system in
order for this to happen. E.g., ice can melt.
ACTIVATION ENERGY
• In some reaction mixtures, the average total energy of the
molecules is too low at the prevailing temperature for a reaction
to take place at a detectable rate. The reaction mixture is said
to be stable.
• For many stable mixtures, the addition of a small amount of
energy starts the reaction which then continues without the
addition of any more stimulus or energy from an outside source.
• The small amount of outside
energy needed to start a
spontaneous processes is
called activation energy.
ACTIVATION ENERGY EXAMPLE
• An ordinary kitchen match provides a good example of these
concepts. The reactants in the match head are stable until the
match is rubbed on a rough surface. The energy of rubbing
provides the necessary activation energy to cause the match
head components to react and the match ignites. Once ignited,
the match continues to burn spontaneously until all the fuel of
the match has reacted with oxygen in the air.
EXOTHERMIC & ENDOTHERMIC DIAGRAMS
• The difference between endothermic and exothermic reactions
is clearly indicated by the following energy diagrams. Note that
in exothermic reactions, the energy is lost as the reaction
occurs. Hence, the products have less energy than the
reactants. The reverse is true for endothermic reactions which
gain energy and cause the products to have more energy than
reactants.
DIFFERENCES IN ACTIVATION ENERGY
• Activation energy differences become quite obvious in energy
diagrams as shown by the following illustrations:
REACTION RATES
• A reaction rate is the speed of a reaction.
• Reaction rates can be determined experimentally by measuring
the change in concentration of a reactant or product and
dividing the change by the time required for the change to
occur, using the following equation:
C C t  C0
Rate 

t
Δt
• In this equation, ∆C is the change in concentration of a reactant
or a product that occurs in a measured amount of time, ∆t. The
value of ∆C is calculated by subtracting the initial concentration,
C0, from the final concentration, Ct.
HOW REACTIONS OCCUR
Collision Theory
• An explanation of how reactions occur is
called a reaction mechanism. A reaction
mechanism is often expressed as a number
of processes that must take place for
reactions to occur. The following
assumptions make up the concept of
“Collision Theory”:
• Assumption 1: Reactant particles must
collide with one another in order for a
reaction to occur.
• Assumption 2: Reactant particles must
collide with at least a certain total energy
if the collision is to result in a reaction.
• Assumption 3: In some cases, colliding
reactant particles must be oriented in a
specific way if a reaction is to occur.
HOW REACTIONS OCCUR (continued)
• WHY COLLISIONS ARE NECESSARY
• Reactant particles must collide if they are to react
(assumption 1). With few exceptions, molecules cannot
react with each other if they do not come in contact. During
collisions, some bonds are broken, atoms are exchanged,
and new bonds form.
• WHY MINIMUM COLLISION ENERGY IS NECESSARY
• The requirement that some bonds of reactant molecules
must break if a reaction is to occur makes the requirement of
minimum collision energies valid. This explains why
hydrogen and oxygen don’t automatically explode without a
spark.
MOLECULAR ORIENTATION
• Orientation effects are related to which side or end of a reacting
particle actually contacts another particle during a collision.
• The orientation of reacting particles is not important in some
reactions such as those between reacting ions in a solution. For
example, Ca2+ ions react with CO32- in solution to form solid
CaCO3, which is insoluble and settles out of the solution. Both
ions can be considered to be spherical charged
particles, so their orientation toward each other
when they collide does not influence the
reaction rate.
• Orientation is generally VERY important in organic and
biological reactions where the molecule structures are very
complex.
FACTORS THAT INFLUENCE
REACTION RATES
• Reaction rates are influenced by a number of different factors,
including:
• the nature of the reactants (includes enthalpy, entropy,
orientation, activation energy)
• the concentration of the reactants,
• the temperature of the reactants,
• the presence of catalysts or inhibitors.
THE NATURE OF THE REACTANTS
• Reactions between oppositely-charged ions in solution occur
almost instantaneously. This is because the ions are strongly
attracted to each other because of their opposite electrical
charges.
THE NATURE OF THE REACTANTS
(continued)
• Reactions between covalently-bonded molecules in which
covalent bonds have to be broken often take place slowly. This
is partially because covalent bonds are relatively strong and
must first be broken (activation energy is high) and partially
because they generally require a specific orientation when
colliding.
• Other characteristics of reactants such as their physical state
(gases, liquids or solids), their molecular sizes, and whether or
not they are polar are also important influences of some
reaction rates.
THE CONCENTRATION OF THE REACTANTS
• The requirement for a collision to occur between reactant
molecules before a reaction can take place accounts for the
reactant concentration influence on reaction rates.
• If a reaction occurs between A and B molecules, and a reaction
mixture contains mostly A molecules, most collisions
participated in by A molecules will be with other A molecules
and the reaction rate will be low.
THE CONCENTRATION OF THE REACTANTS
(continued)
• The reaction between a solid piece of iron and oxygen gas
takes place slowly in part because only iron atoms on the
surface can collide with oxygen molecules. The effective
concentration of iron is low. However, wire wool and finelydivided iron powder rust much more rapidly because the surface
area and effective concentration is much greater.
THE TEMPERATURE OF THE REACTANTS
• The effect of temperature on reaction rates can also be
explained using the concept of molecular collisions.
• An increase in the temperature of the reactants corresponds to
an increase in the velocity and the kinetic energy of the
molecules.
• An increase in velocity will increase the number of molecular
collisions that take place in a fixed amount of time and will thus
increase the reaction rate.
• An increase in the kinetic energy of the colliding molecules will
increase their velocity. Therefore there will be an increase the
strength of collisions and the number of molecules with the
required minimum activation energy.
THE PRESENCE OF CATALYSTS
• Catalysts are substances that speed up chemical
reactions without being used up in the reaction.
• One explanation for catalytic behavior is that
catalysts provide an alternate reaction pathway
that requires less activation energy than the
normal pathway. We could say that when reactants
are bound to the catalysts, that the catalyst
weakens the bonds in the reactant making it easier
to break the bonds.
• Another explanation proposes that solid catalysts
and enzymes provide a surface on which reactant
molecules adsorb with favorable orientations to
each other. Adsorbed molecules with favorable
orientations are located close enough to each other
to react rapidly.
THE PRESENCE OF CATALYSTS (continued)
THE PRESENCE OF INHIBITORS
• An inhibitor is a substance that decreases reaction rates.
• Inhibitors will be introduced in more detail when studying
factors that affect enzyme activity (Chapter 20).
• The characteristic function of many poisons and some
medicines is to inhibit one or more enzymes and to decrease
the rates of the reactions they catalyze.
• Some substances normally found in cells inhibit specific
enzyme-catalyzed reactions and thereby provide a means for
the internal regulation of cellular metabolism.
• Inhibitors are not necessarily bad things. They can help
regulate reactions so they aren’t too fast.
CHEMICAL EQUILIBRIUM
• All chemical reactions can (in principle) go in both directions and
products, located to the right of the arrow, can react to form
reactants, located to the left of the arrow. This condition is
indicated by the use of a double arrow pointing in both
directions as shown below:
H2(g) + I2(g)
2HI(g)
• When the reaction rate toward the right is equal to the reaction
rate toward the left, the reaction is said to be in a state of
equilibrium.
CHEMICAL EQUILIBRIUM (continued)
• When a reaction is in a state of equilibrium, the concentrations
of reactants and products remain constant as time passes.
• The unchanging concentrations of reactants and products in a
reaction at equilibrium are called equilibrium concentrations.
THE POSITION OF EQUILIBRIUM
• The position of equilibrium is an indication of the relative
amounts of reactants and products present in a reaction mixture
at equilibrium.
• The position is said to be to the right when the amount of
product is significantly more than the amount of reactant.
• The position is to the left when more reactant is present than
product.
reactant
left
⇌
product
right
An equilibrium we know about
Alka seltzer and vinegar forms carbon dioxide and water
NaHCO3 + HC2H3O2  H2O + CO2 + NaC2H3O2
HCO3- (aq) + H+ (aq)  H2O(l) + CO2(g)
Position is to the right.
Or the reverse reaction (as in carbonated beverages)
H2O(l) + CO2(g)  HCO3- (aq) + H+ (aq)
Position is to the left.
MATHEMATICAL REPRESENTATION OF
POSITION OF EQUILIBRIUM
• The position of equilibrium can be represented
mathematically by using the concepts of an equilibrium
expression and an equilibrium constant.
• Both concepts will be initially represented using the following
hypothetical equilibrium:
aA + bB ⇆ wW + xX
• In this expression, the lower case letters represent the
stoichiometric coefficients of the reaction and the upper case
letters represent the formulas of the reacting substances.
EQUILIBRIUM EXPRESSIONS
• The equilibrium expression for the general equation on the
previous slide is written as follows:

W  X
K
a
b
A  B
W
X
• In this equation, the brackets,[ ], stand for molar
concentrations of the reactants, A and B, and the products,
W and X. It is seen that each reactant concentration is
raised to a power equal to the stoichiometric coefficient of
that reactant in the equilibrium equation.
EQUILIBRIUM CONSTANT
• The K in an equilibrium expression is called the equilibrium
constant.
• As long as the temperature does not change, it has a constant
value because none of the concentrations used to express it
change with time once equilibrium is established.
• A relatively large value for K indicates that the equilibrium
position is toward the right or products side of the equilibrium.
• A small K indicates an equilibrium position toward the left or
reactant side of the equilibrium.
• We will deal with the equilibrium constant in a general way as
follows.
products
K
reactants
THE RANGE OF K VALUES
• The values for K that have been found
experimentally range between wide
extremes.
• Some, such as, K = 1.1 x 10-36, are so
small that for all practical purposes an
equilibrium mixture would contain only
reactants and the equilibrium position
is extremely far to the left. Reactants are
favored when K << 1.
• Others, such as K = 1.2 x1040, are so
large that for all practical purposes an
equilibrium mixture would contain only
products and the equilibrium position
is extremely far to the right. Products
are favored when K >> 1.
Equilibrium Constants for Acids
•
•
•
•
Acetic Acid HC2H3O2
Sulfuric Acid H2SO4
Carbonic Acid H2CO3
Phosphoric Acid H3PO4
1.8 x 10-5
1.0 x 103
4.2 x 10-7
7.1 x 10-3
• Based on these numbers, where are products (or reactants)
favored?
• Rank these from strongest to weakest (most to least
dissociated).
• Does your answer sound correct in terms of how we defined
Acetic Acid and Sulfuric Acid earlier in that acetic acid was a
weak acid and sulfuric acid was a strong acid?
FACTORS THAT INFLUENCE THE POSITION
OF EQUILIBRIUM
• According to Le Châtelier's principle,
the position of equilibrium shifts in
response to changes made in the
equilibrium.
• The factors that will be considered are:
• concentrations of reactants and
products
• reaction temperature
• catalysts
• In general, Le Châtelier's principle
predicts a shift away from the side to
which something (including heat) is
added and toward the side from which
something is removed.
An equilibrium we know about
Alka seltzer and vinegar forms carbon dioxide and water
NaHCO3 + HC2H3O2  H2O + CO2 + NaC2H3O2
HCO3- (aq) + H+ (aq)  H2O(l) + CO2(g)
What would happen to the equilibrium if we add more vinegar?
H2O(l) + CO2(g)  HCO3- (aq) + H+ (aq)
What would happen to your cola if we allowed carbon dioxide to
escape?
FACTORS THAT INFLUENCE THE POSITION OF
EQUILIBRIUM (continued)
• When the temperature of a reaction is
decreased, heat is removed.
• If the reaction is endothermic, heat is a
reactant and the position of equilibrium
will shift toward the reactants.
• If the reaction is exothermic, heat is a
product and the position of equilibrium
will shift toward the products.
• Consider H2O(s)  H2O(l)
• Catalysts cannot change the position of
equilibrium because they lower the energy
barrier for both the forward and reverse
reactions; therefore, catalysts speed up both
forward and reverse reactions and cannot
change the position of equilibrium.
Le Châtelier's PRINCIPLE EXAMPLE
• Consider the following endothermic reaction at equilibrium:
heat + 4NO2(g) + 6H2O(g)
7O2 (g) + 4NH3(g)
• How would the equilibrium shift under each of the following
situations?
• If an equilibrium mixture was heated
• If some NO2 was added to an equilibrium mixture
• If some NH3 was removed from an equilibrium mixture
• If a catalyst were added
Le Châtelier's PRINCIPLE EXAMPLE
• Consider the following endothermic reaction at equilibrium:
heat + 4NO2(g) + 6H2O(g)
7O2 (g) + 4NH3(g)
• If an equilibrium mixture was heated, the equilibrium
position would shift toward the right to try to use up the
added heat.
• If some NO2 was added to an equilibrium mixture, the
equilibrium position would again shift toward the right to try
to use up the added NO2.
• If some NH3 was removed from an equilibrium mixture, the
equilibrium position would again shift toward the right in an
attempt to replace the NH3 that was removed.
• If a catalyst were added, the equilibrium position would not
change.