Chemical Kinetics
Chapter 14
• Chemical kinetics is concerned
with the speeds or rates of
reaction.
Section 1
• collision theory states that
the greater frequency of
collisions the greater the
rate of reaction.
4 Factors that Affect Reaction Rates
• (based on collision theory)
• Physical state of the reactants. If
the reaction involves a solid the
reaction will have a faster rate if
the solid is in powder form.
( more surface area)
4 Factors that Affect Reaction Rates
• (based on collision theory)
Concentrations of the reactants.
The more concentrated the
reactants the more often
collisions occur
4 Factors that Affect Reaction Rates
• (based on collision theory)
Concentrations of the reactants.
The more concentrated the
reactants the more often
collisions occur
4 Factors that Affect Reaction Rates
• (based on collision theory)
Temperature at which the reaction
occurs.
The greater the temperature the
greater the kinetic energy of
collisions and more often
collisions occur.
4 Factors that Affect Reaction Rates
• (based on collision theory)
Temperature at which the reaction
occurs.
The greater the temperature the
greater the kinetic energy of
collisions and more often
collisions occur.
4 Factors that Affect Reaction Rates
• (based on collision theory)
• Presence of a catalyst- a agent
that increases reaction rate
without being used or
permanently changed in the
process.
Section 2 Reaction Rates
Reaction rates- the change in
concentration of reactants or
products per unit time
Section 2 Reaction Rates
Reaction rates- the change in
concentration of reactants or
products per unit time
Section 2 Reaction Rates
The units for reaction rate is
molarity per unit time
(mol/L) = M.
Section 2 Reaction Rates
• Pg 527 figure 14.3
•A → B
Section 2 Reaction Rates
The reaction rate can be given as
either appearance of product B
or disappearance of reactant A
Average rate with respect to
B = Δ [B]/t
Section 2 Reaction Rates
* The brackets indicate
concentration
* Δ means change in and is always
equal to final minus initial
Section 2 Reaction Rates
• Rates are always expressed as
positive quantities. So if looking
at a disappearing reactant the
rate is the opposite of the change
in concentration per unit time.
• Average rate with respect to A = Δ [A]/t
Sample exercise 14.1
Using figure 14.3 page 527
• Calculate the average rate of
disappearance of A over the time
interval 20s to 40s.
• @ t = 20s [A] = .54M
• @ t = 40s [A] = .31M
Sample exercise 14.1
Using figure 14.3 page 527
• Average rate = - (.3M - .54M) =
( 40s - 20s )
1.2 x 10-2M
s
Practice Exercise pg 528
Calculate the average rate of
appearance of B over the time
interval 0 to 40 s.
@ t = 0 s [B] = 0M
@ t = 40s [B] = .7 M
Practice Exercise pg 528
Calculate the average rate of
appearance of B over the time
interval 0 to 40 s.
• rate = .7M – 0M = 1.8 x 10-2 M
40s - 0s
s
Change of Rate with Time
It is typical for reaction rates to
decrease over time because
concentration of reactants
change over time.
Page 529 figure 14.4 ( notice that
the steepness of the graph
decreases with time.)
• instantaneous rate- rxn rate at
a particular moment in the
reaction.
• Instantaneous rate is
determined from the slope of
the tangent line at the point of
interest.
• The instantaneous rate at t=0 is
called an initial rate.
Sample exercise 14.2 page 530
Looking at 600 s
[.017] @ 800 s
[.042] @ 400s
Instantaneous rate at 600s
017 - .042 = 6.2 x 10-5 M
800s – 400s
s
Reaction Rates and Stoichiometry
• Both examples we have done so
far require that the rate of
appearance of the product = the
rate of disappearance of the
product. The is because the
stiochiometry was 1:1
When stiochiometric relationships
are not one to one:
2HI → H2 + I2
Rate = - ½ Δ[HI]/t = Δ[H2]/t = Δ[I2]/t
In general for the reaction
( lower case letters are coefficients)
aA + bB → cC
+ dD
Rate = -1/a Δ[A]/t =-1/b Δ[B]/t = 1/c Δ[C]/t = 1/d Δ[D]/t
Sample exercise 14.3 page 531
a.)How does the rate the
disappearance of O3 compare to
the rate of appearance of O2 ?
2O3 → 3O2
Rate = -1/2 [O3]/t = 1/3 [O2]/t
b.) – [O3] = 2/3 Δ [O2] = 4.5 10-5 mol O3 /L
s
• 14.5, 14.9 14.11 in class (back of chapter)
14.3 Concentration and Rate
One way to study the effect of
concentration is the consider
how the initial rate reaction
depends on initial
concentration.
NH4+ + NO2- → N2 + 2H2O
• Table 14.2 page 532
• What happens to rate when [NO2] is kept constant and [NH4+] is
doubled? Rate doubles.
The same thing happens to the
rate when [NH4+] is kept constant
and [NO2 ] is doubled.
The rate of the reaction is
proportional to the [NH4+] raised
to first power and to [NO2 ]
raised to the first power.
Rate = k [NH4+][NO2-]
Rate law – an equation that shows
how the rate depends on the
concentration of reactants
In general for:
aA + bB → cC
rate = k [
m
A]
n
[B]
+ dD
• k is the rate constant and changes w/
temp.
• m and n are typically small whole
numbers and are determined by how
the concentration of the reactant
affects rate
• The exponents m and n are called
reaction orders
If the rate is directly proportional to
the concentration of the reactant the
exponent is 1. For example first order
in [NH4+] and first order [NO2-].
Rate=k[NH4+][NO2-].
If the rate is directly proportional to
the concentration of the reactant the
exponent is 1. For example first order
in [NH4+] and first order [NO2-].
Rate=k[NH4+][NO2-].
An exponent of 1 is never written.
An exponent of 0 means that the
concentration of that particular
reactant does not influence rate.
An exponent of 1 means that
the if the concentration is
doubled the rate is doubled. (
directly proportional)
An exponent of 2 means that if
the concentration is doubled the
rate is quadrupled.
note: often the exponents are the
same as the coefficients, but not
always. The exponents must be
determined experimentally.
It is possible to have a reaction
order that is a fraction.
•
•
•
•
•
Sample exercise 14.4
Box 1 :
Box 2 :
Box 3 :
___> ____> ___
•
•
•
•
•
Sample exercise 14.4
Box 1 : k (5)(5)2 = 125 k
Box 2 : k (7)(3)2 = 63 k
Box 3 : k (3)(7)2 = 147 k
3> 1> 2
•
•
•
•
•
Practice exercise
1=
2=
3=
___=____<___
•
•
•
•
•
Practice exercise
1 = k(5)(5) = 25 k
2= k (7)(3) = 21k
3 = k (3)(7) = 21k
2=3<1
Units of Rate Constants
The units of the rate constant
depend on the overall reaction
order of the rate law
Unit of rate const= units of rate/(units of conc.)overall rxn rate
• 1st order
• 2nd order
• 3rd order
= M/s = s-1
M
= M/s = M-1s-1
M2
= M/s = M-2s-1
M3
Sample exercise 14.5
a.)
b.)
c.)
Sample exercise 14.5
a.) 1st order
b.) 3/2 order
c.) 2nd order
Practice exercise 535
Using Initial Rate to Determine Rate
Laws
• The rate law for any experiment
must be determined experimentally.
It is found by observing the effect of
changing initial conc of reactants.
Most reactants have exponents of 0, 1,
or 2.
The rate of a reaction depends on
concentration but rate constant
does not.
Rate constant is affected by
temperature and presence of a
catalyst.
Sample exercise 14.6 Pge 536
Rate law = k [A]m[B]n
Rate law =
k = __________=
rate =
Sample exercise 14.6 Pge 536
Rate law = k [A]m[B]n
Rate law = k [A]2[B]0 = k [A]2
k = 4 X 10-5 = 4 X 10-3 M-1s-1
[.1M]2
rate = (4 X 10-3)(.05 M) = 1X 10 -5 M/s
Practice 536
rate =
k=
Rate =
Practice 536
rate = k [NO]2[H2]
k = 1.23 x 10-3 M/s = 1.23 M-2s-1
(.1M)2 (.1M)
Rate = (1.23 M-2s-1)( .05M)2(.15M)
= 4.61 x 10-4 M/s
Page 565 14.13, 15, 17, 21 and 23
Example 14.6
Rate = k [A]m[B]n
m=____ b/c rate increases by ____ when [ ]
increases by 2
n= ____ b/c rate doesn’t change when conc. is
doubled
Example 14.6
Rate = k [A]m[B]n
m=2 b/c rate increases by 4 when [ ] increases
by 2
n= 0 b/c rate doesn’t change when conc. is
doubled
rate =
k=
c.)
rate = k [A]2
k = rate/[A]2 = 4 x 10-5M/s = 4 x 10-3M-1s-1
(.1M)2
c.) (4 x 10-3)(.05M)2 = 1 x 10 -5 M/s
Practice pg 537
Rate =
k=
Practice pg 537
Rate = k[NO]2[H2]
k= rate/[NO]2[H2] = 1.23 x 10-3 M/s = 1.2 M-2s-1
(.1M)2(.1M)
13.) a.)
b.)
c.) k =
13.) a.) increase by a factor of 4
No the rate constant remains unchanged. When
[A] increases the rate is increased.
b.)second order in A; first order in B; third order
overall.
c.) k = M/s
= M-2s-1
(M2)(M)
15.) a.)
b.)
c.)
15.) a.) Rate = 4.82 x 10-3 s-1[N2O5]
b.) rate = 4.82 x 10-3s-1(.024M)= 1.15 x 10-4M/s
c.) rate is doubled to 2.3 x 10 -4 M/s
17. ) Rate =
a.)
b.)
c.)
17. ) Rate = k [CH3Br][OH-]
.0432 m/s
=k
( 5 x 10-3)(.05)
a.) 1.728 x 102 M-1s-1 =k
b.) M-1s-1
c.) Rate would be tripled
21. a.) rate = k[OCl-][I-]
1.36 x 104 M/s
(1.5 x 10-3M)( 1.5 x10-3M)
b.) 6.0 x 109 M-1s-1 = k
rate = ( 6.0 x 109 M-1s-1) ( 1.0 x 10-3M)(5x10-4M)
= 3 x 103M/s
21. a.) rate =
b.)
rate =
23. a. rate = k[NO2]2[O2]
b.) - k= M/s = M-2s-1
M2 M
c.) - 1.4 x 10-2
= 7.05 x 10 3 M-2s-1
(0.126)2(.0250)
1.13 x10-1
= 7.1 x 10 3 M-2s-1
(.0252)2(.0250)
5.64 x 10-2
(.0252)2(.0125)
= 7.1 x 10 3 M-2s-1
Section 4 The Change of Concentration
with Time
In this section we will use
equations that will tell us the
concentrations of reactants and
products at any given time.
First Order Reactions
Rate depends on the concentration
of a single reactant raised to the
first power.
Rate = - Δ[A]/t= k [A]
using integration we can get an
equation that relates the initial
concentration of A at [A]◦ to the
concentration at any other time
[A]t
ln[A]t – ln [A]◦ = -kt
or
ln [A]t = -kt
[A]◦
or
ln[A]t = -kt + ln [A]◦
( slope intercept form)
These equations can be used to
find [A]t , [A]o , t or k if any three
of those are known.
Example 14.7
b/c
ln[A]t = -kt + ln [A]◦ is
in slope intercept form,
(y=mx +b) the graph of ln[A]t
vs. time will give a straight
line with slope –k and y
intercept ln[A]o
If a reaction is first order this graph
will give a straight line.
some example problems will use
pressure as a unit of
concentration for a gas. Pressure
is directly proportional to the
number of moles per unit volume
and can be used for
concentration.
Second Order Reactions
A reaction whose rate depends
on the reactant concentrations
raised to the second power or
on the concentration of two
reactants to the first power.
For a rxn that is second order in just one
reactant:
Rate = - Δ[A]/t= k [A]2
Rate = - Δ[A]/t= k [A]2
With calculus the rate law can be written:
1 = kt + 1
[A]t
[A]o
This above equation can be used to find [A]t,
[A]o , t or k for second order reactions in one
reactant.
The graph of 1/[A]t vs t will yield a straight line
with slope k and y intercept 1/[A]o
One way to distinguish first order reactions
from second order reactions is to plot ln[A]t
vs. time and 1/[A]t whichever plot gives a
straight line tells you if the rxn is first order or
second order in one reactant. These are the
two of the most common types.
• [A]t – first order
• 1/[A]t- second order
example 14.8
Half-Life
The half life t1/2 is the time
required for the concentration
of a reactant to drop to one
half of its initial value.
For a first order rxn:
[A]t1/2 = ½ [A]o
ln ½ [A]o = -kt1/2
[A]o
ln ½ = -kt1/2
ln ½ = t1/2
-k
.693/k = t1/2
The half life for a first order rxn is
independent of initial conc. of
reactant.
In a first order rxn. The conc. of
the reactant decreases by ½ in
each of a series of regularly
spaced time intervals.
The half life for a second order
and other reactions depends on
the reactant concentrations and
therefore changes as the reaction
progresses.
T1/2 = 1/k[A]o
• Example 14.9
Section 5: Temperature and Rate
• The rates of most chemical
reactions increases as temp
increases
• As temp increases, the rate
constant increases
Collision Model
Molecules must collide to react.
The larger the number of
collisions per second, the greater
the reaction rate. As temperature
increases, the molecules move
faster and collide more often and
with larger KE
Only a tiny fraction of collisions
lead to rxn
•molecules must be oriented
correctly for a collision to result
in a rxn.
•Molecules must have enough
activiation energy (Ea)
activation energy- in order to
react the KE of the molecules
must be large enough to
stretch, bend, and break
bonds. So that the chem. rxn.
can occur.
Activation energy can be
considered a barrier that
represents the energy necessary
to force a molecule through a
relatively unstable intermediate
stage to the final product.
Ea is the difference in energy
between the starting molecule
and the highest energy along
the reaction pathway.
• The most unstable complex is the
activated complex or transition
state.
• Generally the lower the
activation energy the faster the
reaction.
As the temp increases a
greater fraction of molecules
will have the KE necessary to
overcome Ea which leads to a
larger reaction rate.
• The fraction of molecules with enough
energy to overcome Ea is given by :
• f=e-Ea/RT
•
•
f – fraction of molecules with
sufficient energy to overcome Ea
•
T- temperature in Kelvin
•
R – gas constant – 8.314 J/mol • K
• For a temp of 300 K , Ea of 100,000 J/mol
( a typical value)
f= 3.8 x 10-18
moral of the story “ at any given time the
fraction of molecules present with
sufficient energy is extremely small
Arrhenius Equation
• The increase in the rate with increasing
temp is nonlinear.
• Rxn rate is dependent on 3 factors
• fraction of molecules with energy
equal to Ea or more
• # of collisions per second
• fraction of collisions with appropriate
orientation
•
•
•
•
•
•
•
•
Arrhenius Equation
k= Ae-Ea/RT
k – rate constant
Ea – activation energy
R is gas constant ( 8.314 J/mol•K)
T in Kelvin
A is the frequency factor
The frequency factor is constant as temperature is
varied
• Rxn rates decrease as Ea increases
Exercise 4.10
Determining the Activation Energy
• Taking the natural log of both
sides of the Arrhenius equation
gives
•
ln k = - Ea/ RT + ln A
The graph of ln k vs. 1/T will give a
straight line with slope -Ea/R and y
intercept ln A
Or non-graphically
If you know the rate constants at two
temps.
Ln k1/k2 = Ea/R ( 1/T2 - 1/T1)
Or
Ea = -( (ln k1 - lnk2)/ (1/T1- 1/T2))R
Exercise 14.11
T1 = 230 C ( 503K) and k1= 6.3 x 10-4
T2 = 189.7C ( 462.7 K) and k2 = 2.52 x 10-5
Ea = - ((ln _______– ln ________)/ 1/_____ 1/_______)) 8.314
= ___________J/mol
•
•
•
•
•
Exercise 14.11
T1 = 230 C ( 503K) and k1= 6.3 x 10-4
T2 = 189.7C ( 462.7 K) and k2 = 2.52 x 10-5
Ea = - ((ln 6.3 x 10-4 – ln 2.52 x 10-5)/ 1/503 1/462.7)) 8.314
• = 1.6 x 105 J/mol
•
Section 6 Reaction Mechanisms
The process by which a reaction
occurs is called the reaction
mechanism
At the most detailed level a rxn
mechanism will describe
exactly the order in which
bonds are broken and
reformed.
Elementary Steps
When a reaction occurs in a single
step, the process or steps are called
elementary steps.
The number of molecules that
participate as reactants in an
elementary step defines the
molecularity of the step
If a single molecule is involved it is
unimolecular
Elementary steps involving the collision of two
reactant molecules are bimolecular
Elementary steps involving the simultaneous
collision of three molecules are termolecular
The chance that an elementary step involves
the simultaneous collision of 4 molecules is
extremely remote.
Multistep Mechanisms
A multistep mechanism consists of a sequence
of elementary steps. For example:
NO2 + CO → NO + CO2
Actually occurs in two steps
1st step – two NO2 molecules collide to transfer
one O atom
NO2 + NO2 → NO + NO3
2nd step- the NO3 collides with CO and
transfers an O atom
NO3 + CO → NO2 + CO2
The elementary steps always add
together to give the overall chemical
equation
NO2 + NO2 → NO + NO3
NO3 + CO → NO2 + CO2
2NO2 + NO3 +CO → NO2 + NO3 + CO2 + NO
NO2 + CO → NO + CO2
If a reactant cancels so that it
is neither a reactant or a
product in the overall process
it is an intermediate
•
•
•
•
Sample exercise 14.12
O 3 → O2 + O
O3 + O → 2 O 2
1st step involves a single reactant so it is
________________
• 2O3→ 3O2
• O is the ___________
•
•
•
•
Sample exercise 14.12
O 3 → O2 + O
O3 + O → 2 O 2
1st step involves a single reactant so it is
unimolecular
• 2O3→ 3O2
• O is the intermediate
Work practice exercise
Rate Laws for Elementary Steps
Rate laws must be determined
experimentally and cannot be
predicted from coefficients of
balanced equations
The rate law can be determined
from the mechanism
If we know the elementary steps of a
rxn then we know the rate law
Elementary Steps and Rate Laws
Molecularity Elementary steps Rate Law
Unimolecular A→products
rate = k[A]
Bimolecular A + A→products
rate = k [A]2
Bimolecular A+B→products
rate = k[A][B]
Termolecular A+A+A→products rate = k[A]3
Termolecular A+A+B→products rate = k[A]2[B]
Termolecular A+B+C→products rate= k[A][B][C]
The elementary step of a rxn
determines the rate, not the
overall rxn. Therefore, we cannot
determine rate from the
coefficients
Exercise 14.13
most rxns involve more than one
elementary step
often one step is much slower
than the others
b/c the slow step limits the
overall rxn, it is called the rate
determining step or the rate
limiting step.
Each step has its own Ea and rate
constant.
The rate determining step governs
the rate law for the overall rxn.
For example:
NO2 + CO → NO + CO2
Experimentally we know is second order in
NO2 and zero order in CO. The proposed rxn
mechanism is
k1
NO2 + NO2 → NO + NO3
(slow)
k2
NO3 + CO → NO2 + CO2 (fast)
Rate = k1[NO2]2
we know that the rate law cannot
be rate = k[NO2][CO] b/c changing
the conc of CO experimentally
does not change the rate of
reaction.
Mechanisms with an initial fast
step (for example)
2NO + Br2 → 2NOBr
experimentally the rate law
has been determined to be
rate= k[NO]2[Br2]
we would expect a single termolecular step:
NO + NO + Br2→ 2NOBr
rate = k[NO]2[Br2]
However , termolecular processes are
extremely rare.
The actual process has the first
step as the fast step and the
second step as the slow (rate
determining ) step
Step 1 ( fast)
k1
NO + Br2 ↔NOBr2
Step 2 ( slow)
k2
NOBr2 + NO → 2NOBr
It is difficult to determine the
conc. of NOBr2 b/c it is an
intermediate. Intermediates
are unstable and tend to have
a low conc.
b/c the 1st step is fast and the second step is
slow
the [NOBr2] = [NO][Br2]
so looking at step two :
rate = k[NO][Br2][NO] = k [NO]2[Br2]
whenever a fast step precedes a
slow one, we can solve for the
conc. of the intermediate by
assuming that an equilibrium is
established in the first step.
Exercise 14
Section 7 Catalysis
A catalyst lowers the overall
activation energy for a chemical
reaction.
A catalyst is a substance that
changes the speed of a chemical
reaction without undergoing
permanent change itself.
A catalyst that is present in the
same phase as the reacting
molecules is a homogeneous
catalyst
A catalyst usually lowers Ea by
providing a completely different
rxn mechanism
• A heterogeneous catalyst exists in
a different phase than either of
the reactants
• Heterogenous catalysts are often
made of metals or metal oxides
The initial step of heterogeneous
catalysts is usually adsorption in
which the molecules bind to the
surface.
Adsorption usually happens b/c the
atoms at the surface are extremely
reactive.
The places where the reacting
molecules become adsorbed are
called active sites.
The reacting molecules are in
close proximity to one another
increasing the probability of rxn.
Catalytic converters are an
example of heterogeneous
catalysts.
The job of the catalytic converter
is to convert CO and CxHy( unused
hydrocarbons) to CO2 and H2O
and to convert NO and NO2 to N2
and O2
Transition metal oxides and noble
metals are used for the catalytic
converters
Usually the catalysts that are the
most effective for one rxn are
much less effective for the other.
It is necessary to have two
compartments on the catalytic
converter
35% of the Pt, 65% of the Pd and
95% of the Rh used annually is
due to catalytic converters. All of
these metals come from South
Africa and Russia and are more
expensive than gold.
Biological catalysts- enzymes
Enzymes tend to be very large
proteins
Enzymes are very selective for the
reaction they catalyze
The position on the molecule where
the rxn occurs is called the active
site
The substance that undergoes rxn at
the reaction site is the substrate
Lock and Key Model
The substrate fits neatly in to the
active site on the enzyme
As the substrate enters the active
site it is somehow activated and
capable of extreme rapid reaction. (
possibly loss of electron, change of
molecular shape or distribution of
the substrate bonds)
• Once rxn occurs, the substrate
leaves the bond allowing
another substrate to enter.
• An enzyme inhibitor will bond
to the active sit and block the
entry site
The number of individual
catalyzed reactions it called
turnover number
Enzymes have large turnover
numbers meaning they have very
low activation energy.