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1Chemical KineticsTexts: 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
2Overview of kinetics 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
3Rate of reaction symbol:R,v,…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
4Rate Law How does the rate depend upon [ ]s? Find out by experiment
The Rate Law equation R = kn [A] [B] … (for many reactions)
– order, n = + + … (dimensionless)– rate constant, kn (units depend on n)– Rate = kn when each [conc] = unity
5Experimental rate laws?CO + Cl2 COCl2
Rate = k [CO][Cl2]1/2
– Order = 1.5 or one-and-a-half order
H2 + I2 2HI Rate = k [H2][I2]
– Order = 2 or second order
H2 + Br2 2HBr Rate = k [H2][Br2] / (1 + k’ {[HBr]/[Br2]} )
– Order = undefined or none
6Determining 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
7Integration of rate laws Order of reaction
For a reaction aA products the rate law is:
nA
A
n
n
AkdtAdr
akkdefining
AakdtAd
AkdtAd
ar
][][
][][
][][1
rate of change in theconcentration of A
8First-order reaction
)()][]ln([][][
][][
][][
00
0
][
][
1
0
ttkAA
dtkAAd
dtkAAd
AkdtAdr
At
t
A
A
A
A
A
t
9First-order reaction
tkAAttkAA
At
At
0
00
]ln[]ln[)(]ln[]ln[
A plot of ln[A] versus t gives a straightline of slope -kA if r = kA[A]1
10First-order reaction
tkt
tkt
At
At
A
A
eAA
eAA
tkAA
ttkAA
0
0
0
00
][][
][][
][][ln
)(]ln[]ln[
11A Passume that -(d[A]/dt) = k [A]1
0 5 10 151
2
3
4
5
6
7
8
[H2O
2] /
mol
dm
-3
Time / ms
12Integrated rate equationln [A] = -k t + ln [A]0
0 5 10 15
0.2
0.4
0.6
0.8
1.0
ln [H
2O2]
/ m
ol d
m-3
Time / ms
13Half life: first-order reaction
2/10
0
0
][
][21
ln
][][ln
tkA
A
tkAA
A
At
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:
14Half life: first-order reaction
2/12/1
2/1
693.0693.0
693.021ln
tkor
kt
tk
AA
A
15When is a reaction over? [A] = [A]0 exp{-kt}Technically [A]=0 only after infinite time
16Second-order reaction
tA
A
t
A
A
A
dtkA
Ad
dtkA
Ad
AkdtAdr
][
][ 02
2
2
0][
][
][][
][][
17Second-order reaction
tkAA
ttkAA
At
At
0
00
][1
][1
)(][
1][
1
A plot of 1/[A] versus t gives a straightline of slope kA if r = kA[A]2
18Second order test: A + A P
2 4 6 8 1010
12
14
16
18
20
22
24
(1 / [A]0)
1 / [
A]
Time / ms
19Half-life: second-order reaction
2/10
2/10
2/10
0
][1
][1
][1
][2
][1
][1
tAk
ortkA
tkAA
tkAA
AA
Ao
At
20Rate law for elementary reaction 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 + Grate = k3 [A]2 [B]1
21Molecularity of elementary reactions? 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.
22
CO + Cl2 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) ● Decay Cl + CO COCl (2) ● Collisional COCl + Cl2 COCl2 + Cl (3) ● Collisional Cl + Cl Cl2 (4) ● Collisional
– Steps 1 thru 4 comprise the “mechanism” of the reaction.
23- (d[CO]/dt) = k2 [Cl] [CO]If steps 2 & 3 are slow in comparison to 1 & 4then, 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 (?)
24H2 + I2 2 HI Predict: + (1/2) (d[HI]/dt) = k [H2] [I2] But if via:
– I22 I– I + I + H2 2 HI rate = k2 [I]2 [H2]– I + II2
Assume, as before, that 1 & 3 are fast cf. to 2Then: I2 ⇌2 I or K = [I]2 / [I2] Rate = k2 [I]2 [H2] = k2 K [I2] [H2] (identical)Check? I2 + h 2 I (light of 578 nm)
25Problem 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 dm3 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.
26 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]0Complete table:Time, t /mins 0 30 60 90 120ln [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 = + 9.610-3 min-1
27More 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]
28Temperature dependence? C2H5Cl C2H4 + HCl
k/s-1 T/K 6.1 10-5 700 30 10-5 727 242 10-5 765
Conclusion: very sensitive to temperature Rule of thumb: rate doubles for a 10 K rise
29Details of T dependenceHood k = A exp{ -B/T }Arrhenius k = A exp{ - E / RT }A A-factor or
pre-exponential factork at T
E activation energy(energy barrier) J mol -1 or kJ mol-1
R gas constant. T e m pe ra ture
R a te o f rxn
30Arrhenius 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-axisSlope is negative -(E/R); intercept = ln A Experimental Es range from 0 to +400 kJ mol-1
Examples:– H + HCl H2 + Cl 19 kJ mol-1
– H + HF H2 + F 139 kJ mol-1
– C2H5I C2H4 + HI 209 kJ mol-1
– C2H6 2 CH3 368 kJ mol-1
31Practical Arrhenius plot, origin not included
0.0009 0.0010 0.0011 0.0012 0.0013 0.0014 0.0015-8
-6
-4
-2
0
2
4
6
8
Intercept = 27.602 from which A = 1.1 x 1012 dm3 mol-1 s-1
Slope = -22,550 from which E = 188 kJ/mol
ln k
/(dm
3 mol
-1 s
-1)
K / T
32Rate constant expression
RTE
Ak Aexp
212
1
212
1
2
1
2
1
11ln
11exp
)(
)(
exp
TTRE
kk
TTRE
kk
RTE
RTE
AA
kk
A
A
A
A
1
4
202.51314.8
58.6158
1012526.1314.8
693.0
15.3031
15.2931
314.821ln
molkJEE
E
E
AA
A
A
33Photochemical 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 + h S1* Jablonski diagramS* S0 + h fluorescence, phosphorescence
S* + M S0 + M collisional deactivation (quenching)
S* P + Q photochemical reaction
34Example & Jablonski diagram A ruby laser with frequency
doubling to 347.2 nm has an output of 100J with pulse widths of 20 ns.
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 activated?
S 3
S 2
S 1
S 0
T 1
IN T E R N A L C O N V E R SIO N1 0 4 - 1 0 1 2 s -1
IN TE R SY STE M C R O SSIN G1 0 4 - 1 0 1 2 s -1
F L U O R E S C E N C E1 0 6 - 1 0 9 s -1
P H O S P H O R E S C E N C E1 0 -2 - 1 0 4 s -1
35
# of photons = total energy / energy of 1 photon Energy of photon?
# 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 moleculesfraction activated: 1.7467 1020 / 6.022 1020 = 0.29
Jm
msJshc
19
9
1834
10725.5102.347
)103()10626.6(
36Key parameter: quantum yield, = (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 .
Energy of photon? = hc / (6.62610−34 J s)(3.00108 m s−1)/(49010−9 m) = 4.060−19 JPower: 100 Watts = 100 J s-1
Total energy into sample = (100 J s−1)(4560 s)(0.60)= 162 kJ Photons absorbed = (162,000)/(4.060−19) = 4.01023
Molecules reacted? (6.0231023) = 2.07 1023
= 2.07 1023 /4.01023 = 0.52
37Quantum yieldSignificance? = 2.0 for 2HI H2 + I2 reactionHI + h H• + I• (i) primary = 1H• + HI H2 + I• (p)I• + I• I2 (t) For H2 + Cl2 2HCl > 106
Is constant? No, depends on , T, solvent, time. / nm >430 405 400<370 0 0.360.50 1.0
for NO2NO+O
38
Absolute measurement of FA, etc.? No; use relative method. Ferrioxalate actinometer:
C2O42 + 2 Fe3+ 2 Fe2+ + 2 CO2
= 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.
D E T E C T O R
39Rates of photochemical reactions
Br2 + h Br + BrDefinition 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
Rate
1
2
2 2
2
J
J
A A
A
dndt
dn Br
dtF I
n Br V Br
d Brdt
FV
( )
40Stern-Volmer Apply SS approx. to M*:d[M*]/dt = (FA/V) - kF[M*] - kQ[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/)
M + hM* FA / V M* M + h FF / V M* + QM + Q
41Problem 23.8 (Atkins)Benzophenone phosphorescence with triethylamine asquencher in methanol solution. Data is:
[Q] / mol dm-31.0E-3 5.0E-3 10.0E-3FF /(arbitrary) 0.41 0.25 0.16
Half-life of benzophenone triplet is 29 s. Calculate kQ.
42
0.000 0.002 0.004 0.006 0.008 0.0102.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
Y = A + B * X
Parameter Value Error------------------------------------------------------------A 1.96549 0.10995B 424.53279 16.96558
1/F F
[Q] / mol dm-3
43Flash photolysis [RK, Pilling & Seakins, p39 on] Fast burst of laser
light– 10 ns, 1 ps down to
femtosecond High concentrations
of reactive species instantaneously
Study their fate Transition state
spectroscopy J. Phys. Chem. a 4-6-98
X eA R CL A M P
A rFE X C IM E RL A S E R
S S H E A T A B L ER E A C T IO NV E S S E L
44Flash photolysis Adiabatic
– 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 + h CH subsequent CH + O2
45Example [RK, Pilling & Seakins, p48]
CH + O2 products
Excess O2 present
[O2]0 = 8.81014 molecules cm-3
1st order kinetics
Follow [CH] by LIFt / s 20 30 40 60
IF 0.230 0.144 0.088 0.033
Calculate k1 and k2
20 30 40 50 60 70 80-5.0
-4.5
-4.0
-3.5
-3.0
-2.5
-2.0
-1.5
-1.0
ln (I
F)
Time / s
46Problem 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, , 1.1104 dm3/mol/cm.
Calculate the rate constant for recombination.– t / s 0 10 15 25 40 50– Absorbance 0.75 0.58 0.51 0.41 0.32 0.28
How would you determine ?
47Photodissociation [RK, p. 288]
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: 205 fs; separation 600 pm [C & E News 7-Nov-88]
F S L A S E R IC NS A M P L E
M O V A B L E M IR R O R 3 0 m = 1 0 0 fs F R E Q U E N C Y
S U B T R A C T O R
P R O B E P U L S E
P H O T O P U L S E
Beam Splitter
48TS spectroscopy; Atkins p. 834 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.
49Fast flow tubes; 1 m3/s, inert coating, t=d/v In a RF discharge: O2 O + Oorpass H2 over heated
tungsten filament or O3 over 1000oC quartz, etc.
Use non-invasive methods for analysis e.g. absorption, emissionGas titration: add stable NO2 (measurable flow rate) Fast O+NO2 NO+O2 then O+NO NO2
NO2hEnd-point? Lights out when flow(NO2) = flow(O)
N O 2
O 2
50ClO + NO3 J. Phys. Chem. 95:7747 (1991) 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
M S
He
F 2 / He
HC l
He
HNO 3 / HeF 2 / He
SLID ING INJE CTOR
RF
51Problem [RK, Pilling & Seakins, p36]
HO2 + C2H4 C2H5
+ O2 C2H5O2
MS determines LH channel 11%, RH channel 89%
C2H5 signal 6.14 3.95 2.53 1.25 0.70 0.40
Injector d / cm 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
52Flow 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 Cheap to build & operate, sensitive detection available
– Resonance fluorescence– Laser induced fluorescence– Mass spectrometry– Laser magnetic resonance
53Resonance fluorescence 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 + hRxn cell: H + h H* H + h
54LIF; detection of OH 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
v '= 2
v '= 1
v '= 0
v ''= 2
v ''= 1
v ''= 0
2 8 2 n m 3 0 8 n m
55Reactions in shock waves
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”
Diaphragm
Driver TubeDriven Tube
TestRegion
He
ToVacuumSystem
TestMixture
Diaphragm
Driver TubeDriven Tube
TestRegion
He
ToVacuumSystem
TestMixture
56
CH* Chemiluminescence (431 nm) Detected at Endwall and Sidewall
Experiments: Ignition Delay Time
PMT Detector
Lens
Filter (310 nm)
Slit
Endwall
Shock Tube
SidewallO
H*
Time
Ignition
OH
*
TimeIgnition
• Use endwall for ignition
• Use sidewall for profiles
57Mode of action of shock tube 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
D IS T A N C E
T 1
T 2
T 3
T
58Shock Tube Simulation
59Problem 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=9.11012 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. [5.81013 exp(-25,260/T) s-1]
