Chemical
Chemical Kinetics
Kinetics
1
Chapter
Chapter 15
15
Chemical
Chemical Kinetics
Kinetics
•• We
We can
can use
use thermodynamics to
to tell
tell if
if
aa reaction
reaction is
is product
product or
or reactant
reactant favored.
favored.
The sequence of events at the molecular
level that control the speed and
outcome of a reaction.
•• But
But this
this gives
gives us
us no
no info
info on
on HOW FAST
reaction
reaction goes
goes from
from reactants
reactants to
to products.
products.
Br
Br from
from biomass burning destroys
stratospheric
stratospheric ozone.
ozone. (See
(See R.J.
R.J. Cicerone,
Cicerone,
Science,
Science
Science,, volume
volume 263,
263, page
page 1243,
1243, 1994.)
1994.)
Step
Br
Step 1:
1:
Br ++ O
O33 --->
---> BrO + O22
• KINETICS
—
— the
the study
study of
of REACTION
RATES
RATES and their relation to the way the
reaction
MECHANISM.
reaction proceeds,
proceeds, i.e.,
i.e., its
its MECHANISM.
Step
Step 2:
2:
Step
Step 3:
3:
•• The
The reaction
reaction mechanism
mechanism is
is our
our goal!
goal!
NET:
NET:
An automotive catalytic muffler
Reaction
Reaction Rates
Rates
4
5
Determining
Determining aa Reaction
Reaction Rate
Rate
Section
Section 15.1
15.1
Screen 15.2
Dye
Dye Conc
Conc
• Reaction rate = change in
concentration of a reactant or
product with time.
• Three “types” of rates
–initial rate
–average rate
–instantaneous rate
Reaction
Reaction Mechanisms
Mechanisms
2
Blue
Blue dye
dye is
is oxidized
oxidized
with
with bleach.
bleach.
Its
Its concentration
concentration
decreases
decreases with
with time.
time.
The
The rate
rate —
— the
the
change
change in
in dye
dye conc
conc
with
with time
time —
— can
can be
be
determined
determined from
from the
the
plot.
plot.
Time
Time
Page 1
3
Cl
Cl ++ O
O33 --->
---> ClO
ClO + O22
BrO
BrO + ClO + light ---> Br + Cl + O22
22 O
O33 --->
---> 33 O
O22
Factors Affecting Rates
Section
Section 15.2
15.2
• Concentrations
• and physical state of
reactants and products (Screens
15.3-15.4)
• Temperature (Screen 15.11)
• Catalysts (Screen 15.14)
6
Factors Affecting Rates
7
Factors
Factors Affecting
Affecting Rates
Rates
Section
Section 15.2
15.2
• Concentrations
8
Rate with 0.3 M HCl
9
Factors
Factors Affecting
Affecting Rates
Rates
• Physical state of reactants
Catalysts: catalyzed decomp of H2O2
2 H2O2 --> 2 H2O + O2
Rate with 6.0 M HCl
10
Factors
Factors Affecting
Affecting Rates
Rates
• Temperature
11
Concentrations and Rates
To postulate a reaction
mechanism, we study
••reaction
reaction rate and
••its
its concentration
concentration
dependence
Concentrations and Rates
Take
Take reaction
reaction
where
where Cl-- in
in
cisplatin
cisplatin
[Pt(NH
[Pt(NH33))22Cl
Cl33]]
is
is replaced
replaced
by
by H
H22O
O
Rate of change of conc of Pt compd
=
Page 2
Cisplatin
Am't of cisplatin reacting (mol/L)
elapsed time (t)
12
Concentrations and Rates
Cisplatin
13
Cisplatin
Rate of change of conc of Pt compd
=
Concentrations, Rates, &
Rate Laws
14
a A + b B --> x X with
with aa catalyst
catalyst C
C
Rate
Rate == kk [A]
[A]11
If
If [A]
[A] doubles,
doubles, then
then rate
rate goes
goes up
up by
by factor
factor of
of __
__
m[B]n
n[C]p
p
Rate = k [A]m
Rate
Rate of
of reaction
reaction is
is proportional
proportional to
to [Pt(NH
[Pt(NH33))22Cl
Cl22]]
•• If
rxn. is 2nd order in A.
If m
m == 2,
2, rxn.
Rate
Rate == kk [A]
[A]22
The
The exponents
exponents m,
m, n,
n, and
and pp
We
We express
express this
this as
as aa RATE LAW
•
are the reaction order
•are
Rate
Rate of
of reaction
reaction == kk [Pt(NH
[Pt(NH33))22Cl
Cl22]]
Doubling
Doubling [A]
[A] increases
increases rate
rate by
by ________
________
•• If
rxn. is zero order.
If m
m == 0,
0, rxn.
Rate
Rate == kk [A]
[A]00
•
can be 0, 1, 2 or fractions
•can
fractions
where
where k = rate constant
•
must be determined by experiment!
•must
experiment!
kk is
conc. but increases with T
is independent
independent of
of conc.
15
m[B]
Rate
Rate == kk [A]
[A]m
[B]nn[C]
[C]pp
•• If
rxn. is 1st order in A
If m
m == 1,
1, rxn.
In
In general,
general, for
for
Am't of cisplatin reacting (mol/L)
elapsed time (t)
Interpreting Rate Laws
Cisplatin
If
If [A]
[A] doubles,
doubles, rate
rate ________
________
Cisplatin
Deriving Rate Laws
16
Cisplatin
Deriving Rate Laws
Derive rate law and k for
CH3CHO(g) --> CH4(g) + CO(g)
from experimental data for rate of disappearance of
CH3CHO
2
Rate
Rate of
of rxn
rxn == kk [CH
[CH33CHO]
CHO]2
Expt.
Expt
Expt..
[CH
[CH33CHO]
CHO]
((mol/L)
mol/L)
mol/L)
Now
. #3
expt
Now determine
determine the
the value
value of
of k.
k. Use
Use expt.
expt.
#3 data—
data—
11
22
0.10
0.10
0.20
0.20
0.020
0.020
0.081
0.081
33
44
0.30
0.30
0.40
0.40
0.182
0.182
0.318
0.318
Disappear
Disappear of
of CH
CH33CHO
CHO
((mol/L•sec)
mol/L•sec)
mol/L•sec)
Here
.
conc
Here the
the rate
rate goes
goes up
up by
by ______
______ when
when initial
initial conc.
conc.
doubles.
doubles. Therefore,
Therefore, we
we say
say this
this reaction
reaction is
is
_________________
_________________ order.
order.
0.182
/L•s == kk (0.30
/L)22
mol
mol
0.182 mol/L•s
mol/L•s
(0.30 mol/L)
mol/L)
kk =
mol•s)
= 2.0
2.0 (L
(L // mol•s)
Using
. rate
calc
Using kk you
you can
can calc.
calc.
rate at
at other
other values
values of
of [CH
[CH33CHO]
CHO]
at
at same
same T.
T.
Page 3
17
Cisplatin
Concentration/Time Relations
Need
conc. of reactant is as
Need to
to know
know what
what conc.
function
function of
of time.
time. Consider
Consider FIRST
FIRST ORDER
ORDER
REACTIONS
REACTIONS
The
The rate
rate law
law is
is
[A]
= k [A]
time
18
Cisplatin
Concentration/Time Relations
Integrating
Integrating -- (( [A]
[A] //
ln
natural
logarithm
Concentration/Time Relations
19
Sucrose
Sucrose decomposes
decomposes
to
to simpler
simpler sugars
sugars
Rate
Rate of
of disappearance
disappearance
of
of sucrose
sucrose
== kk [sucrose]
[sucrose]
kk == 0.21
0.21 hr
hr-1-1
time)
time) == kk [A],
[A], we
we get
get
[A] = - k t
[A]o
[A]
[A] // [A]
[A]00 =fraction
=fraction remaining
remaining after
after time
time tt
has
has elapsed.
elapsed.
21
Rate of disappear of sucrose = k [sucrose], k = 0.21 hr-1. If
initial [sucrose] = 0.010 M, how long to drop 90% or to
0.0010 M?
Sucrose
Use the first order integrated rate law
ln
Initial
Initial [sucrose]
[sucrose] ==
0.010
0.010 M
M
How
How long
long to
to drop
drop 90%
90%
(to
(to 0.0010
0.0010 M)?
M)?
[A] at time = 0
Concentration/Time Relations
20
0.0010 M
0.010 M
= - (0.21 hr-1) t
ln (0.100) = - 2.3 = - (0.21 hr -1) • time
time = 11 hours
Called the integrated first-order rate law .
Using the Integrated Rate Law
The integrated rate law suggests a way to tell
the order based on experiment.
2 N2O5(g) ---> 4 NO 2(g) + O2(g)
Time (min)
0
1.0
2.0
5.0
[N2O5]0 (M)
1.00
0.705
0.497
0.173
ln [N2O5]0
0
-0.35
-0.70
-1.75
Rate = k [N 2O5]
Using the Integrated Rate Law
22
Using the Integrated Rate Law
23
2 N2O5(g) ---> 4 NO 2(g) + O2(g) Rate = k [N 2O5]
1
Plot
vs
Plot of
of ln
ln [N
[N22O
O55]] vs.
vs.. time
time
is
is aa straight
straight line!
line!
Eqn.
Eqn
Eqn.. for
for straight
straight line:
line:
yy == ax
ax ++ bb
l n [N2 O5 ] vs . t i me
0
[N2 O5 ] vs . t i me
l n [N2 O5 ] vs . t i me
0
ln [N 2O5] = - kt + ln [N2O5]o
-2
0
0
-2
0
ti me
5
Data of conc.
conc. vs.
vs.
time plot do not fit
straight line.
0
ti me
5
Plot of ln [N2 O5] vs.
vs.
time is a straight
line!
Page 4
ti me
5
conc at
time t
rate const
= slope
conc at
time = 0
All 1st order reactions have straight line plot
for ln [A] vs.
vs. time.
(2nd order gives straight line for plot of 1/[A]
vs.
vs. time)
24
25
26
Half-Life
Properties of Reactions
Half-Life
Section
Section 15.4
15.4 && Screen
Screen 15.8
15.8
Screen 15.7
27
HALF-LIFE is the time
it takes for 1/2 a
sample is disappear.
For 1st order reactions,
the concept of HALFLIFE is especially
useful.
• Reaction is 1st order
decomposition of
H2O2.
28
29
30
Half-Life
Half-Life
Half-Life
• Reaction after 654
min, 1 half-life.
• 1/2 of the reactant
remains.
• Reaction after
1306 min, or 2
half-lives.
• 1/4 of the reactant
remains.
• Reaction after 3
half-lives, or 1962
min.
• 1/8 of the reactant
remains.
Page 5
31
Half-Life
Half-Life
Half-Life
Rate = k[sugar] and k = 3.3 x 10-4 sec-1. What is the halflife of this reaction?
Solution
[A] / [A]0 = 1/2 when t = t 1/2
Therefore, ln (1/2) = - k • t1/2
- 0.693 = - k • t 1/2
t1/2
So, for sugar,
= 0.693 / k
t1/2 = 0.693 / k = 2100 sec =
Half-Life
34
Half-Life
Section 15.4 & Screen 15.8
Section 15.4 & Screen 15.8
Rate = k[sugar] and k = 3.3 x 10 -4 sec-1. Half-life
is 35 min. Start with 5.00 g sugar. How much is
left after 2 hr and 20 min?
Solution
2 hr and 20 min = 4 half-lives
Half-life
Time Elapsed Mass Left
1st
35 min
2.50 g
2nd
70
1.25 g
3rd
105
0.625 g
4th
140
0.313 g
Radioactive decay is a first order process.
Tritium ---> electron + helium
3H
0 e
3He
-1
If you have 1.50 mg of tritium, how much is left
after 49.2 years? t 1/2 = 12.3 years
Page 6
33
Section 15.4 & Screen 15.8
Sugar is fermented in a 1st order process (using
an enzyme as a catalyst).
sugar + enzyme --> products
Rate of disappear of sugar = k[sugar]
k = 3.3 x 10 -4 sec-1
What is the half-life of this reaction?
• Reaction after 4
half-lives, or 2616
min.
• 1/16 of the
reactant remains.
Half-Life
32
Section 15.4 & Screen 15.8
35
35 min
Half-Life
Section 15.4 & Screen 15.8
Start with 1.50 mg of tritium, how much is left after 49.2
years? t1/2 = 12.3 years
Solution
ln [A] / [A]0 = -kt
-kt
[A] = ?
[A]0 = 1.50 mg
t = 49.2 mg
Need k, so we calc k from:
k = 0.693 / t 1/2
Obtain k = 0.0564 y -1
Now ln [A] / [A] 0 = -kt
-kt = - (0.0564 y-1) • (49.2 y)
= - 2.77
Take antilog:
antilog: [A] / [A]0 = e-2.77 = 0.0627
0.0627 = fraction remaining
36
Half-Life
Section 15.4 & Screen 15.8
Start with 1.50 mg of tritium, how much is left after 49.2
years? t1/2 = 12.3 years
Solution
[A] / [A]0 = 0.0627
0.0627 is the fraction remaining!
remaining!
Because [A] 0 = 1.50 mg, [A] = 0.094 mg
But notice that 49.2 y = 4.00 half-lives
1.50 mg ---> 0.750 mg after 1 half-life
---> 0.375 mg after 2
---> 0.188 mg after 3
---> 0.094 mg after 4
37
38
39
Half-Lives of Radioactive Elements
Rate of decay of radioactive isotopes given in
terms of 1/2-life.
238U --> 234Th + He
4.5 x 10 9 y
14C --> 14N + beta
5730 y
131I
--> 131Xe + beta
8.05 d
Element 106 - seaborgium
263Sg
Page 7
0.9 s
• Darleane Hoffman of
UC-Berkeley studies
the newest elements.
She is Director of
Seaborg Institute.
• Rec’d ACS Award in
Nuclear Chem, the
Garvan Medal, and
(in 2000) the ACS
highest award, the
Priestley Medal.