Transcript chemical kinetics 3

First-Order Processes
ln [A]t = -kt + ln [A]0
Therefore, if a reaction is first-order, a
plot of ln [A] vs. t will yield a straight
line, and the slope of the line will be -k.
Chemical
Kinetics
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Ex. 14.7
The decomposition of certain insecticide in water
follows first- order kinetics with rate constant of
1.45 /y at 12 c . A quantity of this insecticide is
washed into alake on june 1, leading to
concentration of 5.0 X 10-7 g/cm3. assume that
the average tempreture of the lake is 12
A- what is the concentration of the insecticide on
june of the following year
B- how long will it take for the concentration of the
insecicide to decrease to 3.0 x 10 -7 g/ cm3
Chemical
Kinetics
First-Order Processes
Consider the process in
which methyl isonitrile is
converted to acetonitrile.
CH3NC
CH3CN
Chemical
Kinetics
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First-Order Processes
CH3NC
CH3CN
This data was
collected for this
reaction at 198.9
°C.
Chemical
Kinetics
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First-Order Processes
• When ln P is plotted as a function of time, a
straight line results.
• Therefore,
– The process is first-order.
– k is the negative of the slope: 5.1  10-5 s−1.
Chemical
Kinetics
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• * The reaction 2A → B is first order in A with a rate costant of
2.8x10-2 s -1 at 80 0 C. How long will it take for A to decrease from
0.88 to 0.14 M , then calculate half life time (s)
• Solve: ln[A] = ln[A]0 - kt , ln 0.14 = ln 0.88 - 2.8x10-2 x t
• t = ln 0.14 – ln 0.88 / 2.8x10-2 = 66 s
• Half life time ( t1/2 ) = 0.693/ k = 0.693 / 2.8x10-2 = 24.74 s
..
..
•
• * Given : CH3 - N= N – CH3 (g) → N2(g) + C2H6 (g)
• Time
PN=N-CH3 mmHg
---------------------------• -------------0
284
ln Pt = -kt + ln P0
100
220
ln 220 = -k x 100 + ln 284
150
193
5.394 = - k x 100 + 5.649
200
170
k = - ( 5.394-5.649) /100
______________
2.55x10-3 s-1
Chemical
Kinetics
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Second-Order Processes
Similarly, integrating the rate law for a
process that is second-order in reactant
A, we get
1
1
= kt +
[A]t
[A]0
also in the form
y = mx + b
Chemical
Kinetics
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Second-Order Processes
1
1
= kt +
[A]t
[A]0
So if a process is second-order in A, a
1
plot of [A] vs. t will yield a straight line,
and the slope of that line is k.
Zero order reaction :
Rate ( - ∆[A] / ∆t ) = k [A]0 = k
[A] = [A]0 – k t
,
t1/2 = [A]0 /2 k
Chemical
Kinetics
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Second-Order Processes
The decomposition of NO2 at 300°C is described by
the equation
1
NO2 (g)
NO (g) + 2 O2 (g)
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
Chemical
Kinetics
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Second-Order Processes
• Plotting 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
Chemical
Kinetics
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Second-Order Processes
1
[NO2]
• Graphing ln
vs. t, however,
gives this plot.
Time (s)
[NO2], M
1/[NO2]
0.0
0.01000
100
50.0
0.00787
127
100.0
0.00649
154
200.0
0.00481
208
300.0
0.00380
263
• Because this is a
straight line, the
process is secondorder in [A].
Chemical
Kinetics
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Half-Life
• Half-life is defined
as the time required
for one-half of a
reactant to react.
• Because [A] at t1/2 is
one-half of the
original [A],
[A]t = 0.5 [A]0.
Chemical
Kinetics
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Half-Life
For a first-order process, this becomes
0.5 [A]0
ln
=
−kt
1/2
[A]0
ln 0.5 = −kt1/2
−0.693 = −kt1/2
NOTE: For a first-order
process, then, the half-life
does not depend on [A]0.
0.693
= t1/2
k
Chemical
Kinetics
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Half-Life
For a second-order process,
1
1
= kt1/2 +
0.5 [A]0
[A]0
2
1
= kt1/2 +
[A]0
[A]0
2 − 1 = 1 = kt
1/2
[A]
[A]0
0
1
= t1/2
k[A]0
Chemical
Kinetics
© 2009, Prentice-Hall, Inc.
• Ex 14.9 The reaction of C4H9Cl with
water is first order reaction . Figure shows
How the concentration changes with time at
particular temp.
A- from the graph estamate the t1/2
B- use t1/2 and calculate the rate constant
Chemical
Kinetics