Chapter 13
Chemical Kinetics
The study of reaction rate is called chemical kinetics.
Reaction rate is measured by the change of
concentration (molarity) of reactants or products
per unit time.
Molarity of A: [A], e.g. [NO2]: molarity of NO2
r−
reactantf  reactanti
t f  ti
Δreactant
−
Δt
reactants  products, [reactant]↓ and [product]↑
OR r 
productf  producti
t f  ti
Δproduct

Δt
Unit: mol·L−1·s−1 ≡ M·s−1
2NO2 (g)  2NO (g) + O2 (g)
−
= r(NO2)
rate is a function of time
0 s → 50 s:
r  NO2   
  NO2 
0.0079 M  0.0100 M

 4.2  10-5 M  s1
t
50 s  0 s
50 s → 100 s:
r  NO2   
  NO2 
0.0065 M  0.0079 M

 2.8  10-5 M  s 1
t
100 s  50 s
2NO2 (g)  2NO (g) + O2 (g)
−
= r(NO2)
−
rate is a function of time
50 s → 100 s:
  NO 0.0035 M  0.0021 M
r  NO  

 2.8  10-5 M  s 1
t
100 s  50 s
50 s → 100 s:
r  O2  
 O2  0.0018 M  0.0011 M

 1.4  10-5 M  s 1
t
100 s  50 s
2NO2(g)  2NO(g) + O2(g)
 n NO 2 
Δ[NO2 ]
 Δ


V   Δn NO 2 2
r(NO2 )
 Δ[NO2 ]

Δt





Δ[O2 ]
r(O2 )
Δ[O2 ]
ΔnO2
1
 n O2 
Δ

Δt
 V 
aA+bBcC+dD
r(A) a

r(B) b
r(A) r(B)


,
a
b
r(A) a

r(C) c

r(A) r(C)

a
c
r(A) r(B) r(C) r(D)



r=
a
b
c
d
r does not depend upon the choice of species
2N2O5(g)  4NO2(g) + O2(g)
r(N2O5) = 4.2 x 10−7 M·s−1
What are the rates of appearance of NO2 and O2 ?
Example 13.1. page 567
Consider the following balanced chemical equation:
H2O2(aq) + 3 I–(aq) + 2 H+(aq)  I3–(aq) + 2 H2O(l)
In the first 10.0 seconds of the reaction, the
concentration of I– dropped from 1.000 M to 0.868 M.
(a)Calculate the average rate of this reaction in this
time interval.
(b) Predict the rate of change in the concentration of
H+ (that is, [H+]/t) during this time interval.
Consider the general reaction aA + bB cC and the
following average rate data over some time period Δt:
Δ[A]

 0.0080 M  s 1
Δt
Δ[B]

 0.0120 M  s 1
Δt
Δ[C]
 0.0160 M  s1
Δt
Determine a set of possible coefficients to balance this
general reaction.
−
= r(NO2)
−
All the rates in this table are average rates.
2:00 pm
2:16 pm
Δt
3:16 pm
Δl
0 mile
Barnesville
16 miles
Griffin
56 miles
Atlanta
Average Speed from B to G = 16 miles ÷ 16 min = 1.0 mile/min
Average Speed from G to A = 40 miles ÷ 60 min = 0.7 mile/min
l d l

Instantaneous speed at green spot  lim
t 0 t
dt
Instantaneous speed contains more information
l
Atlanta
56 miles
Δl
Griffin
20 miles
Barnesville
0 mile
0 min
Δt
16 min
t
76 min
Instantaneous speed at the red point = slope of the red solid line =
l
t
Reaction rate is a function of time
Factors that affect reaction rates
State of the reactants
Concentrations of the reactants
Temperature
Catalyst
aA+bBcC+dD
r=k
[A]m
[B]n  
1 Δ[A]
1 d[A]

a Δt
a dt
Differential rate law: how r depends on concentrations
m, n: reaction order, mth order for A, nth order for B
(m+n): overall reaction order
k: rate constant: depends on temperature, but not
concentrations
m and n must be measured from experiments. They can
be different from the stoichiometry.
(1)
2N2O5(g)  4NO2(g) + O2(g)
(2)
CHCl3(g) + Cl2(g)  CCl4(g) + HCl(g)
(3)
H2(g) + I2(g)  2HI(g)
r = k[N2O5]
r = k[H2][I2]
r = k[CHCl3][Cl2]1/2
aA + bB  cC +dD
r = k [A]m [B]n
Units
overall reaction order (m+n) ⇌ unit of k
(1)
2N2O5(g)  4NO2(g) + O2(g)
(2)
CHCl3(g) + Cl2(g)  CCl4(g) + HCl(g)
(3)
H2(g) + I2(g)  2HI(g)
r = k[N2O5]
r = k[H2][I2]
r = k[CHCl3][Cl2]1/2
How to find the rate law by experiment: method of initial rates
NH4+ (aq) + NO2− (aq)  N2(g) + 2H2O(l)
Experiment
Number
Initial [NH4+] (M)
Initial [NO2−] (M)
Initial Rate (M·s−1)
1
0.100
0.0050
1.35 x 10−7
2
0.100
0.010
2.70 x 10−7
3
0.200
0.010
5.40 x 10−7
r = k [NH4+]m [NO2−]n
A very common method to investigate how each factor
affects the whole system:
Change one thing at a time while keep the others constant.
z = f (x,y)
z
 z 
How does the change of x affect z?   
 x  y x
 z 
z
How does the change of y affect z?   
 y  x y
How to find the rate law by experiment: method of initial rates
NH4+ (aq) + NO2− (aq)  N2(g) + 2H2O(l)
Experiment
Number
Initial [NH4+] (M)
Initial [NO2−] (M)
Initial Rate (M·s−1)
1
0.100
0.0050
1.35 x 10−7
2
0.100
0.010
2.70 x 10−7
3
0.200
0.010
5.40 x 10−7
r = k [NH4+]m [NO2−]n
2NO(g) + 2H2(g)  N2(g) + 2H2O(g)
Experiment
Number
Initial [NO] (M)
Initial [H2] (M)
Initial Rate (M·s−1)
1
0.10
0.10
1.23 x 10-3
2
0.10
0.20
2.46 x 10-3
3
0.20
0.10
4.92 x 10-3
(a) Determine the differential rate law
(b) Calculate the rate constant
(c) Calculate the rate when [NO] = 0.050 M and [H2] = 0.150 M
2NO(g) + 2H2(g)  N2(g) + 2H2O(g)
again
Experiment
Number
Initial [NO] (M)
Initial [H2] (M)
Initial Rate (M·s−1)
1
0.10
0.10
1.23 x 10-3
2
0.10
0.30
3.69 x 10-3
3
0.30
0.10
1.11 x 10-2
(a) Determine the differential rate law
(b) Calculate the rate constant
(c) Calculate the rate when [NO] = 0.050 M and [H2] = 0.150 M
A+BC
Experiment
Number
Initial [A] (M)
Initial [B] (M)
Initial Rate (M·s−1)
1
0.100
0.100
4.0 x 10−5
2
0.100
0.200
4.0 x 10−5
3
0.200
0.100
1.6 x 10−4
(a) Determine the differential rate law
(b) Calculate the rate constant
(c) Calculate the rate when [A] = 0.050 M and [B] = 0.100 M
EXAMPLE 13.2 Determining the Order and Rate Constant of a Reaction
NO2(g) + CO(g)  NO(g) + CO2(g)
From the data, determine:
(a) the rate law for the reaction
(b) the rate constant (k) for the reaction
Use the data in table to determine
1) The orders for all three reactants
2) The overall reaction order
3) The value of the rate constant
r = k [BrO3−]m [Br−]n [H+]p
r1 = k (0.10 M)m (0.10 M)n (0.10 M)p = 8.0 x 10−4 M · s−1
r2 = k (0.20 M)m (0.10 M)n (0.10 M)p = 1.6 x 10−3 M · s−1
r3 = k (0.20 M)m (0.20 M)n (0.10 M)p = 3.2 x 10−3 M · s−1
r4 = k (0.10 M)m (0.10 M)n (0.20 M)p = 3.2 x 10−3 M · s−1
one quiz after lab
Relationship among reaction rates as expressed
by different species.
r(A) r(B) r(C) r(D)



r=
a
b
c
d
overall reaction order (m+n) ⇌ unit of k
Method of initial rates:
table of experimental data  rate order, k, rate at other
concentrations.
aA+bBcC+dD
r = k [A]m [B]n
Differential rate law: how r depends on concentrations
Differential rate law: differential equation
[ A]
d [ A]
r

 k [ A]m [ B]n
at
a dt
How concentration changes as a function of time 
integrated rate law
A  Products
First order reaction differential rate law:
Δ[A]
d[A]
r = r(A)  

 k[A]
Δt
dt
First order reaction integrated rate law:
ln[A] kt  ln[A]0
or
[A]0
ln
 kt
[A]
integrated rate law: how concentration changes as a
function of time.
First order reaction integrated rate law:
ln[A]  kt  ln[A]0
[A] is the molarity of A at t
y = mx + b
Plot ln[A] vs. t gives a straight line
Slope = −k,
intercept = ln[A]0
N2O5(g)  2NO2(g) + ½ O2(g)
[N2O5]
(M)
Time (s)
0.1000
0.0707
0.0500
0.0250
0
50
100
200
0.0125
0.00625
300
400
Use these data, verify that the rate law is first order in N2O5,
and calculate the rate constant.
k = 6.93 x 10−3 s−1
Read a similar Example 13.3 on page 575
Using the data given in the previous example, calculate [N2O5]
at 150 s after the start of the reaction.
[N2O5]
(M)
Time (s)
0.1000
0
0.0707
50
0.0500
100
0.0250
200
0.0125
300
0.00625
400
0.0354 M
Practice on Example 13.4 on page 576 and
check your answer
The half-life of a reaction, t1/2, is the time required for a
reactant to reach one-half of its initial concentration.
[N2O5]
(M)
Time (s)
0.1000
0
0.0707
50
0.0500
100
0.0250
200
0.0125
300
0.00625
400
The half-life for first order reaction:
t1 / 2
ln 2

k
The half-life for first order reaction does NOT depend on
concentration.
[N2O5]
(M)
Time (s)
0.1000
0
0.0707
50
0.0500
100
0.0250
200
0.0125
300
0.00625
400
A Plot of [N2O5] versus Time for the Decomposition
Reaction of N2O5
[N2O5 ]  [N2O5 ]0 ekt
A certain first order reaction has a half-life of 20.0 s.
a)Calculate the rate constant for this reaction.
b)How much time is required for this reaction to be 75 %
complete?
t1 / 2
ln 2

k
a) k = 0.0347 s−1
[A]0
ln
 kt
[A]
b) k = 40.0 s
Try Example 13.6 and For Practice 13.6
on page 579 and check your answers
A  Products
Second order reaction differential rate law:
Δ[A]
d[A]
r(A)  

 k[A] 2
Δt
dt
Second order reaction integrated rate law:
1
1
 kt 
[A]
[A]0
integrated rate law: how concentration changes as a
function of time.
Second order reaction integrated rate law:
1
1
 kt 
[A]
[A]0
y = mx + b
Plot 1/[A] vs. t gives a straight line
Slope = k,
intercept = 1/[A]0
The half-life for second order reaction:
t1/ 2
1

k [A]0
The half-life for second order reaction depends on initial
concentration.
A certain reaction has the following general form: A  B
At a particular temperature [A]0 = 2.80 x 10−3 M,
concentration versus time data were collected for this reaction,
and a plot of 1/[A] versus time resulted in a straight line with a
slope value of 3.60 x 10−2 M−1·s−1
a)Determine the (differential) rate law, the integrated rate law,
and the value of the rate constant.
b) Calculate the half-life for this reaction.
c) How much time is required for the concentration of A to
decrease to 7.00 x 10−4 M ?
1
1
 kt 
[A]
[A]0
t1/ 2
1

k [A]0
(show your work, do not copy the question)
For first order reaction, show that
t1 / 2
ln 2

k
from the integrated rate law
ln[A]  kt  ln[A]0
A  Products
Zero order reaction differential rate law:
Δ[A]
d[A]
r(A)  

 k[A] 0  k
Δt
dt
Zero order reaction integrated rate law:
[A]  kt  [A]0
integrated rate law: how concentration changes as a
function of time.
Zero order reaction integrated rate law:
[A]  kt  [A]0
y = mx + b
Plot [A] vs. t gives a straight line
Slope = −k,
intercept = [A]0
The half-life for zero order reaction:
t1/ 2
[ A ]0

2k
The half-life for zero order reaction depends on initial
concentration.
The Decomposition Reaction 2N2O(g)  2N2 (g) + O2 (g)
takes place on a Platinum Surface
The decomposition of ethanol on alumina surface
C2H5OH(g)  C2H4(g) + H2O(g) was studied at 600 K.
Concentration versus time data were collected for this
reaction, and a plot of [C2H5OH] versus time resulted in a
straight line with a slope value of −4.00 x 10−5 M·s−1
a)Determine the (differential) rate law, the integrated rate law,
and the value of the rate constant.
b) If the initial concentration of C2H5OH was 1.25 x 10−2 M,
calculate the half-life of this reaction.
c) How much time is required for all the 1.25 x 10−2 M C2H5OH
to decompose ?
[A]  kt  [A]0
t1/ 2
[ A ]0

2k
Factors that affect reaction rates
State of the reactants
Concentrations of the reactants
Temperature
Catalyst
aA + bB  cC +dD
r = k [A]m [B]n
k: rate constant: depends on temperature, but not
concentrations
A Plot Showing the
Exponential Dependence of
the Rate Constant on
Absolute Temperature
Collision model: molecules must collide to react.
Not all collisions lead to products
T ↑  v ↑  kinetic energy = ½ mv2 ↑
activation energy
Prentice Hall © 2003
Chapter 14
Fraction of molecules whose Ek > Ea is
T↑f↑
f e
Ea ↑  f ↓
 Ea / RT
k  zf  ze
 Ea / RT
k  kexpt
Another factor needs to be taken into account
molecular orientation
Several Possible Orientations for a Collision Between Two
BrNO Molecules
BrNO + BrNO  2NO +Br2
 Ea / RT
k  pze
steric factor p ≤ 1
Arrhenius equation
 Ea / RT
k  Ae
A: frequency factor
T↑k↑
How k depends on T
Ea ↑  k ↓
k1
Ea  1 1 
ln     
k2
R  T1 T2 
At 550 °C the rate constant for
CH4(g) + 2S2(g)  CS2(g) + 2H2S(g)
is 1.1 M·s−1, and at 625 °C the rate constant is
6.4 M·s−1. Using these value, calculate Ea for this reaction.
k1
Ea  1 1 
ln     
k2
R  T1 T2 
Ea = 1.4 x 105 J/mol
Try Example 13.8 on page 585 and
check your answers
h
The ball can climb over the hill only if its kinetic energy is greater
than Ep = mgh, where m is the mass of the ball, h is the height
of the hill, and g is gravitational acceleration.
(a) The Change in Potential as a Function of Reaction Progress
(b) A Molecular Representation of the Reaction
BrNO + BrNO  2NO +Br2
O N Br
exothermic reaction
Reaction Mechanism
Chemical Equation
Reactants  Products
NO2 + CO  NO + CO2
Step 1:
Step 2:
NO2 + NO2  NO3 + NO
NO3 + CO  NO2 + CO2
NO2 + CO  NO + CO2
NO3: intermediate, does not
appear in overall reaction
+
NO2
NO2
NO3
NO
NO3
CO
NO2
CO2
Whole process is called the reaction mechanism.
Each single step is called an elementary reaction/step.
An elementary reaction is a single collision.
For an elementary reaction
aA + bB  cC + dD
r = k[A]a[B]b
Not for overall reaction!
Overall reaction: r = k[A]m[B]n , find m and n by experiments
The number of molecules that react in an elementary
reaction is called the molecularity of that that elementary
reaction (not applicable to overall reaction).
1 ― unimolecular
2 ― bimolecular
3 ― termolecular
What determines the rate of an overall reaction?
Team A
Team B
Slowest step: rate determining step
Which step is the rate determining step? 2nd
exothermic or
endothermic?
Ep
intermediate
Ea, 2
Ea, 1
Reactants
∆E
Products
Reaction progress
Factors that affect reaction rates
State of the reactants
Concentrations of the reactants
Temperature
Catalyst
Catalyst is a substance that speeds up a reaction without
being consumed itself.
Catalyst changes the reaction mechanism through a lower
activation energy pathway.
Energy Plots for a Given Reaction
Ea
Ea
homogeneous
catalyst
heterogeneous
Homogeneous catalyst: same phase as reactants.
3O2(g)  2O3(g)
2NO(g) + O2(g)  2NO2(g)
light
2NO2(g)  2NO(g) + 2O(g)
2O2(g) + 2O(g)  2O3(g)
+
3O2(g)  2O3(g)
NO(g): catalyst; NO2(g), O(g): intermediates
Heterogeneous catalyst: different phase from reactants.
The Decomposition Reaction 2N2O(g)  2N2 (g) + O2 (g)
takes place on a Platinum Surface
These Cookies Contain Partially Hydrogenated
Vegetable Oil
−C−C−
−
−
+ H2 
−
−
C=C
milk sugar = lactose