Chap14-Part4

Chap14-Part4 - Second‐Order
Processes
...

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Unformatted text preview: Second‐Order
Processes
 The
decomposi,on
of
NO2
at
300°C
is
described
by
the
 equa,on
 1
 NO2
(g)
 NO
(g)
+





O2
(g)
 2
 and
yields
data
comparable
to
this:
 Time (s) [NO2], M 0.0 0.01000 50.0 0.00787 100.0 0.00649 200.0 0.00481 300.0 0.00380 2
 Second‐Order
Processes
 •  PloDng
ln
[NO2]
vs.
t
yields
the
 graph
at
the
right.
 •  The
plot
is
not
a
straight
line,
 so
the
process
is
not
first‐ order
in
[A].
 Time (s) [NO2], M ln [NO2] 0.0 0.01000 −4.610 50.0 0.00787 −4.845 100.0 0.00649 −5.038 200.0 0.00481 −5.337 300.0 0.00380 −5.573 3
 Second‐Order
Processes
 •  Time (s) [NO2], M 1/[NO2] 0.0 0.01000 100 50.0 0.00787 127 100.0 0.00649 0.00481 0.00380 263 •  Because
this
is
a
 straight
line,
the
 process
is
second‐ order
in
[A].
 208 300.0 however,
gives
this
 plot.
 154 200.0 1
 Graphing








vs.
t,
 [NO2]
 4
 Sucrose
(a
disaccharide)
reacts
in
acidic
solu,ons
to
form
two
 monosaccharides
(glucose
&
fructose).

At
23oC
in
0.5
M
HCl,

 the
following
data
were
obtained
for
the
disappearance
of
sucrose:
 ,me
(min) 
 
0 
 
39 
 
80 
 
140 
210 
 
 
 
 
 
sucrose
(M)
 
0.316
 
0.274
 
0.238
 
0.190
 
0.146
 Is
the
reac,on
first
or
second
order
with
respect
to
sucrose?
 5
 Sucrose
(a
disaccharide)
reacts
in
acidic
solu,ons
to
form
two
 monosaccharides
(glucose
&
fructose).

At
23oC
in
0.5
M
HCl,

 the
following
data
were
obtained
for
the
disappearance
of
sucrose:
 ,me
(min) 
 
0 
 
39 
 
80 
 
140 
210 
 
 
 
 
 
sucrose
(M)


ln[sucrose]



1/[sucrose]
 
0.316 
0.274 
0.238 
0.190 
0.146 
 
 
 
 
 
 
 
 
 
 
‐1.152 
 
‐1.295 
 
‐1.435 
 
‐1.661 
 
‐1924 
 
 
 
 
 
 
3.16 
 
3.65 
 
4.20
 
5.26
 
6.85
 
 

 

 Is
the
reac,on
first
or
second
order
with
respect
to
sucrose?
First
order
 6
 Temperature
Dependence
of
Reac;on
Rates
 Rate
constants
k
increase
with
temperature.
 Example:

A
chemiluminescent
reac,on.
 At
the
higher
temperature
the
reac,on

 proceeds
faster.


 7
 Temperature
Dependence
of
Reac;on
Rates
 CH3NC
 CH3CN
 gas‐phase
isomeriza,on
 8
 A
first
order
reac,on
is
examined
 for
three
different
condi,ons.
 (a)
Which
two
lines
represent
data
 obtained
at
the
same
temperature?
 (b)
Which
two
lines
represent
data
 obtained
with
the
same
ini,al
condi,on?
 (c)
Which
line
represents
data
taken
 at
the
lower
temperature?
 9
 A
first
order
reac,on
is
examined
 for
three
different
condi,ons.
 (a)
Which
two
lines
represent
data
 obtained
at
the
same
temperature?
 Line
1
and
2:
Same
slope
 Started
at
2
different
ini,al
concentra,ons
 (b)
Which
two
lines
represent
data
 obtained
with
the
same
ini,al
condi,on?
 (c)
Which
line
represents
data
taken
 at
the
lower
temperature?
 10
 A
first
order
reac,on
is
examined
 for
three
different
condi,ons.
 (a)
Which
two
lines
represent
data
 obtained
at
the
same
temperature?
 Line
1
and
2:
Same
slope
 Started
at
2
different
ini,al
concentra,ons
 (b)
Which
two
lines
represent
data
 obtained
with
the
same
ini,al
condi,on?
 Line
2
and
3:
Same
ini,al
concentra,on
 But
2
different
temperatures
 (c)
Which
line
represents
data
taken
 at
the
lower
temperature?
 11
 A
first
order
reac,on
is
examined
 for
three
different
condi,ons.
 (a)
Which
two
lines
represent
data
 obtained
at
the
same
temperature?
 Line
1
and
2:
Same
slope
 Started
at
2
different
ini,al
concentra,ons
 (b)
Which
two
lines
represent
data
 obtained
with
the
same
ini,al
condi,on?
 Line
2
and
3:
Same
ini,al
concentra,on
 But
2
different
temperatures
 (c)
Which
line
represents
data
taken
 at
the
lower
temperature?
 Line
3:
Smaller
slope
 12
 The
Collision
Model
 For
bimolecular
or
higher
order
reac,ons,
the
reactants
must
 collide
for
the
reac,on
to
proceed.
 Higher
temperatures
lead
to
more
frequent
collisions.
 Higher
reactant
concentra,ons
lead
to
more
frequent
collisions.
 Higher
temperatures
lead
to
more
forceful
collisions.
 All
three
of
these
factors
give
higher
reac,on
rates.
 For
collisions
to
be
successful,
the
reactants
must
be
properly
 oriented.
 
Example: 
Cl
+
NOCl
→
Cl2
+
NO
 effec,ve

 ineffec,ve
 Properly
oriented
molecular
collisions
are
important
for
a
reac,on
 Ac;va;on
Energy
 For
a
reac,on
to
proceed,
the
colliding
molecules
must
have
a
total
 kine,c
energy
greater
than
an
energy
called
the
ac,va,on
energy
=
Ea.
 The
value
of
Ea
is
different
for
different
reac,ons.
 The
value
of
Ea
describes
the
ac,va,on
barrier
that
must
be
overcome
 for
the
reac,on
to
proceed.
 Example:

Methyl
Isonitrile
Gas‐Phase
Isomeriza,on
 CH3NC
 CH3CN
 


“intermediate”
 or
“ac,vated
complex”
 or
“transi,on
state”
 Reac,on
Coordinate
Diagrams
 
It
is
helpful
to
 visualize
energy
 changes
throughout
 a
process
on
a
 reac,on
coordinate
 diagram
like
this
one
 for
the
 rearrangement
of
 methyl
isonitrile.
 ©
2009,
Pren,ce‐Hall,
Inc.
 Reac,on
Coordinate
Diagrams
 •  The
diagram
shows
the
 energy
of
the
reactants
 and
products
(and,
 therefore,
ΔE).
 •  The
high
point
on
the
 diagram
is
the
transi,on
 state.
 •  The
species
present
at
the
transi,on
state
is
called
 the
ac,vated
complex.
 •  The
energy
gap
between
the
reactants
and
the
ac,vated
 complex
is
the
ac,va,on
energy
barrier.
 ©
2009,
Pren,ce‐Hall,
Inc.
 The
bond
between
the
H3C
group
and
the
N≡C
group
must
be
stretched
by
the
 incoming
energy,
allowing
the
N≡C
group
to
rotate.

Amerwards,
the
C‐C
bond
 begins
to
form
and
the
energy
drops.

Ea
is
the
ac,va,on
energy
and
ΔE
is
the
 energy
between
the
ini,al
and
final
states.

The
rate
is
determined
by
Ea.
 Maxwell–Boltzmann
Distribu,ons
 •  Temperature
is
 defined
as
a
 measure
of
the
 average
kine,c
 energy
of
the
 molecules
in
a
 sample.
 •  At
any
temperature
there
is
a
wide
distribu,on
 of
kine,c
energies.
 ©
2009,
Pren,ce‐Hall,
Inc.
 Maxwell–Boltzmann
Distribu,ons
 •  As
the
temperature
 increases,
the
curve
 flaqens
and
 broadens.
 •  Thus
at
higher
 temperatures,
a
 larger
popula,on
of
 molecules
has
higher
 energy.
 ©
2009,
Pren,ce‐Hall,
Inc.
 Maxwell–Boltzmann
Distribu,ons
 •  If
the
doqed
line
represents
the
ac,va,on
energy,
 then
as
the
temperature
increases,
so
does
the
 frac,on
of
molecules
that
can
overcome
the
 ac,va,on
energy
barrier.
 • As
a
result,
the
 reac,on
rate
 increases.
 ©
2009,
Pren,ce‐Hall,
Inc.
 Maxwell–Boltzmann
Distribu,ons
 This
frac,on
of
molecules
that
has
an
energy
equal
to
or
greater
than
 Ea
is
given
by,
 f
=
e
 ‐Ea
 RT
 where
R
is
the
gas
constant
and
T
is
the
Kelvin
temperature.
 ©
2009,
Pren,ce‐Hall,
Inc.
 The
ac,va,on
energy
for
the
isomeriza,on
of
methyl
isonitrile
 is
160
kJ/mol.
 a)
Calculate
the
frac,on
of
molecules
with
an
energy
>
Ea
at
500
K.
 b)
Calculate
the
frac,on
of
molecules
with
an
energy
>
Ea
at
510
K.
 c)
Find
the
ra,o.
 The
ac,va,on
energy
for
the
isomeriza,on
of
methyl
isonitrile
 is
160
kJ/mol.
 a)
Calculate
the
frac,on
of
molecules
with
an
energy
>
Ea
at
500
K.
 f
=
e
 ‐Ea
 RT
 Ea
=
160
kJ/mol
=
1.6X105
J/mol
,
T=500K
R=
8.314J/mol‐K

 ‐Ea/RT
=
160000/(500X8.314)
=‐38.5

 f
=
e‐38.5
=

2X10‐17
 b)
Calculate
the
frac,on
of
molecules
with
an
energy
>
Ea
at
510
K.
 Ea
=
160
kJ/mol
=
1.6X105
J/mol
,
T=510K
R=
8.314J/mol‐K

 ‐Ea/RT
=
160000/(510X8.314)
=‐37.7

 f
=
e‐38.5
=

4X10‐17
 c)
Find
the
ra,o.
 f
at
510
/f
at
500
=
2
 An
Increase
of
10K
means
that
twice
more
molecules
having
this
 energy.
 Arrhenius
Equa,on
 
Svante
Arrhenius
developed
a
mathema,cal
 rela,onship
between
k
and
Ea:
 k
=
A
e
 ‐Ea
 RT
 
where
A
is
the
frequency
factor,
a
number
that
 represents
the
likelihood
that
collisions
would
 occur
with
the
proper
orienta,on
for
reac,on.
 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
Ac;va;on
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 ...
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