1

Chemical Kinetics– ‘Elements of Physical Chemistry’, Atkins & de Paula– “Reaction Kinetics” , Pilling & Seakins– “An Examples Course in CK” , Campbell

Rapid revision of second year Kinetics in flow reactors; plug vs stirred Temperature dependence; Arrhenius equation Complex reactions

– chain reactions– polymerisations– explosions; branching chain reactions

2

Revision of 2nd year 5Br- + BrO3

- + 6H+ = 3Br2 + 3H2O

Rate? conc/time or in SI mol dm-3 s-1

– (1/5)(d[Br-]/dt) = – (1/6) (d[H+]/dt) =

(1/3)(d[Br2]/dt) = (1/3)(d[H2O]/dt)

Rate law? Comes from experiment

Rate = k [Br-][BrO3-][H+]2

where k is the rate constant (variable units)

3

Rate 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

4

Rate 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

5

Experimental 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

6

Integration of rate laws

Order of reactionFor a reaction aA products, the rate law is:

nA

A

n

n

Akdt

Adr

akkdefining

Aakdt

Ad

Akdt

Ad

ar

][][

][][

][][1

rate of change in theconcentration of A

7

First-order reaction

)()][]ln([

][

][

][

][

][][

00

0

][

][

1

0

ttkAA

dtkA

Ad

dtkA

Ad

Akdt

Adr

At

t

A

A

A

A

A

t

8

First-order reaction

tkAA

ttkAA

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

9

First-order reaction

tkt

tkt

At

At

A

A

eAA

eA

A

tkA

A

ttkAA

0

0

0

00

][][

][

][

][

][ln

)(]ln[]ln[

10

A Passume that -(d[A]/dt) = k [A]1

0 5 10 151

2

3

4

5

6

7

8

[H2O

2] / m

ol d

m-3

Time / ms

11

Integrated rate equationln [A] = -k t + ln [A]0

0 5 10 15

0.2

0.4

0.6

0.8

1.0

ln [H 2

O2]

/ m

ol d

m-3

Time / ms

12

Half life: first-order reaction

2/10

0

0

][

][21

ln

][

][ln

tkA

A

tkA

A

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:

13

Half life: first-order reaction

2/12/1

2/1

693.0693.0

693.02

1ln

tkor

kt

tk

AA

A

14

When is a reaction over? [A] = [A]0 exp{-kt}

Technically [A]=0 only after infinite time

15

Second-order reaction

tA

A

t

A

A

A

dtkA

Ad

dtkA

Ad

Akdt

Adr

][

][ 02

2

2

0][

][

][

][

][][

16

Second-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

17

Second order test: A + A P

2 4 6 8 1010

12

14

16

18

20

22

24

(1 / [A]0)

1 / [A

]

Time / ms

18

Half-life: second-order reaction

2/10

2/10

2/10

0

][

1

][

1

][

1

][

2

][

1

][

1

tAk

ortkA

tkAA

tkAA

AA

Ao

At

19

Rate 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 + G rate = k3 [A]2 [B]1

20

Molecularity 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.

21

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) Cl + CO COCl (2) COCl + Cl2 COCl2 + Cl (3)

Cl + Cl Cl2 (4)

– Steps 1 thru 4 comprise the “mechanism” of the reaction.

decay collisional collisional collisional

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- (d[CO]/dt) = k2 [Cl] [CO]

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 (?)

23

H2 + 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)

24

Problem 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 8.706.524.893.672.75 Show that the reaction is 1st order in

azomethane & determine the rate constant at this temperature.

25

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 120ln [A] 2.161.881.591.301.01 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

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Proper rate of reaction Define as: (1/i)(dni/dt) mol s-1

where i is a stoichiometric coefficient+ve for products, – ve for reactantsExample: 2H2 + O2 = 2H2O (H2) = – 2, (O2)= – 1 and (H2O)=+2 At constant volume: [A] = n /V or n = V × [A] So: (dn/dt) = V. (d[A]/dt)Old rate of reaction (was rxn rate per unit volume) (d[A]/dt) = (1 / V) (dnA/dt)

27

Flowvs batch reactors [A]=[A]0 exp{-kt}Stirred flow; perfect mixing

spherical shape of volume V

[A]outlet = [A]inlet / { 1 + k (V/)}

‘Plug’ flow; no mixing at all at all cylindrical shape of volume V

[A]outlet = [A]inlet exp{ - k (V/)}

NB (V/) has dimensions (m3 / m3 s-1) = sA ‘residence’ or ‘contact’ time

28

Stirred flow; 1st order-(dnA/dt) = k nA = k V [A]

inflow = outflow + reaction

Assume system at constant volume so that inlet flow rate is equal to outlet flow rate

[A]i = [A]o - (dnA/dt)

[A]i = [A]o + k V [A]o

[A]o = [A]i / {1 + k(V/)}

Test? vary v, measure [A]o &

[A]i, is k the same?

[A ] i

[A ]o

V

29

Plug flow; 1st order rxn-(dnA/dt) = k nA = k V [A]

inflow = outflow + reaction [A] = ([A]+[A]) – (dnA/dt)

–[A] = k V [A] [A]/[A] = – (k/) VAt inlet [A] = [A]i ; at outlet [A] = [A]o

ln [A]o = – k (V/) + ln [A]i

or [A]o = [A]i exp{– k (V/)}

[A ]o[A ] i

Vd V

30

Flow reactor problem Peracetic acid vapour at 2% by volume in nitrogen

carrier gas at a total pressure of 101 kPa and 490 K flows thru 1.4 mm radius teflon tubing and is sampled 1.2 m downstream with the following results:([A]o/[A]i) 0.346 0.578 0.718

/ m3 s-1 3.7110-5 7.2010-5 11.910-5

Show that rxn follows 1st order kinetics & calculate k.

Eqn? [A]o=[A]i exp{- k (V/)} or ln{[A]o/[A]i}= - k(V/)

Reactor vol. V = r2h = 3.14 (0.0014)2 1.2 = 7.39

m

k = - ( / V) ln{[A]o/[A]i} = 5.3208, 5.3410, 5.3349 s-1

31

Temperature 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 10K rise

32

Details 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

33

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-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

34

Practical 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

35

Problem In consecutive rxns the slower step usually

determines the overall rate of rxn. Diethyl adipate (DA) is hydrolysed in 2 steps as:

DA BP The Arrhenius equations are:

first step 1.23106 exp{-5,080/T} s-1

second step 8.80105 exp{-3,010/T} s-1

Which step is rate determining if the rxn is carried out in aqueous solution under atmospheric pressure?

36

0.0000 0.0005 0.0010 0.0015 0.0020 0.0025 0.0030 0.0035 0.0040-6-5-4-3-2-10123456789

1011121314

Comparison of rate constant expressions

ln k

K / T

1.23 x 106 exp(-5080 / T)

8.80 x 105 exp(-3010 / T)

30002750250022502000175015001250 1000 750 500 250

37

Rxn. at equilibrium: ABforward rate constant kF, reverse kR

Kinetics: kF[A]e = kR[B]eso [B]e/[A]e = kF/kR = K

kF = AF exp{-EF/RT} kR = AR exp{-ER/RT}

(kF/kR) = (AF/AR) exp{-(EF- ER)/RT}

Thermodynamics? G0 = - RT ln(K) or K = exp{-G0/RT}G0 = H0 - TS0

K = exp{+S0/R} exp{-H0/RT} (AF/AR) = exp{+S0/R} EF - ER = H0

38

Reaction profile diagram Y-axis?

– Energy– common path via TS

X-axis?Time?Reactant-product identityMulti-dimensional Reaction co-ordinate Schematic diagram

– Concept of energy barrier to rxn H

E F

E R

39

Reaction profile diagramCH3CHO CH4 + CO

EF = 198 kJ/mol Add some iodine as

catalyst EF = 134 kJ/mol– Modest reduction in

barrier– Massive increase in rate

At 773 K catalysed rxn is 20,000 times faster– D(C-C)=340, D(C-H)=420– But D(I-I)=153 kJ/mol

R

P

E RE F

198

kJ/

mol

134

kJ/

mol

40

Thermochemistry from kineticsC2H6 2 CH3

368 kJ mol-1

EF = 368 kJ mol-1 ER = 0 kJ mol-1

HRXN = EF-ER = + 368 kJ mol-1

= 2 Hf(CH3) - Hf(C2H6)

ButHf(C2H6) = - 85 kJ mol-1

Hf(CH3) = (368-85)/2 = +141 kJ mol-1

HRXN = Energy of bond broken = D(H3C-CH3)

Tables of bond dissociation energies

41

ProblemFor the rxn trans cis perfluorobut-2-ene measured kF=3.161013 exp{-31,030/T} s-1

Given that H0=+3.42 kJ mol-1 and S0=-2.03 J K-1 mol-1 Calculate kR at 750 K.

(AF/AR) = exp{+S0/R} AR = AF exp{-S0/R}

EF - ER = H0 ER = EF - H0

-(EF/R) = -31,030 EF = 8.314331,030 J mol-1

EF = 258.0 kJ mol-1 ER = 254.6 kJ mol-1

42

Common mathematical functions

Kinetics, Arrhenius

k = A exp{-E/RT}

Vapour pressure, Clausius-Clayperon

p = p exp{-HV/RT}

Viscosity, Andrade

= A exp{+E/RT}

But re-define as inverse: F= F exp{-E/RT}

43

Data from Keith Laidler, J Chem Educ 49, 343 (1972)T 17.3—25.0 C, chirp rates of 100 to 178 min-1

0.00334 0.00336 0.00338 0.00340 0.00342 0.00344 0.003464.5

4.6

4.7

4.8

4.9

5.0

5.1

5.2

5.3

5.4

5.5

Snowy tree cricket Oecanthus fultoni

Intercept = 26.89 ± 0.19Slope = -6,472 ± 56

Ln (C

hirp

s pe

r s)

K / T

44

Complex (non-elementary) rxnsChain reactions H2 + I2 do not react; I22I then I+H2+I2HI

CH3CHO CH4 + CO in over 90% yield

Traces of C2H6, H2, CH3COCH3, etc.

Rate law rate = k[CH3CHO]3/2

Possible mechanism? CH3CHO CH3

+ HCO initiation

CH3 + CH3CHO CH4 + CH3CO propagation

CH3CO CH3 + CO propagation

CH3 + CH3

C2H6 termination

45

Kinetic analysis? Mathematically difficultSteady-state approximation

– For reactive intermediates assume that rate of formation rate of destruction ie (d[R]/dt) 0

(d[CH3]/dt)= v1 + v3 - v2 - v4

(d[CH3CO]/dt)= v2 - v3 0v2= v3 and v1 = v4

k1[CH3CHO] = 2 k4[CH3]2

S-S conc.[CH3]=(2k1/k4)1/2 [CH3CHO]1/2

(d[CH4]/dt) = k2[CH3][CH3CHO]

= k2(2k1/k4)1/2 [CH3CHO]3/2

So kobs = k2(2k1/k4)1/2

46

Relationships for A and Es kobs = k2(2k1/k4)1/2

Since the A-factor is essentially a rate constant: Aobs = A2(2A1/A4)1/2

The activation energies are related as the logs of the rate constants:

log kobs = log k2 + (1/2){log(k1) - log(k4)}

Eobs = E2 + (1/2){E1 - E4} composite quantity

= 35 + (1/2){340 - 0} = 205 kJ mol-1

205«340; Note how in a chain rxn. the high-energy barrier of 340 kJ mol-1 is circumvented.

47

Chain polymerisationI + M M i Let [R]=total radical conc.

M + M M2 p

M2 + M M3

p

Mn + Mm

Mn+m t

(d[R]/dt)=-2 kT [R]2 + kI[M][I] (neglect rP)SS-approx.: (d[R]/dt)=0 So [R] = (kI/2kT)1/2 [M]1/2 [I]1/2

Rate of propagation of chains: - (d[M]/dt = kP [R][M] (neglect rI) - (d[M]/dt) = kP (kI/2kT)1/2 [I]1/2 [M]3/2

48

Branching chain reactions “On a windy Thursday in the

first week of May in 1937, the Hindenberg airship exploded as it approached its mooring in New Jersey, USA. Newsreel film of the disaster, in which 35 passengers & 1 ground crew died, shows that all 200,000 m3 of H2 in the ship burned in under 45 s.”– Physical Chemistry, Winn, p.1016– http://www.otr.com/hindenburg.html

49

Branching chain reactions

The power of two 1 etc Fifty doublings or 250 1015

€ 1015 for the 380 million citizens of the EU – € 2.6 million each

So reactions in which branching is important can proceed explosively

50

Branching chains MR i

R + MR + P p

RP t

R + MR + R b (d[R]/dt)=vi+kB[R]-kT[R]

If termination>branching [R] {1-exp[-(kT-kB)t]}

If branching>termination [R] {exp[-(kB-kT)t]-1}

[R ]

tim e

51

2 H2 + O2 = 2 H2O

H2 + O2 HO2 + H i

HO2 + H2 H2O + HO p

H + O2 + M HO2 + M p

H + O2 HO + O b

H diffuses towall t

2 H 2 + O 2

1 ba rpre s s u re

7 4 m m dia m e te r

K C l-c o a te d

g la s s s phe re

H 2 O 2

8 0 0 K

52

2 H2 + O2 = 2 H2O 1st limit (sensitive to surface, vessel shape & added inert gas)

– Competition between branching and diffusion to wall of H

– no explosion wall H + O2 HO + Oexplosion 2nd limit (not)

– Competition: – H +O2 + M HO2

+ M no explosion third order – H +O2 HO + Oexplosion second order– hydroperoxyl radical HO2

much less reactive than H HO 3rd limit (sensitive to vessel shape/size)

– rate very fast, high rate of heat release, faster than can be conducted away thermal explosion

– added gases have effect to their heat transport properties