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
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
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
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
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
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
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
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.
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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]
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]
Slope
kslope = -
2.303
-2.303 x slope = k
-4
-3
-2.303 x - 9.0 x 10 = k
2.1 x 10 = k
Slope
slope = - k
3 1
3 1
- (- 2 x 10 s ) = k
2 x 10 s = k
Rate of the Reaction
Rate = k [n-butylchloride]
For the ReactionN2O5 (g) → 2 NO2 (g) + ½ O2 (g)
2 5Rate = k [N O ]
The rate can be used to explain the mechanism
(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
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
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 ]
Factors Affecting the Rate of a Chemical Reaction
• The Physical State of Matter• The Concentration of the Reactants• Temperature• Catalyst
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
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 ]
Rate Constant “k”
• Must be determined experimentally• Its value allows one to find the reaction rate
for a new set of concentrations
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
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
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
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
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
Rate Constant k
m nRate = k A B
2 1Rate = k A B
2
Rate = k
A B
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
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
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
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
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
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
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)
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
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.
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
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))
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
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
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
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
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
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
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
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
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
[ 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
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]
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]
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]
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
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
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
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.
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
Effect of Temperature on the Reaction Rate
Arrhenius Equation
actE- RTk = A e
actEln k = - + ln A
RT
of the form y = mx + b
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
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
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
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
reaction coordinates
PE
intermediate
reactants
products
Example of Heterogeneous Catalyst
H H C C
H H
HH
Finely divided metal
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
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
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
)
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)
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
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
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
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)
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)
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
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-]
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
Rate = [HCN] CH3CH2 C
O
H
k
Revisit the kinetics for
2 NO (g) + O2 (g) → 2 NO2 (g)
22Rate = k [NO] [O ]