F_-_102_F11_Notes

F_-_102_F11_Notes - F)
Chemical
Kinetics


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Unformatted text preview: F)
Chemical
Kinetics
 Petrucci
et
al.,
10th
Edition:
14.1‐14.11
 
 Rate
of
Chemical
Reaction
 • Chemical
kinetics
–
describe
how
fast
a
reaction
occurs
 • Consider
the
reaction:
 
 
 2
SO2(g)
+
O2(g)
→
2
SO3(g)
 
 Rate
of
change
in
concentration
of
reactants
and
products:
 




 
 
 
 Rates
of
Reaction
 
 2
SO2
(g)
+
O2
(g)
→
2
SO3
(g)
 
 Define
the
rate
of
reaction
(r)
as:
 
 
 
 For
a
generic
reaction:
 
 a
A
+
b
B
→
c
C
+
d
D

 
 
 
 
 Question
F1
 For
the
reaction
C2H4Br2
+
3
KI
→
C2H4
+
2
KBr
+
KI3,
when
the
rate
of
reaction
is
2.0
×
10‐5,
 what
is
the
rate
of
disappearance
of
KI?
 ChE102
Fall
2011
Class
Notes
‐
Chemical
Kinetics
 
 
 
 
 
 
 
 F ‐1 
 Measuring
Rates
of
Reaction
 H2O2
(aq)
→
H2O
(l)
+
½
O2
(g)
 • • • Simple
experimental
set‐up
for
determining
the
rate
of
decomposition
of
H2O2
(aq)
 Oxygen
gas
is
trapped
and
its
volume
is
measured
in
the
burette
 Amount
of
H2O2
consumed
and
the
remaining
concentration
of
H2O2
can
be
calculated
 from
the
measured
volume
of
O2(g)

 
 2.5
 [H2O2
],
mol/L
 2
 1.5
 1
 0.5
 0
 0
 400
 800
 1200
 1600
 2000
 2400
 2800
 Time
(s)
 
 
 [see
Petrucci
et
al.,
Table
14.1
&
14.2,
Figure
14.2]
 
 
 Differential
Rate
Laws
 • Rate
Laws:
 • relate
the
_______________________________
to
the
concentration
of
reactants
 • are
generally
empirical
and
cannot
be
derived
from
the
stoichiometry
of
reactions
 (except
for
some
special
cases
that
we
will
discuss
later)
 • Differential
rate
laws
of
a
reaction:
A
+
B
→
products
is
generally
stated
as:
 
 
 
 k:
reaction
rate
constant
(function
of
temperature)

 n:
order
of
reaction
with
respect
to
A
 m:
order
of
reaction
with
respect
to
B
 n
+
m:
overall
order
of
reaction
 ChE102
Fall
2011
Class
Notes
‐
Chemical
Kinetics
 
 
 
 
 
 
 
 F ‐2 
 For
a
given
reaction:
 
 a
A
+
b
B
→
products
 • n
is
not
necessarily
equal
to
a
 • m
is
not
necessarily
equal
to
b
 • n
and
m
must
be
determined
experimentally

 
 For
example,
the
reaction
rate
for:
 
 H2
+
I2
→
2
HI
 is
determined
to
be:

 
 
 r
=
k[H2][I2]
 
 while
the
rate
for
the
very
similar
reaction,
 
 H2
+
Br2
→
2
HBr
 is
found
to
be:
 
 
 r
=

 
 Question
F2
 At
some
point
in
the
reaction


 
 2A
+
B
→
C
+
3D

 [A]
=
0.3629
M.

At
a
time
8.25
minutes
later
[A]
=
0.3187
M.
 What
is
the
average
rate
of
reaction
during
this
time
interval,
expressed
in
M/s?
 
 
 Method
of
Initial
Rates
 In
this
method,
the
instantaneous
rate
of
reaction
is
determined
at
the
_______________________
 of
the
reaction,
before
reactant
concentrations
have
changed
significantly
 H2O2
(aq)
→
H2O
(l)
+
½
O2
(g)
 
 Consider
the
reaction:
 
 NH4+
(aq)
+
NO2–
(aq)
→
N2(g)
+
2
H2O(l)

 The
following
rate
law
is
proposed:

 
 
 
 The
following
set
of
experimental
data
was
measured:
 
 [NH4+]o

 [NO2–]o

 ro
(mol/L·s)

 1

 0.1
M

 0.005
M

 1.35
x
10–7

 2

 0.1
M

 0.01
M

 2.7
x
10–7 3

 0.2
M

 0.01
M

 5.4
x
10–7
 ChE102
Fall
2011
Class
Notes
‐
Chemical
Kinetics
 
 
 
 
 
 
 
 F ‐3 
 
 
 
 Exps
1
&
2:
 
 
 
 
 
 Exps
2
&
3:

 
 
 
 
 The
differential
rate
law
is:
 
 
 An
estimate
of
the
rate
constant
can
be
calculated
by:

 
 
 
 
 Integrated
Rate
Laws
 • Differential
rate
laws
can
be
integrated
in
time
to
generate
integrated
rate
laws

 • The
differential
rate
law
for
the
reaction:


N2O5
→
2
NO2
+
½
O2

 
 is
given
by:

 
 
 
 Therefore:
 d[ N 2O5 ] − = kdt 
 [ N 2O5 ] 
 [ N 2O5 ] t 
 d[ N 2O5 ] − = k dt 
 [ N 2O5 ] [ N 2O 5 ] o 0 
 [N O ] 
 − ln 2 5 = kt [ N 2O5 ]0 
 
 ln[ N 2O5 ] = − kt + ln[ N 2O5 ]0 
 
 First
Order
Reactions
 € The
plot
of
ln
[A]
x
time
is
linear
for
a
first
order
reaction
(with
respect
to
reactant
A),

 • slope
–k

 • intercept
ln
[A]0

 A
→
products
 
 
 ∫ ∫ ChE102
Fall
2011
Class
Notes
‐
Chemical
Kinetics
 
 
 
 
 
 
 
 F ‐4 
 Second
Order
Reaction
 The
plot
of
[A]–1
x
time
is
linear
for
a
2nd
order
reaction
(with
respect
to
reactant
A)
 • slope
k
 • intercept
[A]0–1

 
 
 
 
 
 Zero
Order
Reaction
 The
plot
of
[A]
x
time
is
linear
for
a
zero
order
reaction
(with
respect
to
reactant
A)

 • slope
–k
 • intercept
[A]0
 
 
 
 
 
 Testing
the
Order
of
Reaction
‐
graphically
 A
→
products
 
 [see
Petrucci
et
al.,
Figure
14.7]
 
 Half­Life
of
Reactions
 Half‐life
(t½)
of
a
reactant
is
the
time
required
for
its
concentration
to
decrease
to
½
of
its
 original
value
 • Zero
order
 
 
 • First
order
 
 
 • 
 Second
order
 
 
 ChE102
Fall
2011
Class
Notes
‐
Chemical
Kinetics
 
 
 
 
 
 
 
 F ‐5 
 Summary:
 Order
 Differential
Rate
Law
 Integrated
Rate
Law
 Units
of
k
 t1/2
 0
 r
=
k
 [A]
=
–
kt
+
[A]0
 mol/L·s
 [A]0/2k
 1
 r
=
k[A]
 ln
[A]
=
–
kt
+
ln
[A]0
 1/s
 ln
2/k
 2
 r
=
k[A]2
 [A]
–1
=
kt
+
[A]0–1
 L/mol·s
 1/k[A]0
 
 
 Question
F3
 The
slow
oxidation
of
Fe2+
by
O2
follows
the
differential
rate
law:
 
 
 
 At
35ºC
and
0.5M
HClO4,
k
=
3.7
x
10–3
L/(mol·atm·h).

Assume
PO2
=
0.2
atm.

Calculate
the
 half
life
(t½)
of
a
0.10M
Fe2+
solution
in
0.5M
HClO4
that
is
exposed
to
air.
 
 
 Question
F4
 Time
(s)
 [A]
(M)
 In
the
decomposition
reaction
A
→
products,
the
data
 0
 0.88
 in
the
table
is
obtained.

What
is
the
order
of
this
 25
 0.74
 50
 0.62
 reaction
and
what
is
the
rate
constant,
k?
 75
 0.52
 
 100
 0.44
 
 150
 0.31
 
 200
 0.22
 
 250
 0.16
 Collision
Theory
of
Gaseous
Reactants
 For
the
bimolecular
elementary
reaction,
 
 A
+
B
→
products
 to
occur,
molecules
of
A
and
B
must
collide.
 
 The
rate
of
molecular
collisions
per
volume
can
be
estimated
by
the
expression:
 
 
 
 At
1
atm
and
0ºC,
assuming
ideal
gas
behaviour,
some
typical
values
are:
 Molecular
density:

 
 Molecular
radius:

 Average
speed:

 
 
 
 ChE102
Fall
2011
Class
Notes
‐
Chemical
Kinetics
 
 
 
 
 
 
 
 F ‐6 
 Activation
Energy
 • If
molecular
collisions
were
the
only
controlling
factor
for
chemical
reactions
taking
 place
in
the
gas
phase,
this
very
high
rate
of
molecular
collisions
would
indicate
that
 all
gas
reactions
would
be
extremely
fast.

 • This
does
not
agree
with
experimental
data
on
gas
reactions.
 • In
fact,
reactant
molecules
must
achieve
a
given
energy
level
(activation
energy)
 before
the
reaction
can
take
place.

 
 [see
Petrucci
et
al.,
Figure
14.10]
 
 
 Temperature
Effects
­
Arrhenius
Law
 Arrhenius
law
describes
the
dependency
of
reaction
rate
constants
on
temperature:
 
 
 
 
 A
 pre‐exponential
factor
 
 Ea 
 activation
energy
 
 T
 temperature
in
K
 The
parameters
A
and
Ea
are
positive
constants
that
depend
on
the
reaction
type.

 
 [see
Petrucci
et
al.,
Figure
14.12]
 
 
 Question
F5
 Consider
the
reaction
O
+
O3
→
2
O2.

If
Ea
=
19.1
kJ/mol
and
k
=
5.1
×
106
L/mol·s
at
25°C,
 calculate
k
at
100°C.
 ChE102
Fall
2011
Class
Notes
‐
Chemical
Kinetics
 
 
 
 
 
 
 
 F ‐7 
 Reaction
Mechanisms
 • a
step‐by‐step
description
of
a
chemical
reaction
at
the
molecular
level

 • must
agree
with
the
overall
stoichiometric
equation
for
the
reaction

 • should
agree
with
the
experimentally
determined
rate
laws
for
the
reaction

 • are
proposed
based
on
experimental
data

 • very
difficult,
if
not
impossible,
to
infer
a
reaction
mechanism
from
the
overall
 chemical
equation
alone
 
 ____________________
Reaction
­
each
step
in
a
mechanism
 
 
 Elementary
Reactions
 • Elementary
reactions
are
classified
as:
 • Unimolecular:
 A
→
products
(dissociation)
 • Bimolecular:
 
 A
+
B
→
products
(bimolecular
collision)

 • Termolecular:
 A
+
B
+
C
→
products
(very
rare)
 
 For
elementary
reactions,
the
order
of
the
reaction
with
respect
to
a
given
reagent
equals
 the
stoichiometric
coefficient
of
the
reagent:
 aA
+
bB
→
products
 r
=
k[A]a[B]b

 
 Some
elementary
processes
are
reversible
and
can
be
treated
using
the
laws
of
chemical
 equilibrium
already
covered
in
this
course
 
 Intermediates
‐
chemical
species
produced
in
one
elementary
reaction
that
are
consumed
 in
another
elementary
reaction
 • should
not
appear
in
the
overall
stoichiometric
equation
nor
in
the
final
reaction
 rate
law
 
 Rate
Determining
Step
‐
slowest
elementary
reaction
in
the
mechanism
&
may
determine
 the
rate
of
the
overall
reaction
 
 
 Slow
Step
Followed
by
a
Fast
Step
 Overall
reaction:
 H2
(g)
+
2
ICl
(g)
→
I2
(g)
+
2
HCl
(g)
 
 Rate
of
reaction:
 r
=
k[H2][ICl]

 
 Mechanism:
 H2
(g)
+
ICl
(g)
→
HI
(g)
+
HCl
(g)



(slow)
 
 
 HI
(g)
+
ICl
(g)
→
I2
(g)
+
HCl
(g)
 
(fast)
 
 
 ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
 
 
 H2
(g)
+
2
ICl
(g)
→
I2
(g)
+
2
HCl
(g)
 ChE102
Fall
2011
Class
Notes
‐
Chemical
Kinetics
 
 
 
 
 
 
 
 F ‐8 
 • • 
 The
rate
of
the
overall
reaction,
r
=
k[H2][ICl],
is
equal
to
the
rate
of
the
slow
 elementary
step
(rate
determining
step)

 HI
is
the
reaction
intermediate
for
this
reaction
(it
does
not
appear
in
the
overall
 reaction
equation)
 [see
Petrucci
et
al.,
Figure
14.14]
 
 
 Fast
Reversible
Step
followed
by
a
Slow
Step
 
 Overall
reaction:
 2
NO(g)
+
O2(g)
→
2
NO2(g)
 
 
 
 
 Rate
of
reaction:
 
 termolecular
elementary
reactions
are
highly
improbable
 Mechanism:
 
 2
NO(g)
→
N2O2(g)
 
 
 k1 
 (fast)
 
 
 
 N2O2(g)
→
2
NO(g)
 
 
 k2

 (fast)
 
 
 
 N2O2(g)
+
O2
(g)→
2
NO2(g)

 k3 
 (slow)
 
 
 
 ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
 r
=
k[NO]2[O2]
 (determined
experimentally)
 
 
 
 2
NO(g)
+
O2(g)
→
2
NO2(g)

 
 
 Rate
Limiting
Step
 
 • Rate
limiting‐step:

r
=
k3[N2O2][O2]
 • N2O2
is
an
intermediate

 • Eliminated
from
the
final
rate
of
reaction
equation
using
the
fast
reversible
reaction
 described
in
the
mechanism
 • rate
of
forward
reaction
=
rate
of
reverse
reaction
(only
valid
for
_______
equilibrium):
 


k1[NO]2
=
k2[N2O2]
 
 
 [ N 2O2 ] = 
 
 r = k3 [ N 2O2 ][O2 ] = k3 × 
 k1 [ NO]2 [O2 ] k2 r = k[ NO]2 [O2 ] 
 • € k1 [ NO]2 k2 Note
that
the
expression
for
rate
of
reaction
derived
from
the
reaction
mechanism
 agrees
with
that
obtained
experimentally.
 ChE102
Fall
2011
Class
Notes
‐
Chemical
Kinetics
 
 
 
 
 
 
 
 F ‐9 
 Steady­State
Approximation
 Without
making
any
assumption
regarding
fast
and
slow
steps:
 
 Overall
reaction:
 
 
 
 2
NO(g)
+
O2(g)
→
2
NO2(g)
 
 Rate
of
reaction
determined
experimentally:
 
 
 r
=
k[NO]2[O2]

 
 Mechanism:

 
 
 
 2
NO
(g)
→
N2O2
(g)
 
 
 k1 
 
 
 
 N2O2
(g)
→
2
NO
(g)
 
 
 k2

 
 
 
 N2O2(g)
+
O2
(g)→
2
NO2(g)




 k3 
 

 
 
 ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
 2
NO
(g)
+
O2
(g)
→
2
NO2
(g)

 

 
 • We
know
that
the
rate
of
reaction
will
be
proportional
to
the
rate
of
O2
consumption.
 Therefore,
we
can
write:
 r
=
k3[N2O2][O2]

 
 • We
need
now
to
eliminate
[N2O2]
from
this
equation,
but
we
cannot
use
the
previous
 approach
since
we
do
not
know
if
the
equilibrium
is
fast.
 
 • We
can,
however,
use
the
steady­state
approximation:
the
concentration
of
the
 intermediate
N2O2
should
be
constant
during
the
reaction,
that
is:
 
 
 rate
of
N2O2
formation
=
rate
of
N2O2
consumption
 
 
 
 k1[NO]2
=
k2[N2O2]
+
k3[N2O2][O2]
 
 
 
 
 Final
rate
of
reaction
is:
 
 
 For
fast
equilibrium:
 
 
 
 
 Fast
equilibrium
assumption
is
just
one
solution
of
the
generic
steady‐state
approximation
 ChE102
Fall
2011
Class
Notes
‐
Chemical
Kinetics
 
 
 
 
 
 
 
 F‐10
 Question
F6
 The
decomposition
reaction
N2O5
→
NO3
+
NO2
follows
the
mechanism:
 
 2
N2O5

N2O5*
+
N2O5

 
 N2O5*
→
NO3
+
NO2

 where
N2O5*
is
a
molecule
at
high
energy
level
capable
of
decomposing
into
NO3
and
NO2.

 The
experimentally‐measured
rate
of
reaction
varies
with
N2O5
concentration
such
that
at
 high
[N2O5]
the
reaction
is
1st
order,
but
at
low
[N2O5]
the
reaction
is
2nd
order.

Explain
 why
the
reaction
order
varies
with
[N2O5].
 
 
 Catalysis
 A
Catalyst

 • increases
the
rate
of
reaction
by
providing
a
different
reaction
path
with
lower
 activation
energy
 • participates
in
the
reaction,
but
it
is
not
consumed
by
the
reaction
 • are
classified
as
homogeneous
or
heterogeneous

 
 Therefore,
catalysts
do
not
appear
in
the
overall
chemical
reaction
as
a
reagent
or
product
 
 Homogeneous
Catalysis
 • Homogeneous
catalysts
are
dissolved
in
the
reaction
medium

 • Consider
the
decomposition
of
formic
acid:
 
 

 HCOOH
→
H2O
+
CO
 • Without
a
catalyst,
this
reaction
requires
that
the
H
atom
be
transferred
from
the
C
 atom
to
the
O
atom
connected
to
the
other
H
atom.

 • The
activation
energy
for
this
reaction
is
high
and
the
reaction
rate
is
low.
 • H+
is
a
homogeneous
catalyst
because
it
provides
an
alternative
path
 
 Heterogeneous
Catalysis
 • Heterogeneous
catalysts
are
solids
that
speed
up
the
reaction
rates
of
gas
or
liquid
 reagents
 • The
steps
in
most
heterogeneous‐catalyzed
reactions
are:
 • Adsorption
of
reagents
on
the
catalyst
surface
 • Diffusion
of
reagents
on
the
catalyst
surface
 • Reaction
on
an
active
site
to
form
products
 • Desorption
of
the
product
into
the
gas
or
liquid
phase

 • An
_________________
is
an
atom
located
on
the
surface
of
the
catalyst
that
can
 catalyze
a
given
reaction.
 • Consider
the
reaction:
2CO
+
2NO
→
2CO2
+
N2
 
 Enzymes
 • Enzymes
are
proteins
that
catalyze
bioreactions

 • Enzymes
are
very
specific
&
promote
the
reaction
of
only
a
few
molecule
types
to
 produce
specific
products
 [see
Petrucci
et
al.,
Figure
14.20]
 ChE102
Fall
2011
Class
Notes
‐
Chemical
Kinetics
 
 
 
 
 
 
 
 F‐11
 ...
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This note was uploaded on 12/28/2011 for the course CHE 102 taught by Professor Simon during the Winter '08 term at Waterloo.

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