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.
Prentice Hall © 2003
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.
Prentice Hall © 2003
Chapter 14
Reaction Rates
Prentice Hall © 2003
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
Prentice Hall © 2003
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
Prentice Hall © 2003
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)
Prentice Hall © 2003
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.
Prentice Hall © 2003
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
Prentice Hall © 2003
Chapter 14
Concentration and Rate
• In general rates increase as concentrations increase.
NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l)
Prentice Hall © 2003
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.
Prentice Hall © 2003
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.
Prentice Hall © 2003
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.
Prentice Hall © 2003
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
Prentice Hall © 2003
 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.
Prentice Hall © 2003
Chapter 14
The Change of
Concentration with Time
First Order Reactions
ln At  kt  ln A0
Prentice Hall © 2003
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.
Prentice Hall © 2003
Chapter 14
The Change of
Concentration with Time
Second Order Reactions
1
1
 kt 
At
A0
Prentice Hall © 2003
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
Prentice Hall © 2003
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
Prentice Hall © 2003
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.
Prentice Hall © 2003
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.
Prentice Hall © 2003
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.
Prentice Hall © 2003
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.
Prentice Hall © 2003
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.
Prentice Hall © 2003
Chapter 14
Temperature and Rate
The Orientation Factor
Prentice Hall © 2003
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.
Prentice Hall © 2003
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.
Prentice Hall © 2003
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).
Prentice Hall © 2003
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).
Prentice Hall © 2003
Chapter 14
Temperature and Rate
Activation Energy
Prentice Hall © 2003
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.
Prentice Hall © 2003
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.
Prentice Hall © 2003
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 
Prentice Hall © 2003
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.
Prentice Hall © 2003
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).
Prentice Hall © 2003
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)
Prentice Hall © 2003
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.
Prentice Hall © 2003
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.
Prentice Hall © 2003
Chapter 14
Reaction Mechanisms
Rate Laws for Elementary Steps
Prentice Hall © 2003
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)
Prentice Hall © 2003
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)
Prentice Hall © 2003
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
Prentice Hall © 2003
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.
Prentice Hall © 2003
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.
Prentice Hall © 2003
Chapter 14
End of Chapter 14
Chemical Kinetics
Prentice Hall © 2003
Chapter 14