1Chemical Kinetics

Texts: Atkins, 6th edtn., chaps. 25, 26 & 27“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, ...

Thermochemical 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

3Photochemical activation Initiation of rxn by light absorption; very important

– photosynthesis; rxns in upper atmosphere No. of photons absorbed? Einstein-Stark law: 1 photon

responsible for primary photochemical act (untrue)S0 + h S1* Jablonski diagram

S* S + h fluorescence, phosphorescence

S* + M S + M collisional deactivation (quenching)

S* P + Q photochemical reaction

4Example & 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 is activated?

S 3

S 2

S 1

S 0

T 1

IN TE R N A L C O N V E R SIO N

1 0 4 - 1 0 12 s -1

IN TE R SY STE M C R O SSIN G

1 0 4 - 1 0 12 s -1

F L U O R E S C E N C E

1 0 6 - 1 0 9 s -1P H O S P H O R E S C E N C E

1 0 -2 - 1 0 4 s -1

5No. of photons in 20 ns pulse?Energy of 1 photon? use h & c= so that:

= hc / = (6.62610-34 J s)(3.00108 m s-1)/(347.210-9 m)

= 5.7310-19JEnergy of pulse? is 100JNo. of photons = (100 J) / (5.73J) = 1.751019

No. of molecules? 1 mol dm-3 contains NA per dm-3

So, 10 cm3 of a 0.10 mol dm-3 solution contains:(10/1,000)(1/10)NA = 6.02 1020 molecules No. of photons / no. of molecules = 0.29 or 29%

6Key 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

Mols of photons? (4.01023) /(6.0231023) = 0.66 einsteins = 0.344 / 0.66 = 0.52

7Quantum yield

Significance? = 2.0 for H2 + I2 2HI 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 0.36 0.50 1.0

for NO2NO+O

8

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 rxn in an organic solvent the photoreduction of

anthraquinone in ethanol has a unit quantum yield in the UV.

D E T E C T O R

9Rates of photochemical reactions

Br2 + h 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

Rate

1

2

2 2

2

J

J

A A

A

dn

dt

dn Br

dtF I

n Br V Br

d Br

dt

F

V

( )

10Stern-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; eg, 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

11Problem 26.2 (Atkins) Benzophenone phosphorescence with triethylamine as

quencher in methanol solution. Data:[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.Build table: (1/ FF) 2.44 4.00 6.25

Plot; intercept = (1/ FA) slope = + (1/ FA) (kQ/ kF) Now = ln 2 / k so kF = (0.693 / 2910-6) s-1

Therefore: kQ = 5.1108 dm3 mol-1 s-1

12Problem 26.2 (continued)

Intercept 1.965 ± 0.11

Slope 424.5 ± 17

kQ/kF = 424.5/1.965

= 216

kF = 2.39 104 s-1

kQ = 5.16 108 dm3

mol-1 s-1

0.000 0.002 0.004 0.006 0.008 0.0101

2

3

4

5

6

7

1 /

FF

[Q] / mol dm-3

13Flash 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

14Flash 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– T rise ca. 10 K; quantitative study possible– precursor + h CH subsequent CH + O2

15Example [RK, Pilling & Seakins, p48]

CH + O2 products Excess O2 present, [O2]0 =

8.81014 molecules cm-3, 1st order kinetics, follow [CH] by LIF

t / ms 20 30 40 60IF 0.230 0.144 0.0.88

0.033

Plot ln(IF) versus time t Slope = - k1 k2 = k1 / [O2]0

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

16Problem 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 ?

17Photodissociation [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 YS 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

18TS 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. 27.9 Atkins 6th ed.

19Fast 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 1000C quartz, etc.

Use non-invasive methods for analysis eg absorption, emissionGas titration: add stable NO2 (measurable flow rate) Fast O+NO2 NO+O2 then O+NO NO2

NO2h

End-point? Lights out when flow(NO2) = flow(O)

N O 2

O 2

20ClO + 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

SLIDING INJE CTOR

RF

21Problem [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 15Linear flow velocity was 1,080 cm s-1 at 295 K & 263 Pa.Calculate 1st order rate constant; NB [O2]0>>[C2H5

]0

Either convert d’s into times via flow rate and then plot ln(signal) versus t

Or plot ln(signal) vs d & convert slope

22Flow 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

23Resonance 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: mcwe discharge dissociates H2

H atoms formed in electronically excited state; fluoresce, emitting photon which H-atoms in rxn vessel absorb & re-emit them where they can be detected by PMT

Lamp: H2 H H* H + h

Rxn cell: H + h H* H + h

24LIF; 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

25Reactions 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”

o sc illo sco p eh e liu m

h e liu m

va cu u m

va cu u m va cu u m

d ia p h ra g m s d r ive n se c tio nd r ive r se c tio n

co m p u te r

m o n o ch ro m a to r

co m p o u n d

26Mode 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 (milli-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

27Problem A single-pulse shock tube used to study 1st order rxn

C2H5I C2H4 + HI; to avoid errors in T measurement a comparative study of a standard cpd. S was carried out with C3H7I C3H6 + HI for which kS=9.11012 exp(-21,900/T) s-1. For a rxn time of 220 s 5% decomp. of C3H7I or S occurred.

What was the temp. of the shock wave? [900 K] For C2H5I 0.90% decomp. occurred; evaluate k If at 800 K (k/kS) = 0.102 compute the Arrhenius

equation for k. [5.81013 exp(-25,260/T) s-1]

28Partial solution [X] = [X]0 exp(-kt) k = (1/t) ln {[X]0/[X]}

5% reacted, 95% not reacted so [X]/[X]0=0.95

k =(1/t) ln{[X]0/[X]} = (106/220) ln(1.0526) s-1

k = 233 s-1

k = A exp(-E/RT) T= E/{R (ln A - ln k)}

T = 21,900 / {ln (9.11012) - ln (233)} = 898 K

kS? [S]=[S]0 exp(-kSt) kS=(1/t) ln{[S]0/[S]}

0.90% reacted, 99.1% not reacted so [S]/[S]0=0.991