Chap14-Part3

Chap14-Part3 - LAST
CLASS
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Unformatted text preview: LAST
CLASS
 ‐ Rate
of
a
reac/on
 ‐ Rate
constant
of
a
reac/on
 ‐ Order
of
a
reac/on
 1
 First
Order
Reac/on
 k A→B First
order
reac2on
is
one
whose
rate
depends
on
the
concentra2on
of

a
single
 reactant
raised
to
the
first
power
 Rate
=
‐Δ[A]/Δt
=
k[A]
 Δ[A]/[A]
=
‐kΔt

(differen2al
rate
law)
 Integra2on
on
both
sides,

ln[A]t‐
ln[A]0
=‐kt
(Integrated
rate
law)

 The
concentra2on
of
A
as
a
func2on
of
2me

 [ A]t = [ A]0 exp(−kt ) or ln[ At ] = ln[ A 0] − kt Recall
that
k
has
units
of
2me‐1.
 € 2
 Problem:
The
decomposi2on
of
a
certain
insec2cide
in
water
follows
first
order
 kine2cs
with
a
rate
constant
of
1.45
yr‐1.
If
the
concentra2on
of
the
insec2cide
is
 5.0X10‐7g/cm3,
(a)
what
is
its
concentra2on
aRer

1year
(b)
how
long
will
it
take
for
 the
concentra2on
of
insec2cide
to
decrease
to
3.0X10‐7g/cm3

 3
 Problem:
The
decomposi2on
of
a
certain
insec2cide
in
water
follows
first
order
 kine2cs
with
a
rate
constant
of
1.45
yr‐1.
If
the
concentra2on
of
the
insec2cide
is
 5.0X10‐7g/cm3,
(a)
what
is
its
concentra2on
aRer

1year
(b)
how
long
will
it
take
for
 the
concentra2on
of
insec2cide
to
decrease
to
3.0X10‐7g/cm3

 (a)
k
=
1.45
yr‐1
 [A]0
=
5X10‐7
g/cm3
 ln[A]t=1
yr
=
‐
kt+
ln[A]0
 




=
‐
(1.45
yr‐1)
(1
yr)
+
ln
(5X10‐7)
 




=
‐1.45+
(‐14.51)
=
‐15.96
 
[A]t=1yr
=
exp(‐15.96)
=
1.2X10‐7
g/cm3


 4
 Problem:
The
decomposi2on
of
a
certain
insec2cide
in
water
follows
first
order
 kine2cs
with
a
rate
constant
of
1.45
yr‐1.
If
the
concentra2on
of
the
insec2cide
is
 5.0X10‐7g/cm3,
(a)
what
is
its
concentra2on
aRer

1year
(b)
how
long
will
it
take
for
 the
concentra2on
of
insec2cide
to
decrease
to
3.0X10‐7g/cm3

 (a)
k
=
1.45
yr‐1
 (b)
k
=
1.45
yr‐1
 [A]0
=
5X10‐7
g/cm3
 [A]0
=
5X10‐7
g/cm3
 ln[A]t=1
yr
=
‐
kt+
ln[A]0
 ln[A]t
=
‐
kt+
ln[A]0
 




=
‐
(1.45
yr‐1)
(1
yr)
+
ln
(5X10‐7)
 ln(3X10‐7)=
‐
(1.45
yr‐1)
(t
yr)
+
ln
(5X10‐7)
 




=
‐1.45+
(‐14.51)
=
‐15.96
 




t=
0.35
yr
 
[A]t=1yr
=
exp(‐15.96)
=
1.2X10‐7
g/cm3


 5
 First
Order
Irreversible
Reac/on:

Plots
 1.0 0.8 0.6 B(t)/A(0) B(t)/
A(0)
 .4 0 0.2 0.0 0 1 2 3 4 kt Half
Life:

Half
life
t
½
is
2me
required
for
the
concentra2on
of
a
 reactant
to
reach
one‐half
of
its
ini2al
value
 
[A]t
½

=
½
[A]0

 ln
{[A]t/[A]0}
=
‐kt



ln
{1/2}
=
‐kt
½

 t
½
=
0.693/k
:

The
half
life
does
not
depend
on
the
ini2al
 concentra2on
in
first
order
reac2on
 6
 First
Order
Irreversible
Reac/on:

Plots
 ln[A]t=
‐kt
+
ln[A]0
(Integrated
rate
law)
 Y







=
mX
+
C

 Slope=
‐k
(
k
is
posi2ve)
 [A]
 ln[A]
 t
 t
 7
 Example:
First
Order
Reac/on
 CH3NC
 Pressure
is
a
 measurement
of
 the
concentra2on
 CH3CN
 Gas‐phase
isomeriza2on
 Slope
=
‐5.1X10‐5
s‐1

k=
5.1X10‐5
s‐1

 8
 Half‐Time
 ln
[CH3NC]t
=
‐kt
+
ln
[CH3NC]0
 The
half
life
does
not
depend
on
the
ini2al
concentra2on
in
a
first
order
 9
 reac2on
 First‐Order
Reac/on:

Example
 Americium‐241
is
used
in
smoke
detectors.

It
has
a
first‐order
rate
constant
for
 radioac2ve
decay
of
k
=
1.6
x
10‐3
yr‐1.


 (a)
What
is
the
half‐life?
 (b)
How
much
of
a
1
mg
sample
remains
aRer
three
half‐lives?
 10
 First‐Order
Reac/on:

Example
 Americium‐241
is
used
in
smoke
detectors.

It
has
a
first‐order
rate
constant
for
 radioac2ve
decay
of
k
=
1.6
x
10‐3
yr‐1.


 (a)
What
is
the
half‐life?
 t
½
=
0.693/k
=
0.693/
1.6X10‐3
yr‐1
=
433
yrs

 (b)
How
much
of
a
1
mg
sample
remains
aRer
three
half‐lives?
 11
 First‐Order
Reac/on:

Example
 Americium‐241
is
used
in
smoke
detectors.

It
has
a
first‐order
rate
constant
for
 radioac2ve
decay
of
k
=
1.6
x
10‐3
yr‐1.


 (a)
What
is
the
half‐life?
 t
½
=
0.693/k
=
0.693/
1.6X10‐3
yr‐1
=
433
yrs

 (b)
How
much
of
a
1
mg
sample
remains
aRer
three
half‐lives?
 1)
433
yrs
 2)
433
yrs
 3)
433
yrs
 
1
mg

‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐>
0.5
mg
‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐>
0.25mg
‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐>
0.125mg
 12
 Second
Order
Reac/on
 Second
order
reac2on
is
one
whose
rate
depends
on
the
concentra2on
of

a
 single
reactant
raised
to
the
second
power
 k 2A → B Or








A+
B

Products
 Rate
=
‐Δ[A]/Δt
=
k[A]2
 Δ[A]/[A]2
=
‐kΔt

(differen2al
rate
law)
 Integra2on
on
both
sides,
1/[A]t‐
1/[A]0
=
kt
(Integrated
rate
law)

 The
concentra2on
of
A
as
a
func2on
of
2me
is
 [ A]t = [ A]0 1 + kt[ A]0 or 1 1 = + kt [ A]t [ A]0 Recall
that
k
has
units
of
concentra2on‐12me‐1.
 € 13
 Second
Order
Irreversible
Reac/on:

Plots
 1.0 half‐2me
for
reac2on:
 0.8 0.6 B(t)/
A(0)
 B(t)/A(0) [ A]0 [ A]t = 1 + kt[ A]0 [ A]0 1 / 2[ A]0 = ⇒ (1 + kt1 / 2[ A]0) = 2 1 + kt1/ 2[ A]0 0.4 0.2 0.0 0 1 2 3 kA(0)t t
½
=

1/
k[A]0
 The
half
life
depends
on
the
ini2al
concentra2on
in
a
second
order
 14
 reac2on
 4 Second
Order
Irreversible
Reac/on:

Plots
 1/[A]t
=
kt
+
1/[A]0
(Integrated
rate
law)
 Y







=
mX
+
C

 1/[A]
 [A]
 Slope=
k
 t
 t
 15
 Second‐Order
Processes
 The
decomposi2on
of
NO2
at
300°C
is
described
by
the
 equa2on
 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 16
 Second‐Order
Processes
 •  Plolng
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 17
 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]
 18
 Sucrose
(a
disaccharide)
reacts
in
acidic
solu2ons
to
form
two
 monosaccharides
(glucose
&
fructose).

At
23oC
in
0.5
M
HCl,

 the
following
data
were
obtained
for
the
disappearance
of
sucrose:
 2me
(min) 
 
0 
 
39 
 
80 
 
140 
210 
 
 
 
 
 
sucrose
(M)
 
0.316
 
0.274
 
0.238
 
0.190
 
0.146
 Is
the
reac2on
first
or
second
order
with
respect
to
sucrose?
 19
 Sucrose
(a
disaccharide)
reacts
in
acidic
solu2ons
to
form
two
 monosaccharides
(glucose
&
fructose).

At
23oC
in
0.5
M
HCl,

 the
following
data
were
obtained
for
the
disappearance
of
sucrose:
 2me
(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
reac2on
first
or
second
order
with
respect
to
sucrose?
First
order
 20
 Temperature
Dependence
of
Reac/on
Rates
 Rate
constants
k
increase
with
temperature.
 Example:

A
chemiluminescent
reac2on.
 At
the
higher
temperature
the
reac2on

 proceeds
faster.


 21
 Temperature
Dependence
of
Reac/on
Rates
 CH3NC
 CH3CN
 gas‐phase
isomeriza2on
 22
 A
first
order
reac2on
is
examined
 for
three
different
condi2ons.
 (a)
Which
two
lines
represent
data
 obtained
at
the
same
temperature?
 (b)
Which
two
lines
represent
data
 obtained
with
the
same
ini2al
condi2on?
 (c)
Which
line
represents
data
taken
 at
the
lower
temperature?
 23
 A
first
order
reac2on
is
examined
 for
three
different
condi2ons.
 (a)
Which
two
lines
represent
data
 obtained
at
the
same
temperature?
 Line
1
and
2:
Same
slope
 Started
at
2
different
ini2al
concentra2ons
 (b)
Which
two
lines
represent
data
 obtained
with
the
same
ini2al
condi2on?
 (c)
Which
line
represents
data
taken
 at
the
lower
temperature?
 24
 A
first
order
reac2on
is
examined
 for
three
different
condi2ons.
 (a)
Which
two
lines
represent
data
 obtained
at
the
same
temperature?
 Line
1
and
2:
Same
slope
 Started
at
2
different
ini2al
concentra2ons
 (b)
Which
two
lines
represent
data
 obtained
with
the
same
ini2al
condi2on?
 Line
2
and
3:
Same
ini2al
concentra2on
 But
2
different
temperatures
 (c)
Which
line
represents
data
taken
 at
the
lower
temperature?
 25
 A
first
order
reac2on
is
examined
 for
three
different
condi2ons.
 (a)
Which
two
lines
represent
data
 obtained
at
the
same
temperature?
 Line
1
and
2:
Same
slope
 Started
at
2
different
ini2al
concentra2ons
 (b)
Which
two
lines
represent
data
 obtained
with
the
same
ini2al
condi2on?
 Line
2
and
3:
Same
ini2al
concentra2on
 But
2
different
temperatures
 (c)
Which
line
represents
data
taken
 at
the
lower
temperature?
 Line
3:
Smaller
slope
 26
 ...
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