Chemical Kinetics
Texts: Atkins, 8th edtn., chaps. 22, 23 & 24
Specialist: “Reaction Kinetics” Pilling & Seakins (1995)
Revision
 Photochemical Kinetics
 Photolytic activation, flash photolysis
 Fast reactions
 Theories of reaction rates

– Simple collision theory
– Transition state theory
1
Overview of kinetics

2
Qualitative description
– rate, order, rate law, rate constant, molecularity, elementary,
complex, temperature dependence, steady-state, ...

Reaction dynamics
– H (2S) + ICl (v, J) HI (v´, J´) + Cl (2P1/2)

Modelling of complex reactions C & E News, 6-Nov-89, pp.25-31
– stratospheric O3 tropospheric hydrocarbons H3CCO2ONO2
– combustion
chemical vapour deposition: SiH4 Si films
Rate of reaction {symbol:R,v,…}
3
Stoichiometric equation
m A + n B = p X + q Y
 Rate = - (1/m) d[A]/dt

= - (1/n) d[B]/dt

= + (1/p) d[X]/dt

= + (1/q) d[Y]/dt
– Units: (concentration/time)
– in SI mol/m3/s, more practically mol dm–3 s–1
Rate Law
How does the rate depend upon [ ]s?
 Find out by experiment

The Rate Law equation
 R = kn [A]a [B]b …
(for many reactions)
– order, n = a + b + …
(dimensionless)
– rate constant, kn
(units depend on n)
– Rate = kn when each [conc] = unity
4
Experimental rate laws?

CO + Cl2  COCl2
Rate = k [CO][Cl2]1/2
– Order = 1.5 or one-and-a-half order

Rate = k [H2][I2]
H2 + I2  2HI
– Order = 2 or second order

H2 + Br2  2HBr
Rate = k [H2][Br2] / (1 + k’ {[HBr]/[Br2]} )
– Order = undefined or none
5
Determining the Rate Law

Integration
– Trial & error approach
– Not suitable for multi-reactant systems
– Most accurate

Initial rates
– Best for multi-reactant reactions
– Lower accuracy

Flooding or Isolation
– Composite technique
– Uses integration or initial rates methods
6
Integration of rate laws
Order of reaction
For a reaction aA
the rate law is:

products
1 d [ A]
r k[ A]n
a dt
d [ A]

 - ak[ A]n
dt
defining k A  ak rate of change in the
concentration of A
d [ A]
r
 - k A [ A]n
dt
7
First-order reaction
d [ A]
1
r
 -k A [ A]
dt
d [ A]
 -k A dt
[ A]
[ A]t d [ A]
t
 -k A dt
[ A]0 [ A]
0
ln([ A]t - [ A]0 )  -k A (t - t 0 )


8
First-order reaction
ln[ A]t - ln[ A]0  -k A (t - t0 )
ln[ A]t  ln[ A]0 - k At
A plot of ln[A] versus t gives a straight
line of slope -kA if r = kA[A]1
9
First-order reaction
ln[ A]t - ln[ A]0  - k A (t - t0 )
 [ A]t 
  - k At
ln 
 [ A]0 
[ A]t
- k At
e
[ A]0
[ A]t  [ A]0 e
- k At
10
A P assume that -(d[A]/dt) = k [A]1
11
8
7
[H2O2] / mol dm-3
6
5
4
3
2
1
0
5
10
Time / ms
15
Integrated rate equation
ln [A] = -k t + ln [A]0
1.0
ln [H2O2] / mol dm-3
0.8
0.6
0.4
0.2
0
5
10
Time / ms
15
12
Half life: first-order
reaction

The time taken for [A] to drop to half its original value is called the
reaction’s half-life, t1/2. Setting [A] = ½[A]0 and t = t1/2 in:
 [ A]t 
  - k At
ln 
 [ A]0 
1

 [ A]0 
  - k At1/ 2
ln  2
 [ A]0 




13
Half life: first-order
reaction
1
ln    - k At1/ 2  -0.693
2
0.693
0.693
 t1/ 2 
or k A 
kA
t1/ 2
14
When is a reaction over?
[A] = [A]0 exp{-kt}
Technically [A]=0 only after infinite time

15
Second-order reaction
d [ A]
2
r
 - k A [ A]
dt
d [ A]

k
dt
A
2
[ A]
[ A]t

[ A ]0
d [ A]
[ A]
2
t

 - k A dt
0
16
Second-order reaction

1
1 
  -k A (t - t 0 )
-  [ A]t  [ A]0 
1
1
 k At
[ A]t [ A]0
A plot of 1/[A] versus t gives a straight
line of slope kA if r = kA[A]2
17
Second order test: A + A  P
18
24
22
1 / [A]
20
18
16
14
12
(1 / [A]0)
10
2
4
6
Time / ms
8
10
Half-life: second-order
reaction
1
1
 k At
[ A]t [ A]0
2
1
 k At1/ 2
[ A]o [ A]0
1
1
 k At1/ 2 or
 t1/ 2
[ A]0
k A [ A]0
19
Rate law for elementary reaction
20

Law of Mass Action applies:
– rate of rxn  product of active masses of reactants
– “active mass” molar concentration raised to power
of number of species

Examples:
–
A P + Q
rate = k1 [A]1
– A + B C + D
rate = k2 [A]1 [B]1
– 2A + B E + F + G
rate = k3 [A]2 [B]1
Molecularity of elementary reactions?
21

Unimolecular (decay) A  P
- (d[A]/dt) = k1 [A]

Bimolecular (collision) A + B  P
- (d[A]/dt) = k2 [A] [B]

Termolecular (collision) A + B + C  P
- (d[A]/dt) = k3 [A] [B] [C]

No other are feasible!
Statistically highly unlikely.
CO + Cl2


22
COCl2
Exptal rate law: - (d[CO]/dt) = k [CO] [Cl2]1/2
– Conclusion?: reaction does not proceed as written
– “Elementary” reactions; rxns. that proceed as written at
the molecular level.




Cl2  Cl + Cl
(1)
Cl + CO  COCl
(2)
COCl + Cl2 COCl2 + Cl (3)
Cl + Cl  Cl2
(4)
● Decay
● Collisional
● Collisional
● Collisional
– Steps 1 thru 4 comprise the “mechanism” of the reaction.
- (d[CO]/dt) = k2 [Cl] [CO]
23
If steps 2 & 3 are slow in comparison to 1 & 4
then, Cl2 ⇌ 2Cl or K = [Cl]2 / [Cl2]
So [Cl] = K × [Cl2]1/2
Hence:

- (d[CO] / dt) = k2 × K × [CO][Cl2]1/2
Predict that: observed k = k2 × K
 Therefore mechanism confirmed (?)
H2 + I2 2 HI
Predict: + (1/2) (d[HI]/dt) = k [H2] [I2]
 But if via:

–
I22 I
– I + I + H2 2 HI
–
I + I I2
rate = k2 [I]2 [H2]
Assume, as before, that 1 & 3 are fast cf. to 2
Then: I2 ⇌ 2 I or K = [I]2 / [I2]
 Rate = k2 [I]2 [H2] = k2 K [I2] [H2]
(identical)
Check? I2 + hn 2 I (light of 578 nm)
24
Problem
25
In the decomposition of azomethane, A, at a pressure of
21.8 kPa & a temperature of 576 K the following
concentrations were recorded as a function of time, t:
Time, t /mins
0
30 60 90 120
[A] / mmol dm-3
8.70 6.52 4.89 3.67 2.75
 Show that the reaction is 1st order in azomethane &
determine the rate constant at this temperature.

Recognise that this is a rate law question dealing with
the integral method.
- (d[A]/dt) = k [A]? = k [A]1
Re-arrange & integrate (bookwork)
 Test:
ln [A] = - k t + ln [A]0
Complete table:
Time, t /mins
0
30 60 90 120
ln [A]
2.16 1.88 1.59 1.30 1.01
 Plot ln [A] along y-axis; t along x-axis
 Is it linear? Yes. Conclusion follows.
Calc. slope as: -0.00959 so k = + 9.610-3 min-1
26
More recent questions …

Write down the rate of rxn for the rxn:
C3H8 + 5 O2 = 3 CO2 + 4 H2O
for both products & reactants
[8 marks]
For a 2nd order rxn the rate law can be written:
- (d[A]/dt) = k [A]2
What are the units of k ?
[5 marks]
 Why is the elementary rxn NO2 + NO2  N2O4 referred to
as a bimolecular rxn?
[3 marks]

27
Temperature dependence?

C2H5Cl  C2H4 + HCl
k/s-1
T/K
6.1  10-5
30  10-5
242  10-5
700
727
765
Conclusion: very sensitive to temperature
 Rule of thumb: rate doubles for a 10 K rise

28
Details of T dependence
Hood
 k = A exp{ -B/T }
Arrhenius
 k = A exp{ - E / RT }
Rate
A A-factor or
pre-exponential factor
 k at T 
of
rxn
E activation energy
(energy barrier) J mol -1 or kJ mol-1
R gas constant.
Temperature
29
Arrhenius eqn. k=A exp{-E/RT}
Useful linear form: ln k = -(E/R)(1/T) + ln A
 Plot ln k along Y-axis vs (1/T) along X-axis
Slope is negative -(E/R); intercept = ln A
 Experimental Es range from 0 to +400 kJ mol-1
Examples:
–
–
–
–
H + HCl H2 + Cl
H + HF H2 + F
C2H5I C2H4 + HI
C2H6 2 CH3
19 kJ mol-1
139 kJ mol-1
209 kJ mol-1
368 kJ mol-1
30
Practical Arrhenius plot,
31
origin not included
8
Intercept = 27.602 fromwhich A = 1.1 x 1012 dm3 mol-1 s-1
6
ln k /(dm3 mol-1 s-1)
4
Slope = -22,550 fromwhich E = 188 kJ/mol
2
0
-2
-4
-6
-8
0.0009
0.0010
0.0011
0.0012
K/T
0.0013
0.0014
0.0015
Rate constant expression


k1 A
 exp 

k2 A


(- E A ) 

RT1 
(- E A ) 

RT2 
k1
 - E A  1 1 
 exp 
 - 
k2
 R  T1 T2 
 k1   - E A  1 1 
ln    
 - 
 k 2   R  T1 T2 
32
 - EA 
k  A exp 

 RT 
1 
 1   - E A  1
ln    


 2   8.314  293 .15 303 .15 
 - EA 
- 0.693  
 1.12526  10 - 4
 8.314 


 - EA 
- 6158 .58  
  E A  51 .202 kJ mol -1
 8.314 
Photochemical activation

Initiation of reaction by light absorption; very important
– photosynthesis; reactions in upper atmosphere
No. of photons absorbed? Einstein-Stark law: 1 photon
responsible for primary photochemical act (untrue)
S0 + hn  S1*
Jablonski diagram
S*
 S0 + hn
fluorescence, phosphorescence
S* + M  S0 + M
collisional deactivation (quenching)
S*
 P + Q
photochemical reaction

33
Example & Jablonski diagram


A ruby laser with frequency
S3
doubling to 347.2 nm has
S2
an output of 100J with
pulse widths of 20 ns.
S1
If all the light is absorbed
in 10 cm3 of a 0.10 mol
dm-3 solution of perylene,
what fraction of the
perylene molecules are
S0
activated?
34
INTERNAL CONVERSION
10 4-10 12 s -1
INTERSYSTEM CROSSING
10 4-10 12 s -1
T1
FLUORESCENCE
10 6-10 9 s -1
PHOSPHORESCENCE
10 -2-10 4 s -1
35

# of photons = total energy / energy of 1 photon

Energy of photon?


hc


(6.626 10 -34 Js) (3 10 8 ms -1 )
347 .2 10 -9 m
  5.725  10 -19 J
# of photons = 100 / 5.725  10−19 = 1.7467  1020
# of molecules: 0.1 mol in 1000 cm3, => 1  10−3 mol in 10 cm3
=> 6.022  1020 molecules
fraction activated: 1.7467  1020 / 6.022  1020 = 0.29
Key parameter: quantum yield, F
36
F= (no. of molecules reacted)/(no. of photons absorbed)
 Example: 40% of 490 nm radiation from 100 W source
transmitted thru a sample for 45 minutes; 344 mmol of
absorbing compound decomposed. Find F.
Energy of photon?
 = hc / 
 (6.62610−34 J s)(3.00108 m s−1)/(49010−9 m) = 4.0610−19 J
Power: 100 Watts = 100 J s-1
 Total energy into sample = (100 J s−1)(4560 s)(0.60)= 162 kJ
 Photons absorbed = (162,000)/(4.0610−19) = 4.01023
 Molecules reacted? (6.0231023)  0.344= 2.07 1023
 F = 2.07 1023 / 4.01023 = 0.52
Quantum yield
Significance? F= 2.0 for 2HI  H2 + I2 reaction
HI + hn  H• + I• (i)
primary f= 1
H• + HI  H2 + I• (p)
I• + I•
 I2
(t)
 For H2 + Cl2  2HCl
F> 106
Is F constant? No, depends on , T, solvent, time.
  / nm >430
405
400
<370
 F
0
0.36
0.50
1.0
for NO2NO+O
37
F?
38
DETECTOR


Absolute measurement of FA, etc.? No; use relative method.
Ferrioxalate actinometer:
C2O42- + 2 Fe3+ 2 Fe2+ + 2 CO2


F= 1.25 at 334 nm but fairly constant from 254 to 579 nm
For a reaction in an organic solvent the photo-reduction of
anthraquinone in ethanol has a unit quantum yield in the UV.
Rates of photochemical reactions
39
Br2 + hn  Br + Br
Definition of rate:
 where nJ is stoichiometric
coefficient (+ve for products)
Units: mol s-1
 So FA is moles of photons
absorbed per second
 Finally, the reaction rate per
unit volume in mol s-1 m-3
 or mol m-3 s-1

1  dnJ 
Rate  

n J  dt 
 dn Br2 
-
 dt


 fF fI
A
A


 
n( Br2 )  V Br2
 
 d Br2  f FA

- 

dt
V


Stern-Volmer
 M + hn M*
FA / V
Apply SS approx. to M*:
FF / V
d[M*]/dt = (FA/V) - kF[M*] - kQ[M*][Q]  M* M + hn
 M* + Q M + Q
 Also (FF / V)= kF[M*]
So:
(FA / FF ) = 1 + (kQ /kF) [Q]
And hence:
Plot reciprocal of fluorescent intensity versus [Q]
Intercept is (1/FA) and slope is = (kQ / kF) (1/FA)
 Measure kF in a separate experiment; e.g. measure the
half-life of the fluorescence with short light pulse & [Q]=0
since d[M*]/dt = - kF[M*] then [M*]=[M*]0 exp(-t/t)

40
Problem 23.8 (Atkins)
Benzophenone phosphorescence with triethylamine as
quencher in methanol solution.
Data is:
[Q] / mol dm-3
1.0E-3
5.0E-3
10.0E-3
FF /(arbitrary)
0.41
0.25
0.16
Half-life of benzophenone triplet is 29 s.
Calculate kQ.
41
42
6.5
6.0
5.5
1/FF
5.0
4.5
4.0
Y=A+B*X
3.5
Parameter
Value Error
-----------------------------------------------------------A
1.96549
0.10995
B
424.53279
16.96558
3.0
2.5
2.0
0.000
0.002
0.004
0.006
[Q] / mol dm
0.008
-3
0.010
Flash photolysis

Fast burst of laser
light
– 10 ns, 1 ps down to
femtosecond



43
[RK, Pilling & Seakins, p39 on]
High concentrations
of reactive species
instantaneously
Study their fate
Transition state
spectroscopy
J. Phys. Chem. a 4-6-98
SS HEATABLE
REACTION
VESSEL
Xe
ARC
LAMP
ArF
EXCIMER
LASER
Flash photolysis

Adiabatic
–
–
–
–

44
Light absorbed => heat => T rise
Low heat capacity of gas => 2,000 K
Pyrolytic not photolytic
Study RH + O2 spectra of OH•, C2, CH, etc
Isothermal
– Reactant ca. 100 Pa, inert gas 100 kPa
– T rise ca. 10 K; quantitative study possible
– precursor + hn  CH
subsequent CH + O2 
Example
CH + O2 products
-1.0
-1.5
Excess O2 present
-2.0
molecules
cm-3
1st order kinetics
IF
30
-4.0
-4.5
40
60
0.230 0.144 0.088 0.033
Calculate k1 and k2
-3.0
-3.5
Follow [CH] by LIF
t / s 20
-2.5
ln (IF)
[O2]0 =
8.81014
45
[RK, Pilling & Seakins, p48]
-5.0
20
30
40
50
Time / s
60
70
80
Problem

46
In a flash-photolysis experiment a radical, R, was produced
during a 2 s flash of light and its subsequent decay
followed by kinetic spectrophotometry: R + R  R2

The path-length was 50 cm, the molar absorptivity, ,
1.1104 dm3/mol/cm.

Calculate the rate constant for recombination.
– t / s
– Absorbance
0
10
15
25
40
50
0.75 0.58 0.51 0.41 0.32 0.28
How would you determine ?
Photodissociation
FREQUENCY
SUBTRACTOR
[RK, p. 288]
47
PROBE PULSE
MOVABLE MIRROR
30  m = 100 fs
PHOTOPULSE
FS LASER
Beam Splitter
ICN
SAMPLE
Same laser dissociates ICN at 306 nm & is used to measure
[CN] by LIF at 388.5 nm
 Aim: measure time delay between photolysis pulse and
appearance of CN by changing the timing of the two pulses.
Experimentally: t 20530 fs; separation  600 pm [C & E News 7-Nov-88]

TS spectroscopy;


Changing the wavelength of the
probing pulse can allow not just
the final product, free CN, to be
determined but the intermediates
along the reaction path including
the transition state.
For NaI one can see the
activated complex vibrate at (27
cm-1) 1.25 ps intervals surviving
for 10 oscillations
– see fig. 24.75 Atkins 8th ed.
Atkins p. 834
48
Fast flow tubes; 1 m3/s, inert coating, t=d/v
In a RF discharge: O2 O + O or pass H2 over heated
tungsten filament or O3 over 1000oC quartz, etc.
Use non-invasive methods for analysis e.g. absorption, emission
Gas titration: add stable NO2 (measurable flow rate)
 Fast O+NO2  NO+O2 then O+NO  NO2*NO2 +hn
End-point? Lights out when flow(NO2) = flow(O)

O2
NO2
49
ClO + NO3 J. Phys. Chem. 95:7747 (1991)
50



1.5 m long, 4 cm od, Pyrex tube with sliding injector to vary
reaction time
F + HNO3  NO3 + HF
[NO3] monitor at 662 nm
F + HCl  Cl + HF
followed by Cl + O3  ClO + O2
F2 / He
HNO3 / He

He
MS
SLIDING INJECTOR
HCl
RF
F2 / He
He
Problem
[RK, Pilling & Seakins, p36]
HO2 + C2H4  C2H5 + O2  C2H5O2
MS determines LH channel 11%, RH channel 89%
C2H5 signal
Injector d / cm
6.14 3.95 2.53 1.25 0.70 0.40
3
5
7
10 12 15
Linear flow velocity was 1,080 cm s-1 at 295 K & 263 Pa.
Calculate 1st order rate constant; NB [O2]0>>[C2H5]0
51
Flow tubes; pros & cons
Mixing time restricts timescale to millisecond range
 Difficult to work at pressures > (atm/100)
 Wall reactions can complicate kinetics

– coat with Teflon or halocarbon wax; or vary tube diameter
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Cheap to build & operate, sensitive detection available
–
–
–
–
Resonance fluorescence
Laser induced fluorescence
Mass spectrometry
Laser magnetic resonance
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Resonance fluorescence
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Atomic species (H, N, O, Br, Cl, F) mainly not molecular
 Atomic lines are very narrow; chance of absorption by
another species is highly unlikely
 Resonance lamp: microwave discharge dissociates H2
 H atoms formed in electronically excited state; fluoresce,
emitting photon which H-atoms in reaction vessel absorb
& re-emit them where they can be detected by PMT
Lamp: H2 H H* H + hn
Rxn cell: H + hn H* H + hn
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LIF; detection of OH
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Excitation pulse at 282 nm to
upper state of OH with lifetime
of ns; fluorescence to ground
state at 308 nm
IF  n
relative concentrations not
absolute (drawback).
Right angle geometry
Good candidates:
– CN, CH, CH3O, NH, H, SO
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v'=2
v'=1
v'=0
282 nm
308 nm
v''=2
v''=1
v''=0
Reactions in shock waves
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Wide range of T’s & P’s accessible; 2,000 K, 50 bar routine
Thermodynamics of high-T species eg Ar up to 5,000 K
Study birth of compounds: C6H5CHO CO* + C6H6
Energy transfer rxns.: CO2 + M CO2* + M
Relative rates, use standard rxn as “clock”
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Experiments: Ignition Delay Time
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CH* Chemiluminescence (431 nm) Detected at Endwall and
Sidewall
Shock Tube
OH*
Endwall
Ignition
Slit
Time
OH*
Lens
Filter (310 nm)
PMT Detector
Ignition
Time
Sidewall
• Use endwall for ignition
• Use sidewall for profiles
Mode of action of shock tube
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Fast bunsen-burner (ns)
Shock wave acts as a
piston compressing &
heating the gas ahead of it
Study rxns behind incident
shock wave or reflected
shock wave (ms-s times)
Non-invasive techniques
T & p by computation from
measured shock velocity
P
DISTANCE
T3
T2
T
T1
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Shock Tube Simulation
58
Problem
59
A single-pulse shock tube used to study 1st order reaction
C2H5I  C2H4 + HI; to avoid errors in T measurement
a comparative study was carried out with C3H7I  C3H6
+ HI for which kB=9.11012 exp(-21,900/T) s-1. For
a rxn time of 220 s 5% decomp. of C3H7I occurred.
What was the temp. of the shock wave? [900 K]
 For C2H5I 0.90% decomp. occurred; evaluate kA.
 If at 800 K (kA/kB) = 0.102 compute the Arrhenius
equation for kA. [5.81013 exp(-25,260/T) s-1]
