CHEMISTRY
The Central Science
9th Edition
Chapter 14
Chemical Kinetics
David P. White
Prentice Hall © 2003
Chapter 14
Factors that Affect Reaction
Rates
• Kinetics is the study of how fast chemical reactions
occur.
• There are 4 important factors which affect rates of
reactions:
–
–
–
–
reactant concentration,
temperature,
action of catalysts, and
surface area.
• Goal: to understand chemical reactions at the molecular
level.
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Chapter 14
Reaction Rates
• Speed of a reaction is measured by the change in
concentration with time.
• For a reaction A  B
change in number of moles of B
Average rate 
change in time
moles of B

t
• Suppose A reacts to form B. Let us begin with 1.00 mol
A.
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Chapter 14
Reaction Rates
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Chapter 14
Reaction Rates
– At t = 0 (time zero) there is 1.00 mol A (100 red spheres) and no
B present.
– At t = 20 min, there is 0.54 mol A and 0.46 mol B.
– At t = 40 min, there is 0.30 mol A and 0.70 mol B.
– Calculating,
moles of B 
Average rate 
t
moles of B at t  10  moles of B at t  0 

10 min  0 min
0.26 mol  0 mol
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Chapter 14  0.026 mol/min
10 min  0 min
Reaction Rates
• For the reaction A  B there are two ways of measuring
rate:
– the speed at which the products appear (i.e. change in moles of
B per unit time), or
– the speed at which the reactants disappear (i.e. the change in
moles of A per unit time).
moles of A 
Average rate with respect to A  
t
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Chapter 14
Reaction Rates
Change of Rate with Time
• For the reaction A  B there are two ways of
• Most useful units for rates are to look at molarity. Since
volume is constant, molarity and moles are directly
proportional.
• Consider:
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
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Chapter 14
Reaction Rates
Change of Rate with Time
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
– We can calculate the average rate in terms of the disappearance
of C4H9Cl.
– The units for average rate are mol/L·s or M/s.
– The average rate decreases with time.
– We plot [C4H9Cl] versus time.
– The rate at any instant in time (instantaneous rate) is the slope
of the tangent to the curve.
– Instantaneous rate is different from average rate.
– We usually call the instantaneous rate the rate.
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Chapter 14
Reaction Rates
Reaction Rate and Stoichiometry
• For the reaction
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
we know
C4H9Cl C4H9OH
Rate  

t
t
• In general for
aA + bB  cC + dD
1 A
1 B 1 C 1 D
Rate  



a t
b t
c t
d t
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Chapter 14
Concentration and Rate
• In general rates increase as concentrations increase.
NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l)
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Chapter 14
Concentration and Rate
• For the reaction
NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l)
we note
– as [NH4+] doubles with [NO2-] constant the rate doubles,
– as [NO2-] doubles with [NH4+] constant, the rate doubles,
– We conclude rate  [NH4+][NO2-].
• Rate law:
Rate  k[ NH 4 ][ NO2 ]
• The constant k is the rate constant.
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Chapter 14
Concentration and Rate
•
•
•
•
Exponents in the Rate Law
For a general reaction with rate law
Rate  k[reactant 1]m[reactant 2]n
we say the reaction is mth order in reactant 1 and nth
order in reactant 2.
The overall order of reaction is m + n + ….
A reaction can be zeroth order if m, n, … are zero.
Note the values of the exponents (orders) have to be
determined experimentally. They are not simply related
to stoichiometry.
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Chapter 14
Concentration and Rate
•
•
•
•
Using Initial Rates to Determines Rate
Laws
A reaction is zero order in a reactant if the change in
concentration of that reactant produces no effect.
A reaction is first order if doubling the concentration
causes the rate to double.
A reacting is nth order if doubling the concentration
causes an 2n increase in rate.
Note that the rate constant does not depend on
concentration.
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Chapter 14
The Change of
Concentration with Time
First Order Reactions
• Goal: convert rate law into a convenient equation to give
concentrations as a function of time.
• For a first order reaction, the rate doubles as the
concentration of a reactant doubles.
[A]
Rate  
 k[A]
t
ln A t  ln A 0  kt
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 A t 
   kt
ln 
 A 0 Chapter
 14
The Change of
Concentration with Time
First Order Reactions
• A plot of ln[A]t versus t is a straight line with slope -k
and intercept ln[A]0.
• In the above we use the natural logarithm, ln, which is
log to the base e.
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Chapter 14
The Change of
Concentration with Time
First Order Reactions
ln At  kt  ln A0
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Chapter 14
The Change of
Concentration with Time
Second Order Reactions
• For a second order reaction with just one reactant
1
1
 kt 
At
A0
• A plot of 1/[A]t versus t is a straight line with slope k and
intercept 1/[A]0
• For a second order reaction, a plot of ln[A]t vs. t is not
linear.
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Chapter 14
The Change of
Concentration with Time
Second Order Reactions
1
1
 kt 
At
A0
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Chapter 14
The Change of
Concentration with Time
Half-Life
• Half-life is the time taken for the concentration of a
reactant to drop to half its original value.
• For a first order process, half life, t½ is the time taken for
[A]0 to reach ½[A]0.
• Mathematically,
t1  
2
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2   0.693
ln 1
k
Chapter 14
k
The Change of
Concentration with Time
Half-Life
• For a second order reaction, half-life depends in the
initial concentration:
1
t1  
k A0
2
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Chapter 14
Temperature and Rate
The Collision Model
• Most reactions speed up as temperature increases. (E.g.
food spoils when not refrigerated.)
• When two light sticks are placed in water: one at room
temperature and one in ice, the one at room temperature
is brighter than the one in ice.
• The chemical reaction responsible for
chemiluminescence is dependent on temperature: the
higher the temperature, the faster the reaction and the
brighter the light.
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Chapter 14
Temperature and Rate
The Collision
Model
• As temperature
increases, the rate
increases.
Temperature and Rate
The Collision Model
• Since the rate law has no temperature term in it, the rate
constant must depend on temperature.
• Consider the first order reaction CH3NC  CH3CN.
– As temperature increases from 190 C to 250 C the rate
constant increases from 2.52  10-5 s-1 to 3.16  10-3 s-1.
• The temperature effect is quite dramatic. Why?
• Observations: rates of reactions are affected by
concentration and temperature.
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Chapter 14
Temperature and Rate
•
•
•
•
The Collision Model
Goal: develop a model that explains why rates of
reactions increase as concentration and temperature
increases.
The collision model: in order for molecules to react they
must collide.
The greater the number of collisions the faster the rate.
The more molecules present, the greater the probability
of collision and the faster the rate.
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Chapter 14
Temperature and Rate
The Collision Model
• The higher the temperature, the more energy available to
the molecules and the faster the rate.
• Complication: not all collisions lead to products. In fact,
only a small fraction of collisions lead to product.
The Orientation Factor
• In order for reaction to occur the reactant molecules must
collide in the correct orientation and with enough energy
to form products.
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Chapter 14
Temperature and Rate
The Orientation Factor
• Consider:
Cl + NOCl  NO + Cl2
• There are two possible ways that Cl atoms and NOCl
molecules can collide; one is effective and one is not.
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Chapter 14
Temperature and Rate
The Orientation Factor
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Chapter 14
Temperature and Rate
Activation Energy
• Arrhenius: molecules must posses a minimum amount of
energy to react. Why?
– In order to form products, bonds must be broken in the
reactants.
– Bond breakage requires energy.
• Activation energy, Ea, is the minimum energy required to
initiate a chemical reaction.
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Chapter 14
Temperature and Rate
Activation Energy
• Consider the rearrangement of methyl isonitrile:
H3C N C
H3C
N
C
H3C C N
– In H3C-NC, the C-NC bond bends until the C-N bond breaks
and the NC portion is perpendicular to the H3C portion. This
structure is called the activated complex or transition state.
– The energy required for the above twist and break is the
activation energy, Ea.
– Once the C-N bond is broken, the NC portion can continue to
rotate forming a C-CN bond.
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Chapter 14
Temperature and Rate
•
•
•
•
Activation Energy
The change in energy for the reaction is the difference in
energy between CH3NC and CH3CN.
The activation energy is the difference in energy between
reactants, CH3NC and transition state.
The rate depends on Ea.
Notice that if a forward reaction is exothermic (CH3NC
 CH3CN), then the reverse reaction is endothermic
(CH3CN  CH3NC).
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Chapter 14
Temperature and Rate
Activation Energy
• How does a methyl isonitrile molecule gain enough
energy to overcome the activation energy barrier?
• From kinetic molecular theory, we know that as
temperature increases, the total kinetic energy increases.
• We can show the fraction of molecules, f, with energy
equal to or greater than Ea is
f e
E
 a
RT
where
R is the gas constant
(8.314 J/mol·K).
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Chapter 14
Temperature and Rate
Activation Energy
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Chapter 14
Temperature and Rate
The Arrhenius Equation
• Arrhenius discovered most reaction-rate data obeyed the
Arrhenius equation:
k  Ae
 Ea
RT
– k is the rate constant, Ea is the activation energy, R is the gas
constant (8.314 J/K-mol) and T is the temperature in K.
– A is called the frequency factor.
– A is a measure of the probability of a favorable collision.
– Both A and Ea are specific to a given reaction.
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Chapter 14
Temperature and Rate
Determining the Activation Energy
• If we have a lot of data, we can determine Ea and A
graphically by rearranging the Arrhenius equation:
Ea
ln k  
 ln A
RT
• From the above equation, a plot of ln k versus 1/T will
have slope of –Ea/R and intercept of ln A.
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Chapter 14
Temperature and Rate
Temperature and Rate
Determining the Activation Energy
• If we do not have a lot of data, then we recognize
Ea
Ea
ln k1  
 ln A and ln k2  
 ln A
RT1
RT2
 Ea
  Ea

ln k1  ln k2   
 ln A    
 ln A 
 RT1
  RT2

k1 Ea  1 1 
ln 
  
k2 R  T2 T1 
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Chapter 14
Reaction Mechanisms
• The balanced chemical equation provides information
about the beginning and end of reaction.
• The reaction mechanism gives the path of the reaction.
• Mechanisms provide a very detailed picture of which
bonds are broken and formed during the course of a
reaction.
Elementary Steps
• Elementary step: any process that occurs in a single step.
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Chapter 14
Reaction Mechanisms
Elementary Steps
• Molecularity: the number of molecules present in an
elementary step.
– Unimolecular: one molecule in the elementary step,
– Bimolecular: two molecules in the elementary step, and
– Termolecular: three molecules in the elementary step.
• It is not common to see termolecular processes
(statistically improbable).
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Chapter 14
Reaction Mechanisms
Multistep Mechanisms
• Some reaction proceed through more than one step:
NO2(g) + NO2(g)  NO3(g) + NO(g)
NO3(g) + CO(g)  NO2(g) + CO2(g)
• Notice that if we add the above steps, we get the overall
reaction:
NO2(g) + CO(g)  NO(g) + CO2(g)
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Chapter 14
Reaction Mechanisms
Multistep Mechanisms
• If a reaction proceeds via several elementary steps, then
the elementary steps must add to give the balanced
chemical equation.
• Intermediate: a species which appears in an elementary
step which is not a reactant or product.
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Chapter 14
Reaction Mechanisms
Rate Laws for Elementary Steps
• The rate law of an elementary step is determined by its
molecularity:
– Unimolecular processes are first order,
– Bimolecular processes are second order, and
– Termolecular processes are third order.
Rate Laws for Multistep Mechanisms
• Rate-determining step: is the slowest of the elementary
steps.
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Chapter 14
Reaction Mechanisms
Rate Laws for Elementary Steps
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Chapter 14
Reaction Mechanisms
Rate Laws for Multistep Mechanisms
• Therefore, the rate-determining step governs the overall
rate law for the reaction.
Mechanisms with an Initial Fast Step
• It is possible for an intermediate to be a reactant.
• Consider
2NO(g) + Br2(g)  2NOBr(g)
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Chapter 14
Reaction Mechanisms
Mechanisms with an Initial Fast Step
2NO(g) + Br2(g)  2NOBr(g)
• The experimentally determined rate law is
Rate = k[NO]2[Br2]
• Consider the following mechanism
k1
NOBr2(g)
Step 1: NO(g) + Br2(g)
(fast)
k-1
Step 2: NOBr2(g) + NO(g)
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Chapter 14
k2
2NOBr(g) (slow)
Reaction Mechanisms
Mechanisms with an Initial Fast Step
• The rate law is (based on Step 2):
Rate = k2[NOBr2][NO]
• The rate law should not depend on the concentration of
an intermediate (intermediates are usually unstable).
• Assume NOBr2 is unstable, so we express the
concentration of NOBr2 in terms of NOBr and Br2
assuming there is an equilibrium in step 1 we have
k1
[ NOBr2 ] 
[ NO][Br2 ]
k1
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Chapter 14
Reaction Mechanisms
Mechanisms with an Initial Fast Step
• By definition of equilibrium:
k1[ NO][Br2 ]  k1[ NOBr2 ]
• Therefore, the overall rate law becomes
k1
k1
Rate  k2
[ NO][Br2 ][ NO]  k2
[ NO]2[Br2 ]
k1
k1
• Note the final rate law is consistent with the
experimentally observed rate law.
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Chapter 14
Catalysis
• A catalyst changes the rate of a chemical reaction.
• There are two types of catalyst:
– homogeneous, and
– heterogeneous.
• Chlorine atoms are catalysts for the destruction of ozone.
Homogeneous Catalysis
• The catalyst and reaction is in one phase.
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Chapter 14
Catalysis
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Chapter 14
Catalysis
Homogeneous Catalysis
• Hydrogen peroxide decomposes very slowly:
2H2O2(aq)  2H2O(l) + O2(g)
• In the presence of the bromide ion, the decomposition
occurs rapidly:
–
–
–
–
2Br-(aq) + H2O2(aq) + 2H+(aq)  Br2(aq) + 2H2O(l).
Br2(aq) is brown.
Br2(aq) + H2O2(aq)  2Br-(aq) + 2H+(aq) + O2(g).
Br- is a catalyst because it can be recovered at the end of the
reaction.
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Chapter 14
Catalysis
Homogeneous Catalysis
• Generally, catalysts operate by lowering the activation
energy for a reaction.
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Chapter 14
Catalysis
Catalysis
•
•
•
•
Homogeneous Catalysis
Catalysts can operate by increasing the number of
effective collisions.
That is, from the Arrhenius equation: catalysts increase k
be increasing A or decreasing Ea.
A catalyst may add intermediates to the reaction.
Example: In the presence of Br-, Br2(aq) is generated as
an intermediate in the decomposition of H2O2.
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Chapter 14
Catalysis
Homogeneous Catalysis
• When a catalyst adds an intermediate, the activation
energies for both steps must be lower than the activation
energy for the uncatalyzed reaction. The catalyst is in a
different phase than the reactants and products.
Heterogeneous Catalysis
• Typical example: solid catalyst, gaseous reactants and
products (catalytic converters in cars).
• Most industrial catalysts are heterogeneous.
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Chapter 14
Catalysis
Heterogeneous Catalysis
• First step is adsorption (the binding of reactant molecules
to the catalyst surface).
• Adsorbed species (atoms or ions) are very reactive.
• Molecules are adsorbed onto active sites on the catalyst
surface.
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Chapter 14
Catalysis
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Chapter 14
Catalysis
Heterogeneous Catalysis
• Consider the hydrogenation of ethylene:
C2H4(g) + H2(g)  C2H6(g), H = -136 kJ/mol.
– The reaction is slow in the absence of a catalyst.
– In the presence of a metal catalyst (Ni, Pt or Pd) the reaction
occurs quickly at room temperature.
– First the ethylene and hydrogen molecules are adsorbed onto
active sites on the metal surface.
– The H-H bond breaks and the H atoms migrate about the metal
surface.
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Chapter 14
Catalysis
Heterogeneous Catalysis
– When an H atom collides with an ethylene molecule on the
surface, the C-C  bond breaks and a C-H  bond forms.
– When C2H6 forms it desorbs from the surface.
– When ethylene and hydrogen are adsorbed onto a surface, less
energy is required to break the bonds and the activation energy
for the reaction is lowered.
Enzymes
• Enzymes are biological catalysts.
• Most enzymes are protein molecules with large molecular
masses (10,000 to 106 amu).
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Chapter 14
Catalysis
Enzymes
• Enzymes have very specific shapes.
• Most enzymes catalyze very specific reactions.
• Substrates undergo reaction at the active site of an
enzyme.
• A substrate locks into an enzyme and a fast reaction
occurs.
• The products then move away from the enzyme.
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Chapter 14
Catalysis
Enzymes
• Only substrates that fit into the enzyme lock can be
involved in the reaction.
• If a molecule binds tightly to an enzyme so that another
substrate cannot displace it, then the active site is blocked
and the catalyst is inhibited (enzyme inhibitors).
• The number of events (turnover number) catalyzed is
large for enzymes (103 - 107 per second).
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Chapter 14
Catalysis
Enzymes
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Chapter 14
End of Chapter 14
Chemical Kinetics
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Chapter 14