Chemical Kinetics PDF PowerPoint Presentation

1. 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

2. Rate Data for A + B → C
time (seconds) Concentration A Concentration B, Concentration C,
mol/L mol/L mol/L
0 0.76 0.38 0
1 0.31 0.16 0.20
2 0.13 6.5 x 10-2 0.40
3 5.2 x 10-2 2.6 x 10-2 0.58
4 2.1 x 10-2 1.1 x 10-2 0.73
5 8.8 x 10-3 4.4 x 10-3 0.86
6 3.6 x 10-3 1.8 x 10-3 0.95
7 1.4 x 10-3 7.0 x 10-4 1.02
8 6.1 x 10-4 3.1 x 10-4 1.07
9 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

3. A + B→C
1.2
C
1
0.8
Concentration (mol/L)
0.6
A
0.4 B
0.2
0
0 2 4 6 8 10 12
time (seconds)

4. The Rate of a Chemical Reaction
• The speed of a reaction can be examined by the
decrease in reactants or the increase in products.
• aA +bB →cC + d D
Rate = k  A  B
m n
Where m and n are determined experimentally, and not necessarily
Equal to the stiochiometry of the reaction

5. Reaction A → 2 B
A A A B B B B
B B B B
A A A
B B B B
A A A B B B B
A B B B B
A = 6.022 x 1022 molecules B = 6.022 x 1022 molecules
in a 1.00 L container in a 1.00 Liter container
1 mol/L 2 mol/L
Δ  A 1 Δ  B
- =
t 2 t

6. Average Rate
Δ A
• Rate of A disappearing is - t
• Let’s suppose that after 20 seconds ½ half of A
disappears.
• Then
Δ  A A f  Ai 0.50 mol/L - 1.00 mol/L mol
- =- =- = - 2.5 x 10-2 M/s or - 2.5 x 10-2
t tf  ti 20 s - 0 s L-s
1 Δ  B
• And Rate of B appearing is
2 t
• Then
1 Δ  B Bf -Bi 1 1.00 mol/L - 0.00 mol/L mol
= = x = 2.5 x 10-2 M/s or 2.5 x 10-2
2 t tf - ti 2 20 s - 0 s L-s

7. Average Rate Law for the General Equation
aA+bB→cC + dD
1 Δ  A 1 Δ  B 1 Δ C 1 Δ  D
- x = - x = x = x
a Δt b Δt c Δt d Δt
For Example:
N2O5 (g) → 2 NO2 (g) + ½ O2 (g)
Δ  N 2O5 (g) 
  1 Δ  NO2 (g) 
  = 2 x Δ O2 (g) 
 
- = x
Δt 2 Δt Δt

8. 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.

9. Consider the Reaction:
CH3CH2CH2CH2Cl (aq) + H2O (l) → CH3CH2CH2CH2OH (aq) + HCl (aq)
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

10. mol
Average Rates, L-s
time, seconds [CH3CH2CH2CH2Cl] Average Rate, mol
L-s
[C4 H9Cl]

t
0.0 0.1000 1.90 x 10-4
50.0 0.0905
time, seconds [CH3CH2CH2CH2Cl] Average Rate, mol
L-s
[C4 H9Cl]

t
50.0 0.0905 1.70 x 10-4
100.0 0.0820

11. mol
Average Rates, L-s
time, seconds [CH3CH2CH2CH2Cl] Average Rate, mol
L-s
[C4 H9Cl]

t
100.0 0.0820 1.58 x 10-4
150.0 0.0741
time, seconds [CH3CH2CH2CH2Cl] Average Rate, mol
L-s
[C4 H9Cl]

t
150.0 0.0741 1.74 x 10-4
200.0 0.0671

12. mol
Average Rates, L-s
time, seconds [CH3CH2CH2CH2Cl] Average Rate, mol
L-s
[C4 H9Cl]

t
200.0 0.0671 1.22 x 10-4
300.0 0.0549
time, seconds [CH3CH2CH2CH2Cl] Average Rate, mol
L-s
[C4 H9Cl]

t
300.0 0.0549 1.01 x 10-4
400.0 0.0448

13. mol
Average Rates, L-s
time, seconds [CH3CH2CH2CH2Cl] Average Rate, mol
L-s
[C4 H9Cl]

t
400.0 0.0448 8.00 x 10-5
500.0 0.0368
time, seconds [CH3CH2CH2CH2Cl] Average Rate, mol
L-s
[C4 H9Cl]

t
500.0 0.0368 5.60 x 10-5
800.0 0.0200

14. 0.12
0.1
0.08
Concentration (mol/L)
Instantaneous Rate or initial rate at t=0 s
0.06
Instantaneous Rate at t = 500 s
0.04
0.02
0
0 100 200 300 400 500 600 700 800 900
time (seconds)

15. 0.10 M - 0.060 M
Instaneous Rateat 0 s =
190 s - 0 s
0.040 M
Instaneous Rateat 0 s = = 2.1 x 10-4 M s
190 s
0.042 M - 0.020 M
Instaneous Rateat 500 s =
800 s - 400 s
0.022 M
Instaneous Rateat 500 s = = 5.5 x 10-5 M s
400 s

16. 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

17. Data from the hydrolysis of n-butyl chloride
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

18. IF Zero Order
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

19. 60
50
40
[n-butyl chloride]
30
20
10
0
0 100 200 300 400 500 600 700 800 900
time (seconds)
Therefore, the reaction is not zero order

20. 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

21. 60
50
40
1/[n-butyl chloride]
30
20
10
0
0 100 200 300 400 500 600 700 800 900
time (seconds)
Therefore, the reaction is not second order

22. IF First Order Reaction
time log [C4H9Cl] ln[C4H9Cl]
(seconds)
0 -1 -2.3
50 -1.04 -2.4
100 -1.09 -2.51
150 -1.13 -2.60
200 -1.17 -2.69
300 -1.26 -2.90
400 -1.35 -3.11
500 -1.43 -3.29
800 -1.7 -3.92

23. First Order Plot
0
0 100 200 300 400 500 600 700 800 900
-0.2
-0.4
-0.6
log [n-butyl chloride]
-0.8
-1
-1.2 y = -0.0009x - 0.9985
-1.4
-1.6
-1.8
time (seconds)

24. First Order Plot
0
0 100 200 300 400 500 600 700 800 900
-0.5
-1
-1.5
ln[C4H9Cl]
-2
-2.5
-3 y = -0.002x - 2.2987
-3.5
-4
-4.5
time (seconds)

25. Slope
k
slope = -
2.303
-2.303 x slope = k
-2.303 x - 9.0 x 10-4 = k
2.1 x 10-3 = k

26. Slope
slope = - k
3 1
- (- 2 x 10 s ) = k
2 x 103 s 1 = k

27. Rate of the Reaction
Rate = k [n-butylchloride]

28. For the Reaction
N2O5 (g) → 2 NO2 (g) + ½ O2 (g)
Rate = k [N2O5 ]
The rate can be used to explain the mechanism

29. (1) Slow Step
.. .. ..
.. ..
:O O: :O ..
+ O
.. +
slow .+
.O
.. + O:
N N N + N
:O : :O : :O :
.. .. .. :O :
- - ..
- -
NO2 NO3
N2O5
(2) Fast Step
.. ..
.. :O
.O + O:
.. fast .+ 1/2 O2
N N +
:O : :O :
.. ..
- -
NO3 NO2

30. Sum of the two steps:
.. .. ..
.. :O :O
:O
+ O: .+
+ O
.. .+ N
N N N + + 1/2 O2
:O : :O :
:O : :O : .. ..
.. .. -
- - -
NO2 NO2
N2O5
N2O5 → 2 NO2 + ½ O2
or
2 N2O5 → 4 NO2 + O2

31. Application
Mechanism of a Chemical Reaction
(a)
Suggest a possible mechanism for
NO2 (g) + CO (g) → NO (g) + CO2 (g)
Given that
Rate = k [NO2(g) ]2
(b)
Suggest a possible mechanism for
2 NO2 (g) + F2 (g) → 2 NO2F (g)
Given that
Rate = k [NO2 (g) ] [F2 (g) ]

32. Factors Affecting the Rate of a Chemical
Reaction
• The Physical State of Matter
• The Concentration of the Reactants
• Temperature
• Catalyst

33. 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

34. A Second Order Reaction
H2O2 (aq) + I (aq)  H2O (l) + O2 (g)
-
-
Rate = k [H2O2(aq) ] [I(aq) ]

35. Rate Constant “k”
• Must be determined experimentally
• Its value allows one to find the reaction rate
for a new set of concentrations

36. 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

37. From Experiments 1 and 2
Rate = k  A  B
m n
Solution A:
(1) 5.5 x 10-6 M/s = k 0.1000 M 0.1000 M
m n
k 0.2000 M 0.1000 M
m n
(2) 2.2 x 10-5 M/s =
Divide equation (1) into equation (2)
k  0.2000 M  0.1000 M 
-5 m n
2.2 x 10 M/s
=
k  0.1000 M   0.1000 M 
-6 m n
5.5 x 10 M/s
4   2
m
2=m

38. Solution B:
(1) log (5.5 x 10-6 ) = log k + m log 0.1000 + n log 0.1000
(2) log (2.2 x 10-5 ) = log k + m log 0.2000 + n log 0.1000
Subtract equation (2) from equation (1)
log (5.5 x 10-6 ) -log (2.2 x 10-5 ) = m [log  0.1000 -log  0.2000]
-5.3 - (-4.7) = m [-1 - (-0.7)]
-0.6 = m [-0.3]
-0.6
=m
-0.3
2=m

39. From Experiments 4 and 5
Solution A:
(1) 1.65 x 10 M/s = k 0.1000 M 0.3000 M
-5 m n
(2) 3.3 x 10-5 M/s = k 0.1000 M 0.6000 M
m n
Divide equation (1) into equation (2)
k  0.1000 M  0.6000 M 
-5 m n
3.3 x 10 M/s
=
k  0.1000 M   0.3000 M 
-6 m n
1.65 x 10 M/s
2   2
n
1=n

40. Solution B:
(1) log (1.65 x 10-5 ) = log k + m log 0.1000 + n log 0.3000
(2) log (3.3 x 10-5 ) = log k + m log 0.1000 + n log 0.6000
Subtract equation (2) from equation (1)
log (1.65 x 10-5 ) -log (3.3 x 10-5 ) = n [log  0.3000 -log  0.6000]
-4.78 - (-4.5) = n [-0.5227 - (-0.2218)]
-0.3 = n [-0.3]
-0.3
=n
-0.3
1=n

41. Rate Constant k
Rate = k  A  B
m n
Rate = k  A  B
2 1
Rate
=k
 A   B
2

42. Rate Constant k
From Experiment 3
Rate
=k
 A   B
2
M
8.800 x 10-5
s =k
0.4000 M  0.1000 M 
2
1-3
5.500 x 10 2
=k
M s
L2
5.500 x 10-3 2
=k
mol s

43. Rate Constant k
From Experiment 1
Rate
=k
 A   B
2
M
5.500 x 10-5
s =k
0.1000 M  0.1000 M 
2
-3 1
5.500 x 10 2
=k
M s
L2
5.500 x 10-3 2
=k
mol s

44. Your Understanding of this Process
Consider the Data for the Following Reaction:
O O
_
CH3 C + OH CH3 C + CH3OH
_
OCH3 O
Experiment [CH3CO2CH3] [-OH] Initial Rate,
M M M/s
1 0.050 0.050 0.00034
2 0.050 0.100 0.00069
3 0.100 0.100 0.00137
Determine the Rate Law Expression and the value of k consistent
With these data.

45. From Experiments 1 and 2
n
Rate = k [CH3CO2CH3 ] m
 - OH 
 
Solution :
(1) 3.4 x 10 M/s = k 0.050 M 0.050 M
-4 m n
(2) 6.9 x 10-4 M/s = k 0.50 M 0.100 M
m n
Divide equation (1) into equation (2)
k  0.050 M  0.100 M 
-4 m n
6.9 x 10 M/s
=
k  0.050 M   0.050 M 
-5 m n
3.4 x 10 M/s
2   2
n
1=n

46. From Experiments 2 and 3
Solution :
(1) 6.9 x 10 M/s = k 0.050 M 0.050 M
-4 m n
(2) 1.37 x 10-3 M/s = k 0.100 M 0.100 M
m n
Divide equation (1) into equation (2)
k  0.100 M  0.100 M 
-3 m n
1.37 x 10 M/s
=
k  0.050 M   0.100 M 
-5 m n
6.9 x 10 M/s
2   2
m
1=m

47. Rate Expression
Rate = k [CH3CO2CH3 ] [ - OH]
Rate
-
=k
[CH3CO2CH3 ] [ OH]
M
0.00137
s =k
[0.100 M] [0.100 M]
1
0.137 =k
Ms
L
0.137 =k
mol s

48. Assignment
Determine the Rate Law for the following reaction from the
given data:
2 NO (g) + O2 (g) → 2 NO2 (g)
Experiment [NO (g)] [O2 (g)] Initial Rate,
M M M/s
1 0.020 0.010 0.028
2 0.020 0.020 0.057
3 0.020 0.040 0.114
4 0.040 0.020 0.227
5 0.010 0.020 0.014

49. Relationship Between Concentration and Time
A  product
First Order Reaction
ao  x x
dx
= k (a o -x)
dt -ln (a o -x) = kt + C
dx at x = o and t = o, C = -ln a o
= k dt
a o -x -ln (a o -x) = kt - ln a o
dx
 a o -x = k  dt ln a o - ln (a o -x) = kt
ao
let s = a o -x ln = kt
a o -x
ds = -dx or
ds
 = k  dt log
ao
=
kt
s a o -x 2.303
-ln s = kt + C

50. A plot of
ao
ln versus t
a o -x
Gives a Straight line
The Following Reaction is a First Order Reaction:
H H H
H C H H C H
C C C
 H C
H H H H
Plot the linear graph for concentration versus time
and obtain the rate constant for the reaction.

51. Data for the Transformation of cylcpropane to propene
ao X ao –x Ln[ao]/[ ao –x] t
M M M seconds
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

52. 0.2
0.18
y = 2E-05x - 4E-17
0.16
ln (ao /(ao – x))
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
0 2000 4000 6000 8000 10000 12000 14000
-0.02
time (seconds)
slope = 2 x 10-5 s-1

53. Data for the Transformation of cylcpropane to propene
ao X ao –x Ln [ ao –x] t
M M M seconds
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

54. -2.98
0 2000 4000 6000 8000 10000 12000 14000
-3
-3.02
-3.04
-3.06
y = -2E-05x - 2.9957
Ln (ao – x)
-3.08
-3.1
-3.12
-3.14
-3.16
-3.18
-3.2
time (seconds)
-slope = -2 x 10-5s-1
Slope = 2 x 10-5 s-1

55. Relationship Between Concentration and Time
A  product
Second Order Reaction
ao  x x
dx
= k (a o -x) 2
dt 1
dx = kt + C
2
= k dt a o -x
(a o -x)
dx
1
at x = o and t = o, C =
 (a o -x)2 = k  dt ao
let s = a o -x 1 1
= kt +
ds = -dx a o -x ao
ds
 = k  dt 1 1
s 2
- = kt
1 a o -x a o
= kt + C
s

56. A plot of
1
versus t
a o -x
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 time
and obtain the rate constant for the reaction.

57. Data for the Transformation of hydrogen iodide gas to hydrogen and iodine
ao X ao –x 1/[ ao –x] t
M M M M-1 Minutes
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

58. 2000
1800
1600
1400
y = 30x + 100
1/(ao – x)
1200
1000
800
600
400
200
0
0 10 20 30 40 50 60 70
time (minutes)
slope = 30. L mol-1 min-1

59. 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 = 0
or y = ao – x ; x = t; slope = -k and the intercept = ao

60. Zero Order Reaction
dx a o -x t
 dx =  dt
=k
dt
dx = k dt ao 0
 dx = k  dt a o - (a o -x) = kt
x = kt + C - (a o -x) = kt - a o
at t = 0 and x =0; C = 0
(a o -x) = - kt + a o
x = kt

61. Application of the Graphical Method for Determining
the Order of a Reaction
N2O5 (g) → 2 NO2 (g) + ½ O2 (g)
[ N 2O 5 ] t
M minutes
2.08 3.07
1.67 8.77
1.36 14.45
0.72 31.28
Tabulate the data so that each order may be tested

62. Data tabulation to determine which order will give a linear graph
[ N2O5 ] t ln[ N2O5 ] 1/[ N2O5 ]
M minutes (first order) M-1
(zero order) (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

63. Test for zero order reaction
2.5
2
1.5
[N2O5]
1
0.5
0
0 5 10 15 20 25 30 35
time (minutes)
Not linear; therefore, the reaction is not zero order

64. Test for first order reaction
0.8
0.6
y = -0.0376x + 0.8462
0.4
ln[N2O5]
0.2
0
0 5 10 15 20 25 30 35
-0.2
-0.4
time (minutes)
Linear; therefore, the reaction is first order

65. Test for second order reaction
1.6
1.4
1.2
1
1/[N2O5]
0.8
0.6
0.4
0.2
0
0 5 10 15 20 25 30 35
time (minutes)
Non-linear; therefore, the reaction is not second order

66. Half Life for a First Order Reaction
ao
ln = kt 1
1 2
x ao
2
ln 2= k t 1
2
0.693 = k t 1
2
0.693
= t1
k 2

67. Half Life for a Second Order Reaction
1 1
- = kt 1
1
ao ao
2
2
2 1
- = kt 1
ao ao 2
1
= kt 1
ao 2
1
= t1
k ao 2

68. Half Life for a Zero Order Reaction
1
a o = - kt 1 + a o
2 2
1
a o - a o = - kt 1
2 2
1
 a o = - kt 1
2 2
ao
= t1
2k 2

69. 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.

70. Half-Life Fraction of cyclopropane Fraction of cyclopropane
Remaining after 18.0 hours Remaining after 51.5 hours
ao ao
0.693 ln = kt ln = kt
= t1 a o -x a o -x
0.054/h 2
a o -x a o -x
12.8 h = t 1 ln = - kt ln = - kt
2
ao ao
a o -x a o -x
= e- kt = e- kt
ao ao
a o -x a o -x
= e- (0.054/h) x 18.0 h = e- (0.054/h) x 51.5 h
ao ao
a o -x a o -x
= e- 0.972 = e- 2.8
ao ao
a o -x a o -x
= 0.38 = 0.061
ao ao

71. Effect of Temperature on the Reaction Rate
Arrhenius Equation
Eact
-
RT
k=Ae
Eact
ln k = - + ln A
RT
of the form y = mx + b

72. Use the Following Data to Determine the Eact
for
2 N2O (g)  2 N2 (g) + O2 (g)
T k Ln k 104(1/T)
K M-1/s
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

73. 5
0
0 2 4 6 8 10 12 14 16 18 20
-5
ln k
-10
-15
-20
y = -3.0712x + 29.721
-25
-30
104(1/T)
E act
slope = -3.07 x 104 K = -
R
-3.07 x 104 K x - R = E act
J
-3.07 x 104 K x - 8.314 = E act
K mol
2.55 x 105 J/mol = E act
255 kJ/mol = Eact

74. 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

75. Example of a Homogeneous Catalyst
1 1
1. H 2O2 (aq) + Br-

(aq) Br2 (aq) + H 2O (l) + O 2 (g)
2 2
1 1
2. H 2O2 (aq) + Br2 (aq)  Br(aq) + H 2O (l) + O 2 (g)
-
2 2

76. PE
intermediate
reactants
products
reaction coordinates

77. Example of Heterogeneous Catalyst
H H
H H C C
H H
Finely divided metal

78. 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.
.. H H
O:
C C
CH3CH2 C + H C N: CH3CH2 C N : + :N C CH2CH3
H
:OH
.. HO :
..

79. 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
∞

80. Check to determine first order in HCN
-2.35
0 2 4 6 8 10 12 14 16 18
-2.4
-2.45
-2.5
ln([HCN]-x)
-2.55
-2.6
-2.65
-2.7
-2.75
time (minutes)

81. Check to determine first order in propionaldehyde
0
0 2 4 6 8 10 12 14 16 18
-0.5
time (minutes)
-1
ln ([propionaldehyde] -x)
-1.5
-2
-2.5
-3
-3.5
-4

82. So close; therefore, let’s take another approach. Let [HCN] = ao and
[propionaldehyde] = bo
Then,
dx
= k (a o -x) (b o -x)
dt
dx
= k dt
(a o -x) (b o -x)
dx
 (a o -x) (bo -x) = k  dt
Solution:
1 (a o -x) 1 bo
ln = kt - ln
(a o -bo ) b o -x (a o -b o ) a o

83. 1 (a o -x) 1 0.0566
ln = kt - ln
0.0424 M b o -x 0.0424 M 0.0990
-1 (a o -x)
23.6 M ln = kt - 23.6 M -1 ln 0.572
bo -x
-1 (a o -x)
23.6 M ln = kt - 23.6 M -1 (-0.559)
bo -x
-1 (a o -x)
23.6 M ln = kt + 13.2 M -1
bo -x

84. 1 (a o -x) 1 0.0566
ln = kt - ln
0.0424 M b o -x 0.0424 M 0.0990
-1 (a o -x)
23.6 M ln = kt - 23.6 M -1 ln 0.572
bo -x
(a o -x)
23.6 M -1 ln = kt - 23.6 M -1 (-0.559)
bo -x
(a o -x)
23.6 M -1 ln = kt + 13.2 M -1
bo -x

85. Let’s construct the data in a different format
time, minutes [HCN] - x [CH3CH2CHO]-x ([HCN]-x)
(23.6) ln
([CH3CH2 CHO]-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

86. 30
25
20
([HCN] - x) y = 0.6778x + 13.184
23.6 ln
([CH3CH 2CHO]-x)
15
10
5
0
0 2 4 6 8 10 12 14 16 18
time, minutes
Slope = 0.678; therefore, k = 0.678 M-1min-1

87. Rate = k [HCN] [CH3CH2CHO]
Mechanism:
k1
+ -
1. HCN + H2O H3O + CN
k -1
+
O OH
k2
2. + + H2O
CH3CH2 C + H3O CH3CH2 C
k-2
H H
+ H
3. OH
k3
- C
CH3CH2 C + CN
CH3CH2 CN
H HO

88. Steps 1 and 2 are fast equilibrium steps; and step 3 is the rate determining step
+
OH
Rate = k3 CH3CH2 [ CN- ]
C
H
k 1 [HCN] [H2O] = k -1 [H3O+] [CN-]
k 1 [HCN] [H2O]
= [CN-]
k -1 [H3O+]

89. O
CH3CH2 C +
[H3O ]
k +
2
H OH
= CH3CH2 C
k [H2O] H
-2
O
CH3CH2 C
+
k3 k1 k2 [HCN] [H2O] H [H3O ]
Rate =
+
k -1 k [H3O ] [H2O]
-2

90. O
Rate = k [HCN] CH3CH2 C
H

91. Revisit the kinetics for
2 NO (g) + O2 (g) → 2 NO2 (g)
Rate = k [NO]2 [O2 ]