Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Unit 7 (Chp 14):
Chemical Kinetics
(rates)
John D. Bookstaver
St. Charles Community College
St. Peters, MO
 2006, Prentice Hall, Inc.
Chemical Kinetics
Chemical Kinetics is the study of:
1) Rates at which reactants are consumed
or products are produced in a chemical rxn.
2) Factors that affect the rate of a reaction
according to Collision Theory (temperature,
concentration, surface area, & catalyst).
3) Mechanisms, or sequences of steps, for
how a reaction actually occurs.
4) Rate Laws (equations) used to calculate
rates, rate constants, concentrations, & time.
Reaction Rates
Reaction rates are described by the change in
concentration (DM) of reactants (consumed)
or of products (produced) per change in time.
Reaction Rates
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
[C4H9Cl]
measured at
various times
recall:
[ ] brackets represent
concentration in molarity (M)
Reaction Rates
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
D[C4H9Cl]
Dt
“change in”
Reaction Rates
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
average
rate slows
WHY?
fewer
reactant
collisions
Reaction Rates
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
D[C4H9Cl] (rise)
(run)
Dt
• The slope of a
line tangent to
the curve at any
point is the
instantaneous
rate at that time.
Reaction Rates
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
• All reactions slow
down over time.
• The initial rate
of reaction is
commonly chosen
for analysis and
comparison.
Reaction Rates and Stoichiometry
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
rate of consumption
of reactant
=
rate of production
of product.
(IF 1:1 mol ratio)
Rate:
–D[C4H9Cl] D[C4H9OH]
=
Dt
Dt
↓ reactant = ↑ product
production of +D[NO]
products per time Dt
Are equalized to a
stoich. coefficient of 1
D[NO]
D[O2]
–D[NO2]
=
=
2 Dt
Dt
2 Dt
change during rxn
concentration (M)
Reaction Rates:
change in conc.
slope =
change in time
consumption of –D[NO2]
reactants per time Dt
2 NO2  2 NO + O2
D[NO2]
Dt
2 NO
1 O2
time (t)
Rate Ratios
aA + bB
HW p. 619 #20
cC + dD
1 D[A]
1 D[B]
Rate = −
=
= −
a Dt
b Dt
1 D[C]
1 D[D]
=
c Dt
d Dt
2 HI(g)  H2(g) + I2(g)
Initial rate of production of H2 is 0.050 M∙s–1.
What is the rate of consumption of HI?
=
0.050 M∙s–1 H2 x 2 mol HI = 0.10 M∙s–1 HI ?
mol
L∙s
1 mol H2
mol ratio
–0.10 M∙s–1 HI
Rate Laws
Rate equations (or rate laws) have the form:
rate = k[A]x[B]y[C]z
rate
constant
order with respect to reactants
A, B, & C
…or…
number of each particle involved
in collision that affects the rate
overall order of reaction = x + y +…
Example:
overall order (4 particles
–
–
rate = k[BrO3 ][Br ][H+]2 = ___
4th order in collision)
The Collision Model
• reactant bonds break, then product bonds form.
• Reaction rates depend on collisions
between reactant particles with:
activation energy:
1) greater frequency
minimum E required
2) enough energy(Eact )
to start reaction
3) proper orientation
unsuccessful
successful
reactants
Collision
products
Reaction Coordinate Diagram
transition state
Potential Energy

demo
Eact
…aka…
Energy
Profile
∆E
Reaction progress

Factors That Affect Reaction Rates
4 Factors that Affect Reaction Rates:
1)
2)
3)
4)
Concentration
Temperature
(exposed) Surface Area (particle size)
Catalyst
Factors That Affect Reaction Rates
1)
Concentration of Reactants
 ↑ concentration, ↑ collision frequency
 increase pressure of gases
Fe(s) + O2(g)  Fe2O3(s)
20%
of air
is
O2(g)
100%
O2(g)
Factors That Affect Reaction Rates
2) Temperature
↑ Temp, ↑ rate
collisions of…
greater frequency
greater energy
Eact at higher Temp
unsuccessful collisions
(bounce off)
more particles
over Eact
successful collisions
(react)
Temperature and k (rate constant)
↑ Temp, ↑ rate
rate = k [A]x
k is temp. dependent
(k changes with temp)
Factors That Affect Reaction Rates
3)
Surface Area (particle size)
 smaller pieces, more exposed
surface area for collision.
Factors That Affect Reaction Rates
Catalyst ↑ rate by changing the reaction
mechanism by…
demo
potential energy
4)
Uncatalyzed
Catalyzed
2 H2O2
+ 2 Br– + 2 H+
–
…lowering the Eact .
 consumed,
then produced
(not used up)
Br2
(intermediate)
H+
2 H2O2 Br ,
2 H2O + O2
reaction progress
2 H2O + O2
+ 2 Br– + 2 H+
Surface Catalysts
H2 + H2C=CH2
Catalysts can orient reactants
to help bonds break and form.
H2 + H2C=CH2  H3C–CH3
CH3CH3
Enzymes
• biological catalysts in living systems.
• A substrate fits into the active site of the
enzyme much like a key fits into a lock.
(IMAFs work here)
HW p. 621
#50,51,64
recall… Rate Laws
Rate equations (or rate laws) have the form:
rate = k[A]x[B]y[C]z
rate
constant
order with respect to reactants
A, B, & C
…or…
number of each particle involved
in collision that affects the rate
overall order of reaction = x + y +…
Example:
overall order (4 particles
–
–
rate = k[BrO3 ][Br ][H+]2 = ___
4th order in collision)
Reaction Mechanisms
The sequence of molecular collisions and
changes by which reactants become products
is called the reaction mechanism.
• Rxns may occur in separate elementary steps.
• The overall reaction occurs only as fast as the
slowest, rate-determining step. (RDS)
Slow First Step
NO2 (g) + CO (g)  NO (g) + CO2 (g)
• A proposed mechanism for this reaction is:
Step 1: NO2 + NO2  NO3 + NO (slow)
Step 2: NO3 + CO  NO2 + CO2 (fast)
• NO3 intermediate is produced then consumed.
• Rate law depends on the slow 1st step:
rate = k [NO2]2
• CO is not involved in the slow RDS,
so it does not appear in the rate law.
(OR…the order w.r.t. CO is ___)
0th
Slow Second Step
2 NO(g) + Br2(g)  2 NOBr(g)
• A proposed mechanism is:
Step 1: NO + Br2
NOBr2
(fast)
Step 2: NOBr2 + NO  2 NOBr
(slow)
• NOBr2 intermediate is produced then consumed.
• Rate law depends on the slow 2nd step:
rate = k2 [NOBr2] [NO]
• But cannot include intermediate [NOBr2] (b/c
it’s difficult to control conc.’s of intermediates).
Slow Second Step
Step 1: NO + Br2
NOBr2
Step 2: NOBr2 + NO  2 NOBr
(fast)
(slow)
From RDS Step 2: rate = k2 [NOBr2] [NO]
b/c step 1 is in equilibrium
From Step 1: rateforward = ratereverse
k1 [NO] [Br2] = k−1 [NOBr2]
solve for
[NOBr2]
substitute for
k1
[NO]
[Br
]
=
[NOBr
]
2
2
[NOBr2]
k−1
Slow Second Step
Step 1: NO + Br2
NOBr2
Step 2: NOBr2 + NO  2 NOBr
HW p. 625
#92, 94
WS Reaction
Mechanisms
(fast)
(slow)
rate = k2 [NOBr2] [NO]
k
k
2
1 [NO] [Br ] [NO]
rate =
2
k−1
rate = k [NO]2 [Br2]
substitute for
k1
[NO]
[Br
]
=
[NOBr
]
2
2
k−1
[NOBr2]
recall… Rate Laws
Rate equations (or rate laws) have the form:
rate = k[A]x[B]y[C]z
rate
constant
order with respect to reactants
A, B, & C
…or…
number of each particle involved
in collision that affects the rate
overall order of reaction = x + y +…
Example:
overall order (4 particles
–
–
rate = k[BrO3 ][Br ][H+]2 = ___
4th order in collision)
Orders from Experimental Data:
Rate varies with Concentration
Initial Concentrations
1
0.0050
0.25
0.30
Rate in M
per unit
time
10
2
0.010
0.25
0.30
20
3
0.010
0.50
0.30
40
4
0.010
0.50
0.60
160
Experiment [BrO3–] , M
[Br–] , M
[H+] , M
Orders from Experimental Data:
Rate varies with Concentration
Initial Concentrations
1
0.0050
0.25
0.30
Rate in M
per unit
time
10
2
0.010
0.25
0.30
20
3
0.010
0.50
0.30
40
4
0.010
0.50
0.60
160
Experiment [BrO3–] , M
[Br–] , M
[H+] , M
In experiments A & B:
rate = k [BrO3–]x
• doubling [BrO3–] doubles the rate.
2 = [2]x
(other conc.’s kept constant)
1=x
• rate is 1st order w.r.t. [BrO3–]
“w.r.t.” (with respect to)
Orders from Experimental Data:
Rate varies with Concentration
Initial Concentrations
1
0.0050
0.25
0.30
Rate in M
per unit
time
10
2
0.010
0.25
0.30
20
3
0.010
0.50
0.30
40
4
0.010
0.50
0.60
160
Experiment [BrO3–] , M
[Br–] , M
[H+] , M
In experiments B & C:
• doubling [Br–] doubles the rate
(other conc.’s kept constant)
• rate is 1st order w.r.t. [Br–]
rate = k [Br–]y
2 = [2]y
1=y
Orders from Experimental Data:
Rate varies with Concentration
Initial Concentrations
1
0.0050
0.25
0.30
Rate in M
per unit
time
10
2
0.010
0.25
0.30
20
3
0.010
0.50
0.30
40
4
0.010
0.50
0.60
160
Experiment [BrO3–] , M
[Br–] , M
[H+] , M
In experiments C & D:
• doubling [H+] quadruples the rate
(other conc.’s kept constant)
• rate is 2nd order w.r.t. [H+]
rate = k [H+]z
4 = [2]z
2=z
Orders from Experimental Data:
Rate varies with Concentration
Initial Concentrations
1
0.0050
0.25
0.30
Rate in M
per unit
time
10
2
0.010
0.25
0.30
20
3
0.010
0.50
0.30
40
4
0.010
0.50
0.60
160
Experiment [BrO3–] , M
[Br–] , M
[H+] , M
Orders found by experiment are:
x=1 , y=1 , z=2
Rate law is:
–
–
rate = k [BrO3 ] [Br ] [H+]2
Orders in Rate Laws
1)…only found experimentally (from data).
2)…do NOT come from the coefficients of
reactants of an overall reaction.
3)…represent the number of reactant particles
(coefficients) in the RDS of the mechanism.
4)…zero order reactants have no effect on rate
b/c they do not appear in the RDS of the
mechanism (coefficient of 0 in RDS).
5)…typically 0, 1, 2, but can be any # or fraction
Units of k (rate constant)
Units of k give info about order.
HW p. 618
#22,24,28
Rate is usually (M∙s–1), or (M∙min–1), etc.
1st
2nd
3rd
Order
0th
rate =
rate =
rate =
Rate rate =
k[A]
k[A]0
k[A]2 k[A]2[B]
Law
M=?
M = ?∙M M = ?∙M2 M = ?∙M3
s
s
s
s
k Units
M
s
1
s
M∙s–1
s–1
1
M∙s
M–1∙s–1
1
M2∙s
M–2∙s–1
HW p. 619 #28
Determine the rate law for the reaction
(from experimental data).
(a)
rate = k [ClO2]x [OH–]y
rate1 = (0.0248) = k (0.060)x(0.030)y
rate2 = (0.00276) = k (0.020)x(0.030)y
x 0.030 y
0.060
(0.0248) =
(0.00276) = 0.020 0.030
9=
(3)x
(1)y
9 = (3)x x = 2
HW p. 619 #28
Determine the rate law for the reaction
(from experimental data).
(a)
rate = k [ClO2]2 [OH–]y
rate3 = (0.00828) = k (0.020)2(0.090)y
rate2 = (0.00276) = k (0.020)2(0.030)y
3 = (3)y
y=1
rate = k [ClO2]2 [OH–]1
OR rate = k [ClO2]2 [OH–]
HW p. 619 #28
Calculate the rate constant (with units).
(b)
rate = k [ClO2]2 [OH–]
Exp 1: (0.0248) = k (0.060)2(0.030)
(0.0248)
k=
(0.060)2(0.030)
k = 230 M–2∙s–1
M = ?∙M2∙M
s
HW p. 619 #28
Calculate the rate when [ClO2] = 0.010 M
and [OH–] = 0.025 M.
(c)
rate = k [ClO2]2 [OH–]
rate = (230)(0.010)2(0.025)
rate = 0.00058 M∙s–1
Rate Laws
Differential rate laws express (reveal) the
relationship between the concentration
of reactants and the rate of the reaction.
The differential rate law is usually
just called “the rate law.”
Integrated rate laws express (reveal)
the relationship between
concentration of reactants and time.
Integrated Rate Laws
Using calculus to integrate a first-order
rate law gives us:
[A]t
ln
= −kt
[A]0
[A]0 = initial conc. of A at t = 0 .
[A]t = conc. of A at any time, t .
given on exam
ln [A]t – ln [A]0 = –kt
ln [A]t = –kt + ln [A]0
y
= mx + b
First-Order Processes
ln [A]t = –kt + ln [A]0
y
first-order
m = –k
ln [A]
t
= mx + b
If a reaction is
first-order, a plot
of ln [A] vs. t is a
straight line, and
the slope of the
line will be –k.
First-Order Processes
CH3NC
CH3CN
at 198.9 oC
First-Order Processes
• When ln P is plotted as a function of time, a
straight line results.
• Therefore,
 The process is first-order.
 k is the negative slope: 5.1  10–5 s−1.
Second-Order Processes
Similarly, integrating the rate law for a
process that is second-order in reactant
A, we get…
given
1
1
on
–
= kt
[A]t [A]0
exam
1
1
= kt +
[A]t
[A]0
y = mx + b
Second-Order Processes
1
1
= kt +
[A]t
[A]0
y = mx + b
second-order
If a reaction is
m=k
second-order, a
plot
of
1/[A]
vs.
t
1
is a straight line,
[A]
and the slope of
the line is k.
t
Second-Order Processes
The decomposition of NO2 at 300°C is described by
the equation
NO2 (g)
NO (g) + 1/2 O2 (g)
and yields the following data:
Time (s)
0.0
50.0
100.0
200.0
300.0
[NO2], M
0.01000
0.00787
0.00649
0.00481
0.00380
Second-Order Processes
• Graphing ln [NO2] vs. t
yields:
• The plot is NOT a straight
line, so the process is
NOT first-order in [A].
Time (s)
0.0
50.0
100.0
200.0
300.0
[NO2], M
0.01000
0.00787
ln [NO2]
−4.610
−4.845
0.00649
0.00481
0.00380
−5.038
−5.337
−5.573
Second-Order Processes
• Graphing 1/[NO2] vs. t,
however, gives this plot.
• Because this IS a
straight line, the process
is second-order in [A].
Time (s)
0.0
50.0
100.0
200.0
300.0
[NO2], M
0.01000
0.00787
1/[NO2]
100
127
0.00649
0.00481
0.00380
154
208
263
Zero-Order Processes
zero-order
[A]
t
If a reaction is
zero-order, a plot
of [A] vs. t is a
straight line, with
a slope = rate.
Summary of
Integrated Rate Laws
and
Straight-Line Graphs
zero-order
[A]
HW p. 620 #33, 38, 43
ln [A]
first-order
m = –k
t
ln [A]t = –kt + ln [A]0
1
[A]
t
second-order
m=k
t
1
1
= kt +
[A]t
[A]0
Half-life (t1/2):
• time at which half of initial amount remains.
[A]t = 0.5 [A]0
initial
concentration
at time, t
concentration
at time, t = 0
Half-life (t1/2) is constant
for 1st order only.
Half-life (t1/2) depends on k:
For a 1st order process:
[A]t1/2 = 0.5 [A]0
ln [A]t – ln [A]0 = –kt
ln 0.5 [A]0 – ln [A]0 = –kt1/2
0.5 [A]0
ln
= −kt1/2
[A]0
ln 0.5 = −kt1/2
given
on
exam
−0.693 = −kt1/2
0.693
= t1/2
k
Half-life & Radiometric Dating
A wooden object from an archeological site is
subjected to radiometric dating by carbon-14.
The activity of the sample due to 14C is
measured to be 11.6 disintegrations per second
(current amount of C-14).
The activity of a carbon sample of equal mass
from fresh wood is 15.2 disintegrations per
second (assumed as original amount of C-14).
The half-life of 14C is 5715 yr.
What is the age of the archeological sample?
Half-life & Radiometric Dating
A wooden object from an archeological site is
subjected to radiometric dating by carbon-14.
The activity of the sample due to 14C is
[A]t = 11.6 current
dis/s
measured to be 11.6 disintegrations
per second
at time, t
(current amount of C-14).
The activity of a carbon sample of equal mass
from fresh wood is 15.2 disintegrations
perdis/s
[A]0 = 15.2 initial
second (assumed as originalat
amount
time, t0of=C-14).
0s
The half-life of 14C is 5715 yr. t1/2 = 5715 yr
What is the age of the archeological sample?
t = ? yr
Half-life & Radiometric Dating
First, determine the rate constant, k.
ln Nt − ln N0 = –kt
easy way
(given: t1/2 = 5715 yr)
ln 0.5N0 − ln N0 = –k(5715)
0.5 N0
ln
= −k(5715)
N0
ln 0.5 = −k(5715)
–0.693 = −k(5715)
–0.693
=k
–5715
0.693
= t1/2
k
0.693
= 5715 yr
k
0.693
=k
5715 yr
k = 1.21  10−4 yr−1
Half-life & Radiometric Dating
Now we can determine t:
ln Nt − ln N0 = –kt
ln(11.6) – ln(15.2) = −(1.21  10−4) t
–0.270 = −(1.21  10−4) t
2230 yr = t
Summary
0th Order
1st Order
2nd Order
Rate Law
Rate = k
Rate = k[A]
Rate = k[A]2
ln[A] = –kt + ln[A]0
1
1
 kt 
[ A]
[ A]0
Integrated
Rate Law
[A] = –kt + [A]0
y
Linear plot [A]
k & slope
of line
Units of k
Half-Life
= mx + b
ln[A]
t
t
y = mx + b
1
[A]
t
Slope = –k
Slope = –k
Slope = k
M=?
s
M = ?∙M
s
M = ?∙M2
s
M∙s–1
s–1
0.693
t1/ 2 
k
M–1∙s–1
depends on [A]0
HW p. 620 #34,36,40,32
depends on [A]0