Chemical Kinetics Chapter 17 Chemical Kinetics Aka Reaction Rates.
GC Chemical Kinetics
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Transcript of GC Chemical Kinetics
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Chemical Kinetics
• Study of speed with which a chemical reaction occurs and the factors affecting that speed
• Provides information about the feasibility of a chemical reaction
• Provides information about the time it takes for a chemical reaction to occur
• Provides information about the series of elementary steps which lead to the formation of product
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time (seconds) Concentration A mol/L
Concentration B, mol/L
Concentration C, mol/L
0 0.76 0.38 01 0.31 0.16 0.202 0.13 6.5 x 10-2 0.403 5.2 x 10-2 2.6 x 10-2 0.584 2.1 x 10-2 1.1 x 10-2 0.735 8.8 x 10-3 4.4 x 10-3 0.866 3.6 x 10-3 1.8 x 10-3 0.957 1.4 x 10-3 7.0 x 10-4 1.028 6.1 x 10-4 3.1 x 10-4 1.079 2.5 x 10-4 1.3 x 10-4 1.07
10 1.0 x 10-4 5.0 x 10-5 1.07
Rate Data for A + B → C
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A + B → C
0 2 4 6 8 10 120
0.2
0.4
0.6
0.8
1
1.2
time (seconds)
Conc
entr
ation
(mol
/L)
A
B
C
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The Rate of a Chemical Reaction
• The speed of a reaction can be examined by the decrease in reactants or the increase in products.
• a A + b B → c C + d D
m nRate = k A B
Where m and n are determined experimentally, and not necessarilyEqual to the stiochiometry of the reaction
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Reaction A → 2 B
A A A
A A
A A A
A
A
B B
B B B
B B
B B
B B
B
B B
B B
B B
BB
1 mol/L 2 mol/L
A = 6.022 x 1022 molecules B 6.022 x 1022 molecules=
in a 1.00 L container in a 1.00 Liter container
Δ A Δ B1- =
t 2 t
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Average Rate
• Rate of A disappearing is• Let’s suppose that after 20 seconds ½ half of A
disappears.• Then
• And Rate of B appearing is • Then
Δ A-
t
Δ B1
2 t
-2 -2f i
f i
Δ B B -B1 1 1.00 mol/L - 0.00 mol/L mol = = x = 2.5 x 10 M/s or 2.5 x 10
2 t t - t 2 20 s - 0 s L-s
-2 -2f i
f i
Δ A A A 0.50 mol/L - 1.00 mol/L mol- = - = - = - 2.5 x 10 M/s or - 2.5 x 10
t t t 20 s - 0 s L-s
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Average Rate Law for the General Equationa A + b B → c C + d D
Δ A Δ B Δ C Δ D1 1 1 1- x = - x = x = x
a Δt b Δt c Δt d Δt
For Example:N2O5 (g) → 2 NO2 (g) + ½ O2 (g)
2 5 (g) 2 (g) 2 (g)Δ N O Δ NO Δ O1- = x = 2 x
Δt 2 Δt Δt
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Determination of the Rate Equation
• Determined Experimentally• Can be obtained by examining the initial rate
after about 1% or 2% of the limiting reagent has been consumed.
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Consider the Reaction:CH3CH2CH2CH2Cl (aq) + H2O (l) → CH3CH2CH2CH2OH (aq) + HCl (aq)
time (seconds) Concentration n-butyl chloride
mol/L
0 0.1050 9.05 x 10-2
100 8.2 x 10-2
150 7.41 x 10-2
200 6.71 x 10-2
300 5.49 x 10-2
400 4.48 x 10-2
500 3.68 x 10-2
800 2.00 x 10-2
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Average Rates, mol
L-s
time, seconds [CH3CH2CH2CH2Cl] Average Rate,
0.0 0.1000 1.90 x 10-4
50.0 0.0905
mol
L-s
4 9[C H Cl]
t
time, seconds [CH3CH2CH2CH2Cl] Average Rate,
50.0 0.0905 1.70 x 10-4
100.0 0.0820
mol
L-s
4 9[C H Cl]
t
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Average Rates, mol
L-s
time, seconds [CH3CH2CH2CH2Cl] Average Rate,
100.0 0.0820 1.58 x 10-4
150.0 0.0741
mol
L-s
4 9[C H Cl]
t
time, seconds [CH3CH2CH2CH2Cl] Average Rate,
150.0 0.0741 1.74 x 10-4
200.0 0.0671
mol
L-s
4 9[C H Cl]
t
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Average Rates, mol
L-s
time, seconds [CH3CH2CH2CH2Cl] Average Rate,
200.0 0.0671 1.22 x 10-4
300.0 0.0549
mol
L-s
4 9[C H Cl]
t
time, seconds [CH3CH2CH2CH2Cl] Average Rate,
300.0 0.0549 1.01 x 10-4
400.0 0.0448
mol
L-s
4 9[C H Cl]
t
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Average Rates, mol
L-s
time, seconds [CH3CH2CH2CH2Cl] Average Rate,
400.0 0.0448 8.00 x 10-5
500.0 0.0368
mol
L-s
4 9[C H Cl]
t
time, seconds [CH3CH2CH2CH2Cl] Average Rate,
500.0 0.0368 5.60 x 10-5
800.0 0.0200
mol
L-s
4 9[C H Cl]
t
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0 100 200 300 400 500 600 700 800 9000
0.02
0.04
0.06
0.08
0.1
0.12
time (seconds)
Conc
entr
ation
(mol
/L)
Instantaneous Rate or initial rate at t=0 s
Instantaneous Rate at t = 500 s
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at 0 s
-4at 0 s
0.10 M - 0.060 MInstaneous Rate =
190 s - 0 s0.040 M
Instaneous Rate = = 2.1 x 10 M s190 s
at 500 s
-5at 500 s
0.042 M - 0.020 MInstaneous Rate =
800 s - 400 s0.022 M
Instaneous Rate = = 5.5 x 10 M s400 s
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Order of Reaction
• Zero order –independent of the concentration of the reactants, e.g, depends on light
• First order - depends on a step in the mechanism that is unimolecular
• Pseudo first order reaction – one of the reactants in the rate determining step is the solvent
• Second order – depends on a step in the mechanism that is bimolecular
• Rarely third order – depends on the step in the mechanism that is termolecular
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time (seconds) Concentration n-butyl chloride
mol/L
0 0.10
50 9.05 x 10-2
100 8.2 x 10-2
150 7.41 x 10-2
200 6.71 x 10-2
300 5.49 x 10-2
400 4.48 x 10-2
500 3.68 x 10-2
800 2.00 x 10-2
Data from the hydrolysis of n-butyl chloride
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time (seconds) [C4H9Cl]
0 0.10
50 9.05 x 10-2
100 8.2 x 10-2
150 7.41 x 10-2
200 6.71 x 10-2
300 5.49 x 10-2
400 4.48 x 10-2
500 3.68 x 10-2
800 2.00 x 10-2
IF Zero Order
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0 100 200 300 400 500 600 700 800 9000
10
20
30
40
50
60
time (seconds)
[n-b
utyl
chlo
ride]
Therefore, the reaction is not zero order
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If Second Order
time (seconds) 1/[C4H9Cl]
0 10
50 11.0
100 12.2
150 13.5
200 14.9
300 18.2
400 22.3
500 27.2
800 50
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0 100 200 300 400 500 600 700 800 9000
10
20
30
40
50
60
time (seconds)
1/[n
-but
yl ch
lorid
e]
Therefore, the reaction is not second order
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time (seconds)
log [C4H9Cl] ln[C4H9Cl]
0 -1 -2.350 -1.04 -2.4
100 -1.09 -2.51150 -1.13 -2.60200 -1.17 -2.69300 -1.26 -2.90400 -1.35 -3.11500 -1.43 -3.29800 -1.7 -3.92
IF First Order Reaction
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First Order Plot
0 100 200 300 400 500 600 700 800 900
-1.8
-1.6
-1.4
-1.2
-1
-0.8
-0.6
-0.4
-0.2
0
f(x) = − 0.000873532550693703 x − 0.99846318036286
time (seconds)
log
[n-b
utyl
chlo
ride]
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First Order Plot
0 100 200 300 400 500 600 700 800 900
-4.5
-4
-3.5
-3
-2.5
-2
-1.5
-1
-0.5
0
f(x) = − 0.00201664887940235 x − 2.29870864461046
time (seconds)
ln[C
4H9C
l]
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Slope
kslope = -
2.303
-2.303 x slope = k
-4
-3
-2.303 x - 9.0 x 10 = k
2.1 x 10 = k
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Slope
slope = - k
3 1
3 1
- (- 2 x 10 s ) = k
2 x 10 s = k
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Rate of the Reaction
Rate = k [n-butylchloride]
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For the ReactionN2O5 (g) → 2 NO2 (g) + ½ O2 (g)
2 5Rate = k [N O ]
The rate can be used to explain the mechanism
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(1) Slow Step
(2) Fast Step
NO
N
OO
O O....
: :.. ..
: : : :.. ..- -
slow
--.... ::::
....::
..
.. OO
O O
NO
N .+ +
++
+
N2O5NO2
NO3
-..
::
..:O
O
N .++
NO2NO3
+ON
O
O....
:..
: :..-
fast 1/2 O2
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Sum of the two steps:
NO
N
OO
O O....
: :.. ..
: : : :.. ..- - -
..::
..:O
O
N .+ + ++
N2O5NO2
1/2 O2
+.N
O
O:
: :..-
NO2
+
N2O5 → 2 NO2 + ½ O2
or2 N2O5 → 4 NO2 + O2
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ApplicationMechanism of a Chemical Reaction
Suggest a possible mechanism forNO2 (g) + CO (g) → NO (g) + CO2 (g)
Given that 2
2(g)Rate = k [NO ]
(a)
(b)
Suggest a possible mechanism for2 NO2 (g) + F2 (g) → 2 NO2F (g)
Given that
2 (g) 2 (g)Rate = k [NO ] [F ]
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Factors Affecting the Rate of a Chemical Reaction
• The Physical State of Matter• The Concentration of the Reactants• Temperature• Catalyst
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For A Reaction to Occur
• Molecules Must Collide• Molecules must have the Appropriate
Orientation• Molecules must have sufficient energy to
overcome the energy barrier to the reaction-• Bonds must break and bonds must form
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A Second Order Reaction
-2 2 (aq) (aq) 2 (l) 2 (g)H O + I H O + O
-2 2(aq) (aq)Rate = k [H O ] [I ]
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Rate Constant “k”
• Must be determined experimentally• Its value allows one to find the reaction rate
for a new set of concentrations
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The following data were collected for the rate of the reaction Between A and B, A + B → C , at 25oC. Determine the rate law for the reaction and calculate k.
Experiment [A], moles/L [B], moles/L Initial Rate, M/s
1 0.1000 0.1000 5.500 x 10-6
2 0.2000 0.1000 2.200 x 10-5
3 0.4000 0.1000 8.800 x 10-5
4 0.1000 0.3000 1.650 x 10-5
5 0.1000 0.6000 3.300 x 10-5
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Solution A:
From Experiments 1 and 2
m nRate = k A B
m n-6(1) 5.5 x 10 M/s = k 0.1000 M 0.1000 M
m n-5(2) 2.2 x 10 M/s = k 0.2000 M 0.1000 M
Divide equation (1) into equation (2)
m n-5
m n-6
m
k 0.2000 M 0.1000 M2.2 x 10 M/s =
5.5 x 10 M/s k 0.1000 M 0.1000 M
4 2
2 = m
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Solution B:
Subtract equation (2) from equation (1)
-6 -5log (5.5 x 10 ) -log (2.2 x 10 ) = m [log 0.1000 -log 0.2000 ]
-5.3 - (-4.7) = m [-1 - (-0.7)]
-0.6 = m [-0.3]
-0.6 = m
-0.32 = m
-6(1) log (5.5 x 10 ) = log k + m log 0.1000 + n log 0.1000
-5(2) log (2.2 x 10 ) = log k + m log 0.2000 + n log 0.1000
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Solution A:
From Experiments 4 and 5
Divide equation (1) into equation (2)
m n-5(1) 1.65 x 10 M/s = k 0.1000 M 0.3000 M
m n-5(2) 3.3 x 10 M/s = k 0.1000 M 0.6000 M
m n-5
m n-6
n
k 0.1000 M 0.6000 M3.3 x 10 M/s =
1.65 x 10 M/s k 0.1000 M 0.3000 M
2 2
1 = n
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Solution B:
Subtract equation (2) from equation (1)
-5(1) log (1.65 x 10 ) = log k + m log 0.1000 + n log 0.3000
-5(2) log (3.3 x 10 ) = log k + m log 0.1000 + n log 0.6000
-5 -5log (1.65 x 10 ) -log (3.3 x 10 ) = m [log 0.3000 -log 0.6000 ]
-4.78 - (-4.5) = m [-0.5227 - (-0.2218)]
-0.3 = m [-0.3]
-0.3 = m
-0.31 = m
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Rate Constant k
m nRate = k A B
2 1Rate = k A B
2
Rate = k
A B
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Rate Constant kFrom Experiment 3
2
Rate = k
A B
-5
2
M8.800 x 10
s = k 0.4000 M 0.1000 M
-32
2-3
2
15.500 x 10 = k
M s
L5.500 x 10 = k
mol s
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Rate Constant kFrom Experiment 1
2
Rate = k
A B
-32
2-3
2
15.500 x 10 = k
M s
L5.500 x 10 = k
mol s
-5
2
M5.500 x 10
s = k 0.1000 M 0.1000 M
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Your Understanding of this ProcessConsider the Data for the Following Reaction:
Experiment [CH3CO2CH3]M
[-OH]M
Initial Rate, M/s
1 0.050 0.050 0.000342 0.050 0.100 0.000693 0.100 0.100 0.00137
Determine the Rate Law Expression and the value of k consistentWith these data.
CH3 C
O
OCH3
+ OH_
CH3 C
O_
O
+ CH3OH
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Solution :
From Experiments 1 and 2
Divide equation (1) into equation (2)
nm -3 2 3Rate = k [CH CO CH ] OH
m n-4(1) 3.4 x 10 M/s = k 0.050 M 0.050 M
m n-4(2) 6.9 x 10 M/s = k 0.50 M 0.100 M
m n-4
m n-5
n
k 0.050 M 0.100 M6.9 x 10 M/s =
3.4 x 10 M/s k 0.050 M 0.050 M
2 2
1 = n
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Solution :
From Experiments 2 and 3
Divide equation (1) into equation (2)
m n-4(1) 6.9 x 10 M/s = k 0.050 M 0.050 M
m n-3(2) 1.37 x 10 M/s = k 0.100 M 0.100 M
m n-3
m n-5
m
k 0.100 M 0.100 M1.37 x 10 M/s =
6.9 x 10 M/s k 0.050 M 0.100 M
2 2
1 = m
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Rate Expression
-3 2 3Rate = k [CH CO CH ] [ OH]
-3 2 3
Rate = k
[CH CO CH ] [ OH]
M0.00137
s = k[0.100 M] [0.100 M]
10.137 = k
M sL
0.137 = kmol s
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AssignmentDetermine the Rate Law for the following reaction from the given data:
Experiment [NO (g)]M
[O2 (g)]M
Initial Rate, M/s
1 0.020 0.010 0.0282 0.020 0.020 0.0573 0.020 0.040 0.1144 0.040 0.020 0.2275 0.010 0.020 0.014
2 NO (g) + O2 (g) → 2 NO2 (g)
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Relationship Between Concentration and Time
First Order Reactiono
A product
a xx
o
o
o o
o o
o
o
o
o
-ln (a -x) = kt + C
at x = o and t = o, C = -ln a
-ln (a -x) = kt - ln a
ln a - ln (a -x) = kt
aln = kt
a -x
or
a ktlog =
a -x 2.303
o
o
o
o
dx = k (a -x)
dtdx
= k dta -x
dx = k dt
a -x
let s = a -x
ds = -dx
ds = k dt
s-ln s = kt + C
![Page 50: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/50.jpg)
A plot of o
o
aln versus t
a -x
Gives a Straight line
The Following Reaction is a First Order Reaction:
C
C C
H
H
H H
H
H
C
CC
H
H
H
H
H
H
Plot the linear graph for concentration versus timeand obtain the rate constant for the reaction.
![Page 51: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/51.jpg)
Data for the Transformation of cylcpropane to propene
ao M
XM
ao –xM
Ln[ao]/[ ao –x] tseconds
0.050 0 0.050 0 0
0.050 0.0004 0.0496 9.0 x 10-3 600
0.050 0.0009 0.0491 0.0180 1200
0.050 0.0015 0.0485 0.0300 2000
0.050 0.0022 0.0478 0.045 3000
0.050 0.0036 0.0464 0.075 5000
0.050 0.0057 0.0443 0.120 8000
0.050 0.0070 0.0430 0.150 10000
0.050 0.0082 0.0418 0.180 12000
![Page 52: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/52.jpg)
slope = 2 x 10-5 s-1
0 2000 4000 6000 8000 10000 12000 140000
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0.2
f(x) = 0.000015 x
time (seconds)
ln (a
o /(
ao –
x))
![Page 53: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/53.jpg)
Data for the Transformation of cylcpropane to propene
ao M
XM
ao –xM
Ln [ ao –x] tseconds
0.050 0 0.050 -2.996 0
0.050 0.0004 0.0496 -3.0038 600
0.050 0.0009 0.0491 -3.014 1200
0.050 0.0015 0.0485 -3.026 2000
0.050 0.0022 0.0478 -3.0407 3000
0.050 0.0036 0.0464 -3.070 5000
0.050 0.0057 0.0443 -3.1167 8000
0.050 0.0070 0.0430 -3.1466 10000
0.050 0.0082 0.0418 -3.1748 12000
![Page 54: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/54.jpg)
0 2000 4000 6000 8000 10000 12000 14000
-3.2
-3.15
-3.1
-3.05
-3
-2.95
-2.9
f(x) = − 1.50112367675131E-05 x − 2.99568114479088
time (seconds)
Ln (a
o –
x)
-slope = -2 x 10-5s-1
Slope = 2 x 10-5 s-1
![Page 55: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/55.jpg)
Relationship Between Concentration and Time
Second Order Reactiono
A product
a xx
2o
2o
2o
o
2
dx = k (a -x)
dtdx
= k dt(a -x)
dx = k dt
(a -x)
let s = a -x
ds = -dx
ds = k dt
s1
= kt + Cs
o
o
o o
o o
1 = kt + C
a -x
1at x = o and t = o, C =
a
1 1 = kt +
a -x a
1 1 - = kt
a -x a
![Page 56: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/56.jpg)
A plot of
Gives a Straight line
The Following Reaction is a Second Order Reaction:
2 HI (g) → H2 (g) + I2 (g)
Plot the linear graph for concentration versus timeand obtain the rate constant for the reaction.
o
1 versus t
a -x
![Page 57: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/57.jpg)
ao
MXM
ao –xM
1/[ ao –x]M-1
tMinutes
0.0100 0 0.0100 100 0.00
0.0100 0.0060 0.00400 250 5.00
0.0100 0.0075 0.00250 400 10.0
0.0100 0.0086 0.00143 700 20.0
0.0100 0.0090 0.0010 1000 30.0
0.0100 0.0099 0.00077 1300 40.0
0.0100 0.0094 0.00063 1600 50.0
0.0100 0.0095 0.00053 1900 60.0
Data for the Transformation of hydrogen iodide gas to hydrogen and iodine
![Page 58: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/58.jpg)
0 10 20 30 40 50 60 700
200
400
600
800
1000
1200
1400
1600
1800
2000
f(x) = 30 x + 100
time (minutes)
1/(a
o –
x)
slope = 30. L mol-1 min-1
![Page 59: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/59.jpg)
Graphical Method for Determining the Order of a Reaction
• First Order Reaction: y = ln (ao – x); x = t; slope = -k; and the intercept is ln ao
or y = ln (ao /(ao – x)); x = t; slope = k; and the intercept =0
• Second Order Reaction: y = 1/ (ao – x); x = t; slope = k; and the intercept = 1/ ao
• Zero Order Reaction: y = x; x = t; slope = k and the intercept = 0or y = ao – x ; x = t; slope = -k and the intercept = ao
![Page 60: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/60.jpg)
Zero Order Reaction
dx = k
dtdx = k dt
dx = k dt
x = kt + C
at t = 0 and x =0; C = 0
x = kt
o
o
a -x t
a 0
o o
o o
o o
dx = dt
a - (a -x) = kt
- (a -x) = kt - a
(a -x) = - kt + a
![Page 61: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/61.jpg)
Application of the Graphical Method for Determining the Order of a Reaction
N2O5 (g) → 2 NO2 (g) + ½ O2 (g)
[ N2O5 ]M
tminutes
2.08 3.071.67 8.771.36 14.450.72 31.28
Tabulate the data so that each order may be tested
![Page 62: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/62.jpg)
[ N2O5 ]M
(zero order)
tminutes
ln[ N2O5 ](first order)
1/[ N2O5 ]M-1
(second order)2.08 3.07 0.732 0.481
1.67 8.77 0.513 0.599
1.36 14.45 0.307 0.735
0.72 31.28 -0.329 1.390
Data tabulation to determine which order will give a linear graph
![Page 63: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/63.jpg)
Test for zero order reaction
Not linear; therefore, the reaction is not zero order
0 5 10 15 20 25 30 350
0.5
1
1.5
2
2.5
time (minutes)
[N2O
5]
![Page 64: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/64.jpg)
Test for first order reaction
Linear; therefore, the reaction is first order
0 5 10 15 20 25 30 35
-0.4
-0.2
0
0.2
0.4
0.6
0.8
f(x) = − 0.0375520327162226 x + 0.846217630868234
time (minutes)
ln[N
2O5]
![Page 65: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/65.jpg)
Test for second order reaction
Non-linear; therefore, the reaction is not second order
0 5 10 15 20 25 30 350
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
time (minutes)
1/[N
2O5]
![Page 66: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/66.jpg)
Half Life for a First Order Reaction
o1
2
o
12
12
12
aln = kt
1 x a
2ln 2= k t
0.693 = k t
0.693 = t
k
![Page 67: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/67.jpg)
Half Life for a Second Order Reaction
12
oo
12
o o
12
o
12
o
1 1 - = kt
1 aa2
2 1 - = kt a a
1 = kt
a
1 = t
k a
![Page 68: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/68.jpg)
Half Life for a Zero Order Reaction
o 1 o2
o o 12
o 12
o1
2
1 a = - kt + a
21
a - a = - kt 2
1 a = - kt
2a
= t 2 k
![Page 69: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/69.jpg)
Application of Half Life
• The rate constant for transforming cyclopropane into propene is 0.054 h-1
• Calculate the half-life of cyclopropane.• Calculate the fraction of cyclopropane
remaining after 18.0 hours.• Calculate the fraction of cyclopropane
remaining after 51.5 hours.
![Page 70: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/70.jpg)
12
12
0.693 = t
0.054/h12.8 h = t
Half-Life
o
o
o
o
- kto
o
- (0.054/h) x 51.5 ho
o
- 2.8o
o
o
o
aln = kt
a -x
a -xln = - kt
a
a -x= e
a
a -x= e
a
a -x= e
a
a -x= 0.061
a
Fraction of cyclopropaneRemaining after 18.0 hours
Fraction of cyclopropaneRemaining after 51.5 hours
o
o
o
o
- kto
o
- (0.054/h) x 18.0 ho
o
- 0.972o
o
o
o
aln = kt
a -x
a -xln = - kt
a
a -x= e
a
a -x= e
a
a -x= e
a
a -x= 0.38
a
![Page 71: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/71.jpg)
Effect of Temperature on the Reaction Rate
Arrhenius Equation
actE- RTk = A e
actEln k = - + ln A
RT
of the form y = mx + b
![Page 72: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/72.jpg)
Use the Following Data to Determine the Eact
for
2 (g) 2 (g) 2 (g)2 N O 2 N + O
TK
kM-1/s
Ln k 104(1/T)
1125 11.5900 2.450 8.890
1053 1.6700 0.510 9.50
1001 0.3800 -0.968 9.99
838 0.0010 -6.810 11.9
![Page 73: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/73.jpg)
8.5 9 9.5 10 10.5 11 11.5 12 12.5
-8
-6
-4
-2
0
2
4
f(x) = − 3.07115033064349 x + 29.7212674570598
104(1/T)
ln k
act255 kJ/mol = E
4 act
4act
4act
5act
Eslope = -3.07 x 10 K = -
R
-3.07 x 10 K x - R = E
J-3.07 x 10 K x - 8.314 = E
K mol
2.55 x 10 J/mol = E
![Page 74: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/74.jpg)
Effect of a Catalyst on the Rate of a Reaction
• Lowers the energy barrier to the reaction via lowering the energy of activation
• Homogeneous catalyst- in the same phase as the reacting molecules
• Herterogeneous catalyst – in a different phase from the reacting molecules
![Page 75: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/75.jpg)
Example of a Homogeneous Catalyst
-2 2 (aq) (aq) 2 (aq) 2 (l) 2 (g)
-2 2 (aq) 2 (aq) (aq) 2 (l) 2 (g)
1 11. H O + Br Br + H O + O
2 21 1
2. H O + Br Br + H O + O2 2
![Page 76: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/76.jpg)
reaction coordinates
PE
intermediate
reactants
products
![Page 77: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/77.jpg)
Example of Heterogeneous Catalyst
H H C C
H H
HH
Finely divided metal
![Page 78: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/78.jpg)
An interesting problem:
The reaction between propionaldehyde and hydrocyanic acid have been observed by Svirbely and Roth and reported in the Journal of the American Chemical Society. Use this data to ascertain the order of the reaction and the value of the rate constant for this reaction.
C
H
O
+ C
OHC N
H
::
..
H C N :
:..
+C
H
HON C:
..:
CH3CH2 CH3CH2 CH2CH3
![Page 79: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/79.jpg)
time, minutes [HCN] [CH3CH2CHO]
2.78 0.0990 0.0566
5.33 0.0906 0.0482
8.17 0.0830 0.0406
15.23 0.0706 0.0282
19.80 0.0653 0.0229
∞ 0.0424 0.0000
![Page 80: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/80.jpg)
Check to determine first order in HCN
0 2 4 6 8 10 12 14 16 18
-2.8
-2.7
-2.6
-2.5
-2.4
-2.3
-2.2
time (minutes)
ln([H
CN]-x
)
![Page 81: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/81.jpg)
Check to determine first order in propionaldehyde
0 2 4 6 8 10 12 14 16 18
-4
-3.5
-3
-2.5
-2
-1.5
-1
-0.5
0
time (minutes)
ln ([
prop
iona
ldeh
yde]
-x)
![Page 82: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/82.jpg)
So close; therefore, let’s take another approach. Let [HCN] = ao and [propionaldehyde] = bo
Then,
o o
o o
o o
o o
o o o o o o
dx = k (a -x) (b -x)
dtdx
= k dt(a -x) (b -x)
dx = k dt
(a -x) (b -x)
Solution:
(a -x) b1 1 ln = kt - ln
(a -b ) b -x (a -b ) a
![Page 83: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/83.jpg)
o
o
-1 -1o
o
-1 -1o
o
-1 -1o
o
(a -x)1 1 0.0566 ln = kt - ln
0.0424 M b -x 0.0424 M 0.0990
(a -x)23.6 M ln = kt - 23.6 M ln 0.572
b -x
(a -x)23.6 M ln = kt - 23.6 M (-0.559)
b -x
(a -x)23.6 M ln = kt + 13.2 M
b -x
![Page 84: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/84.jpg)
o
o
-1 -1o
o
-1 -1o
o
-1 -1o
o
(a -x)1 1 0.0566 ln = kt - ln
0.0424 M b -x 0.0424 M 0.0990
(a -x)23.6 M ln = kt - 23.6 M ln 0.572
b -x
(a -x)23.6 M ln = kt - 23.6 M (-0.559)
b -x
(a -x)23.6 M ln = kt + 13.2 M
b -x
![Page 85: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/85.jpg)
time, minutes [HCN] - x [CH3CH2CHO]-x
2.55 0.0906 0.0482 14.9
5.39 0.0830 0.0406 16.9
12.76 0.0706 0.0282 21.7
17.02 0.0653 0.0229 24.8
Let’s construct the data in a different format
3 2
([HCN]-x)(23.6) ln
([CH CH CHO]-x)
![Page 86: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/86.jpg)
0 2 4 6 8 10 12 14 16 180
5
10
15
20
25
30
f(x) = 0.67777080987964 x + 13.183621262835
time, minutes
Slope = 0.678; therefore, k = 0.678 M-1min-1
3 2
([HCN] - x)23.6 ln
([CH CH CHO]-x)
![Page 87: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/87.jpg)
3 2Rate = k [HCN] [CH CH CHO]
Mechanism:
HCN + H2Ok
k
1
2
-1
H3O+ + CN
-1.
CH3CH2 C
O
H
2. + H3O+
k
k-2
CH3CH2 C
OH
H
+
+ H2O
3.
CH3CH2 C
H
OH+
+ CN- C
H
HO
CH3CH2 CN
k3
![Page 88: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/88.jpg)
Steps 1 and 2 are fast equilibrium steps; and step 3 is the rate determining step
Rate = CH3CH2 C
OH
H
+
[ CN- ]k3
k 1 [HCN] [H2O] = k -1 [H3O+] [CN
-]
k 1 [HCN] [H2O]=
k-1 [H3O+]
[CN-]
![Page 89: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/89.jpg)
CH3CH2 C
OH
H
+CH3CH2 C
O
H
[H3O+]
k2
-2[H2O]k
=
Rate = k3 k1 [HCN] [H2O]
k -1
[H3O+]
CH3CH2 C
O
H
[H3O+]
k2
-2 [H2O]k
![Page 90: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/90.jpg)
Rate = [HCN] CH3CH2 C
O
H
k
![Page 91: GC Chemical Kinetics](https://reader033.fdocuments.us/reader033/viewer/2022061119/546a71a0af7959842c8b6742/html5/thumbnails/91.jpg)
Revisit the kinetics for
2 NO (g) + O2 (g) → 2 NO2 (g)
22Rate = k [NO] [O ]