Chap14-Part5

Chap14-Part5 - REVIEW
SESSION‐6:
SATURDAY



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Unformatted text preview: REVIEW
SESSION‐6:
SATURDAY

 CHAPMAN
211‐
10.30
am
to
Noon
 CHEMICAL
KINETICS

 Send
me
your
Ques-ons
on
Ch‐13
and
Ch‐14
 Arrhenius
Equa-on
 
Taking
the
natural
 logarithm
of
both
sides,
 the
equa-on
becomes
 Ea
 1
 ln
k
=
‐








(




)
+
ln
A
 T
 R 
 y


=




m





x




+


b
 Therefore,
if
k
is
determined
experimentally
at
 several
temperatures,
Ea
can
be
calculated
from
the
 1
 slope
of
a
plot
of
ln
k
vs.


 T
 Sample
Exercise
14.10
 Assume
that
all
three
reac-ons
have
equivalent
frequency
factors
A.
 Rank
the
reac-ons
in
order
of
slowest
to
fastest.
 2
is
the
slowest
and
1
is
the
fastest 
 

 Determining
the
AcHvaHon
Energy
 Ea k = A exp(− ) RT ln( xy ) = ln( x) + ln( y ) Ea ⇒ ln(k ) = ln( A) + ln[exp(− )] RT Ea ⇒ ln(k ) = ln( A) − RT plot
ln(k)
vs.
1/T
 intercept
=
ln(A)



&



A
=
exp[intercept]
 slope
=
‐Ea/R



&



Ea
=
‐
(R)(slope)
 Sample
Exercise
14.11
 temperature
(oC) 
rate
constant
k
(sec‐1)
 189.7 198.9 230.2 251.2 
 
 
 
 
 
 
 
 
2.52
x
10‐5
 
5.25
x
10‐5
 
6.30
x
10‐4
 
3.16
x
10‐3
 (a)  Calculate
ln(k)
and
1/T
(in
1/K).
 (b)  Determine
the
ac-va-on
energy.
 (c)  Determine
the
frequency
factor.
 (d)
Find
the
rate
constant
at
430.0
K.
 Determining
Rate
Constants
without
Frequency
Factors
 Ea k1 = A exp(− ) RT1 ⇒ & Ea k 2 = A exp(− ) RT2 Ea A exp(− ) Ea Ea k2 RT2 = = exp( − ) Ea k1 RT1 RT2 A exp(− ) RT1 Ea 1 1 ⇒ k 2 = k1 exp[ ( − )] R T1 T2 
 
 
ReacHon
Mechanisms
 A
balanced
equa-on
for
a
chemical
reac-on
shows
the
reactants
 and
products
in
the
proper
stoichiometry.
 This
equa-on
provides
no
direct
informa-on
about
the
mechanism
 of
the
reac-on
(e.g.,
intermediates).
 An
equa-on
showing
all
of
the
steps,
including
intermediates,
 indicates
the
elementary
mechanism.
 If
the
elementary
mechanism
is
known,
its
form
can
be
used
to
 write
the
rate
laws
rather
than
determining
them
empirically.
 
 
 
ReacHon
Mechanisms
 The
number
of
reactants
in
a
given
step
of
an
elementary
reac-on

 defines
the
molecularity
of
the
step.
 
one
reactant
–
unimolecular
 
two
reactants
–
bimolecular
 
three
reactants
–
termolecular 
 
etc.
 Unimolecular
steps
do
not
require
collisions.
 Bimolecular
steps
do
require
collisions.
 Termolecular
steps
are
rare,
requiring
three‐fold
collisions.
 MulHstep
Mechanisms
 
 
 

 Reac-on
mechanisms
can
be
broken
down
into
elementary
reac-ons
called

 steps
in
the
reac-on.

For
example:
 Overall: 
 
NO2(g)
+
CO(g)
→
NO(g)
+
CO2(g)
 1st
Step: 
 
NO2
(g)
+
NO2(g)
→
NO3(g)
+
NO(g)
 2nd
Step:
 
NO3(g)
+
CO(g)
→
NO2(g)
+
CO2(g)
 Sum: 
2
NO2(g)
+
NO3(g)
+
CO(g)
→
NO3(g)
+
NO(g)
+
NO2(g)
+
CO2(g)
 Simplify:

2
NO2(g)
+
NO3(g)
+
CO(g)
→
NO3(g)
+
NO(g)
+
NO2(g)
+
CO2(g)
 Net:
 
NO2(g)
+
CO(g)
→
NO(g)
+
CO2(g)
 NO3(g)
is
an
intermediate
because
it
is
neither
a
reactant
nor
a
product.
 MulHstep
Mechanisms
 Reac-on
mechanisms
can
be
broken
down
into
elementary
reac-ons
called

 steps
in
the
reac-on.

For
example:
 Overall: 
 
2NO
(g)
+
Br2(g)
→
2
NOBr(g)
 1st
Step: 
 
NO(g)
+
Br2(g)
→
NOBr2(g)
 2nd
Step:
 
NOBr2(g)
+
NO(g)
→
2NOBr(g)
 Sum: 

 
2
NO(g)
+
Br2(g)
+
NOBr2(g)
→
NOBr2(g)
+
2NOBr(g)

 Net:
 
2
NO(g)
+
Br2(g)
→
2
NOBr(g)
 NOBr2(g)
is
an
intermediate
because
it
is
neither
a
reactant
nor
a
product.
 Sample
Exercise
14.12
 It
has
been
proposed
that
the
conversion
of
ozone
(O3)
into
molecular
oxygen
 (O2)
proceeds
by
the
following
two‐step
mechanism:
 
 
 
O3(g)
→
O2(g)
+
O(g)
 
O3(g)
+
O(g)
→
2
O2(g)
 (a)  Describe
the
molecularity
of
each
elementary
reac-on.
 
 

 (b)
Write
the
equa-on
for
the
overall
reac-on.
 (c)
Find
the
intermediate. 
 

 Sample
Exercise
14.12
 It
has
been
proposed
that
the
conversion
of
ozone
(O3)
into
molecular
oxygen
 (O2)
proceeds
by
the
following
two‐step
mechanism:
 
 
 
O3(g)
→
O2(g)
+
O(g)
 
O3(g)
+
O(g)
→
2
O2(g)
 (a)  Describe
the
molecularity
of
each
elementary
reac-on.
 
 
1
and
2
 (b)
Write
the
equa-on
for
the
overall
reac-on.
 
2
O3
(g)
3
O2
(g)
 (c)
Find
the
intermediate. 
O
(Nascent
Oxygen,
[O])

 

 Determining
Rate
Laws
for
Elementary
ReacHons
 What
is
the
molecularity
of
the
following
elementary
reac-ons?
 Write
the
rate
law
for
each.
 
 
(a)
2
NO(g)
→
N2O2(g)
 
 

 
 

 
 
(b)
SO3(g)
→
SO2(g)
+
O(g)
 
 

 What
is
the
molecularity
of
the
following
elementary
reac-ons?
 Write
the
rate
law
for
each.
 
 
(a)
2
NO(g)
→
N2O2(g)
 
 
 
 
Molecularity
=2

 
rate
=
k[NO]2
 

 
 
(b)
SO3(g)
→
SO2(g)
+
O(g)
 
 
 
 
Molecula-ty=1
 
Rate
=
k[SO3]
 

 Rate
Determining
Step
 Many
chemical
reac-ons
have
mul-ple
steps.
 Each
step
has
its
own
ac-va-on
energy
and
rate
constant.
 If
one
step
is
much
slower
than
the
others,
the
overall
reac-on
 cannot
proceed
faster
than
this
slow
step.
 In
this
case,
the
slow
step
is
called
the
rate‐determining
step,
or
 the
rate‐limi-ng
step.
 The
rate‐limi-ng
step
determines
the
overall
reac-on
rate.
 Rate‐LimiHng
Step
 The
rate
of
traffic
flow
from
place

 1
to
place
3
is
determined
by
the
 rate
of
movement
through
plaza
A.
 The
rate
of
traffic
flow
from
place

 1
to
place
3
is
determined
by
the
 rate
of
movement
through
plaza
B.
 Two‐Step
Mechanisms
with
a
Slow
First
Step
 Example: 
NO2(g)
+
CO(g)
→
NO(g)
+
CO2(g)
 1st
Step: 
 
NO2
(g)
+
NO2(g)
→
NO3(g)
+
NO(g) 
(k1,
slow)
 2nd
Step: 
 
NO3(g)
+
CO(g)
→
NO2(g)
+
CO2(g) 
(k2,
fast)
 The
overall
reac-on
rate
is
controlled
by
the
first
step.

Thus,
 Rate = k1 [ NO2 ] 2 Two‐Step
Mechanisms
with
a
Fast
First
Step

 Example: 
2NO
(g)
+
Br2(g)
→
2
NOBr(g)
 1st
Step: 
 
NO(g)
+
Br2(g)
↔
NOBr2(g) 
(k1,k‐1
fast)
 2nd
Step: 
 
NOBr2(g)
+
NO(g)
→
2NOBr(g) 
(k2,
slow)
 Rate = k 2 [ NOBr2 ][ NO ] This
equa-on
is
not
too
useful
because
it
depends
on
the
intermediate
concentra-on.
 Assume
the
first
step
is
in
rapid
equilibrium.

Then
 k1 [ NO ][ Br2 ] = k −1 [ NOBr2 ] ⇒ k1 [ NO ][ Br2 ] = [ NOBr2 ] k −1 k1 [ NO ][ Br2 ] k1 k 2 ⇒ Rate = (k 2 [ NO ])( )=( )[ Br2 ][ NO ] 2 k −1 k −1 Consider
the
diagram
to
the
 right,
represen-ng
two
steps
 inan
overall
reac-on.
 red
=
O,
blue
=
N,
green
=

F
 (a)  Write
the
equa-on
for
the
overall
reac-on.
 (c)
Find
the
rate
if
the
first
 step
is
slow.
 (b)
Iden-fy
the
intermediate. 

 Consider
the
diagram
to
the
 right,
represen-ng
two
steps
 in
an
overall
reac-on.
 red
=
O,
blue
=
N,
green
=

F
 (a)  Write
the
equa-on
for
the
overall
reac-on.
 NO2
+
F2

NO2F+
F
 NO2+F

NO2F
 (b)
Iden-fy
the
intermediate. 

 (c)
Find
the
rate
if
the
first
 step
is
slow.
 Consider
the
diagram
to
the
 right,
represen-ng
two
steps
 in
an
overall
reac-on.
 red
=
O,
blue
=
N,
green
=

F
 (a)  Write
the
equa-on
for
the
overall
reac-on.
 NO2
+
F2

NO2F+
F
 NO2+F

NO2F
 (b)
Iden-fy
the
intermediate. 
F
is
the
intermediate
 (c)
Find
the
rate
if
the
first
 step
is
slow.
 Consider
the
diagram
to
the
 right,
represen-ng
two
steps
 in
an
overall
reac-on.
 red
=
O,
blue
=
N,
green
=

F
 (a)  Write
the
equa-on
for
the
overall
reac-on.
 NO2
+
F2

NO2F+
F
 NO2+F

NO2F
 (c)
Find
the
rate
if
the
first
 step
is
slow.
 Rate
=
k[NO2][F]
 (b)
Iden-fy
the
intermediate. 
F
is
the
intermediate
 Catalysis
 A
catalyst
is
a
substance
that
changes
the
rate
of
a
chemical
 reac-on
without
undergoing
a
chemical
change
itself.
 A
homogeneous
catalyst
is
present
in
the
same
phase
as
the
 reactants
and
products.
 Example
(slow): 
2
H2O2(aq)
→
2
H2O(l)
+
O2(g)
 With
bromine
as
a
catalyst
(fast):

 
step
1: 
2
Br‐(aq)
+
H2O2(aq)
+
2
H+(aq)
→
Br2(aq)
+
2H2O(l)
 
step
2: 
Br2(aq)
+
H2O2(aq)
→
2Br‐(aq)
+
2H+(aq)
+
O2(g)
 
sum: 

2
H2O2(aq)
→
2
H2O(l)
+
O2(g)
 A
catalyst
usually
speeds
up
the
reac-on
rate
by
providing
a
different
 pathway
for
the
reac-on
having
a
lower
ac-va-on
barrier.

The
energies
 of
the
reactants
and
products
are
not
changed,
only
the
speed
of
the
 reac-on.
 The
diagram
above
represents
an
imaginary
two‐step
mechanism.
The
orange,

 green
and
blue
spheres
represent
elements
A,
B
&
C,
respec-vely.

 (a)  Write
the
equa-on
for
the
net
reac-on
that
is
occurring.
 (b)
Iden-fy
the
intermediate. 

 (c)
Iden-fy
the
catalyst. 
 

 The
diagram
above
represents
an
imaginary
two‐step
mechanism.
The
orange,

 green
and
blue
spheres
represent
elements
A,
B
&
C,
respec-vely.

 (a)  Write
the
equa-on
for
the
net
reac-on
that
is
occurring.
 A2+
AB+
AC
BA2
+
A
+
AC
 BA2+
A
+
AC

A2+
BA2+
C
 Net:
AB+AC
BA2+
C
 (b)
Iden-fy
the
intermediate. 
A
 (c)
Iden-fy
the
catalyst.
A2
is
the
catalyst 
 

 Heterogenous
Catalysis
 A
heterogeneous
catalyst
exists
in
a
different
phase
than
the
 reactants
and
products.
 Usually
the
catalyst
is
a
solid
and
the
reactants
and
products

 are
in
a
gas
or
liquid
phase.
 The
ini-al
step
is
usually
the
adsorp-on
of
the
reactants
to
the
 surface
of
the
catalyst.
 Example
‐‐

the
reac-on
of
hydrogen
gas
with
ethylene
gas
to
 form
ethane
gas
 
 
C2H4(g)
+
H2(g)
→
C2H6(g)
 (a)  (b)  (c)  (d)  The
hydrogen
and
ethylene
adsorb
to
a
metal
surface.
 The
H‐H
bond
is
broken
to
give
hydrogen
atoms.
 The
hydrogen
atoms
move
to
the
adsorbed
ethylene
and
react.
 Fully
formed
ethane
molecules
desorb
and
return
to
the
gas
phase.
 Enzymes
 The
rates
of
most
biochemical
reac-ons
are
highly
regulated
 by
catalysts
called
enzymes.
 Most
enzymes
are
proteins
(polymers
of
amino
acids).
 These
proteins
have
large
molar
masses
(e.g.,
100,000
gm/mol)
 and
very
well‐defined
three‐dimensional
structures.
 Enzymes
are
generally
highly
specific
for
the
reac-on
they
catalyze.
 The
ac-vi-es
of
enzymes
themselves
are
highly
regulated.
 An
enzyme
from
the
liver
called
catalase
causes
hydrogen
 peroxide
to
rapidly
decompose
into
water
and
oxygen.

 The
reac-on
catalyzed
by
an
enzyme
occurs
at
a
small
por-on
 of
the
enzyme
called
the
ac-ve
site.
 The
reactants
are
called
substrates.
 The
specificity
is
explained
in
part
by
the
lock‐and‐key
model.
 An
enzyme
called
lysozyme.
 Lysozyme
with
its
bound
substrate.
 ...
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