Chemical kinetics or dynamics

3 lectures leading to one exam question

Texts: “Elements of Physical Chemistry”Atkins & de Paula

Specialist text in Hardiman Library– “Reaction Kinetics” by Pilling & Seakins,

1995These notes available On NUI Galway web pages at

http://www.nuigalway.ie/chem/degrees.htm

What is kinetics all about?1

Academic? Ozone chemistry Ozone; natural formation (l » 185-240 nm)

– O2 + hu ® 2O – O + O2 ® O3

Ozone; natural destruction (l » 280-320 nm) Thomas Midgely– O3 + hu ® O + O2 1922 TEL;

1930 CFCs– O + O3 ® 2O2

‘Man-made’ CCl2F2 + hu ® Cl + CClF2

– Cl + O3 ® Cl ̶ O + O2

– Cl ̶ O + O ® Cl + O2– -----------------------------– Net result is: O + O3 ® 2 O2

1995 Nobel for chemistry: Crutzen, Molina & Rowland 1996 CFCs phased out by Montreal protocol of 19872

Chemical kinetics

Thermodynamics– Direction of

change Kinetics

– Rate of change

– Key variable: time

What times?– 1018 s age of

universe– 10-15 s atomic

nuclei– 108 to 10-14 s

Ideal theory of kinetics?– structure, energy– calculate fate

Now?– compute rates of

elementary reactions

– most rxns not elementary

– reduce observed rxn. to series of elementary rxns.

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4

Reaction Timescale K

H2O (aq) = H+ + OH– seconds 10–14

H+ + OH– = H2O (aq) microseconds 1014

2 H2 (g) + O2 (g) = 2 H2O (l) years 1041

4 Al (s) + 3 O2 (g) = 2 Al2O3 (s) decades 10277

Kinetics determines the rate at which change occurs

Thermodynamics vs kinetics

Pressure vs CAD in an engine

5

⇌Kinetics and equilibrium

kinetics equilibrium

6

CH 4

H 2-O 2

CH 2O

C3 S pecies

C2H 6-C2H 4-C2H 2

CO

CH 3O H

C2H 5O H

Hierarchical structure

C.K. Westbrook and F.L. DryerProg. Energy Combust. Sci., 10 (1984) 1–57.

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Reaction Af nf Eaf Ar nr Ear

k = A Tn exp(-Ea/RT)Reaction mechanism

c2h5oh = c2h4+h2o 1.25E+14 0.1 2.00E+04 1.11E+07 1.77 8.08E+03

c2h5oh = ch2oh+ch3 2.00E+23 -1.68 9.64E+04 8.38E+14 -0.22 7.02E+03

c2h5oh = c2h5+oh 2.40E+23 -1.62 9.95E+04 9.00E+15 -0.24 4.65E+03

c2h5oh = ch3cho+h2 7.24E+11 0.1 9.10E+04 4.91E+07 0.99 7.50E+04

c2h5oh+o2 = pc2h4oh+ho2 2.00E+13 0 5.28E+04 2.19E+10 0.28 4.43E+02

c2h5oh+o2 = sc2h4oh+ho2 1.50E+13 0 5.02E+04 1.95E+11 0.09 4.88E+03

c2h5oh+oh = pc2h4oh+h2o 1.81E+11 0.4 7.17E+02 4.02E+08 0.92 1.79E+04

c2h5oh+oh = sc2h4oh+h2o 6.18E+10 0.5 -3.80E+02 1.63E+09 0.83 2.39E+04

c2h5oh+oh = c2h5o+h2o 1.50E+10 0.8 2.53E+03 7.34E+09 0.91 1.72E+04

c2h5oh+h = pc2h4oh+h2 1.88E+03 3.2 7.15E+03 3.93E-01 3.83 9.48E+03

c2h5oh+h = sc2h4oh+h2 8.95E+04 2.53 3.42E+03 2.21E+02 2.97 1.28E+04

c2h5oh+h = c2h5o+h2 5.36E+04 2.53 4.41E+03 2.47E+03 2.74 4.19E+03

c2h5oh+ho2 = pc2h4oh+h2o2 2.38E+04 2.55 1.65E+04 2.88E+03 2.48 2.83E+03

c2h5oh+ho2 = sc2h4oh+h2o2 6.00E+12 0 1.60E+04 8.59E+12 -0.26 9.42E+03

c2h5oh+ho2 = c2h5o+h2o2 2.50E+12 0 2.40E+04 6.66E+13 -0.48 7.78E+03

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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

9

10

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)

Rate of reaction {symbol: R,n, ..}

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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

12

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

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

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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

15

First-order reaction

)()][]ln([

][

][

][

][

][][

00

0

][

][

1

0

ttkAA

dtkA

Ad

dtkA

Ad

Akdt

Adr

At

t

A

A

A

A

A

t

16

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

17

First-order reaction

tkt

tkt

At

At

A

A

eAA

eA

A

tkA

A

ttkAA

0

0

0

00

][][

][

][

][

][ln

)(]ln[]ln[

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A ® P assume that -(d[A]/dt) = k [A]1

0 200 400 600 800 1000

0.1

0.2

0.3

0.4

0.5

0.6

[A] (

mo

l dm

-3)

Time (s)

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

0 200 400 600 800 1000-2.2

-2.0

-1.8

-1.6

-1.4

-1.2

-1.0

-0.8

-0.6

-0.4ln[A] versus Time

Y = A + B * X

Parameter Value Error------------------------------------------------------------A -0.55841 0.01861B -0.00149 3.82077E-5------------------------------------------------------------

ln[A

]

Time (s)

Slope = -k1st Order reaction

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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:

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Half life: first-order reaction

2/12/1

2/1

693.0693.0

693.02

1ln

tkor

kt

tk

AA

A

22

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

Technically [A]=0 only after infinite time

23

Second-order reaction

tA

A

t

A

A

A

dtkA

Ad

dtkA

Ad

Akdt

Adr

][

][ 02

2

2

0][

][

][

][

][][

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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

25

Second order test: A + A ® P

0 50 100 150 200 250 300

10

20

30

40

50

Y = A + B * X

Parameter Value Error------------------------------------------------------------A 9.62071 0.23913B 0.13737 0.00134

1/[A

rSO

2H

] / L

mo

l-1

time / min

Slope = k2nd Order reaction

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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

Kinetics and equilibrium

kinetics equilibrium

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Initial Rate Method

5 Br- + BrO3- + 6 H+ ® 3 Br2 + 3 H2O

General example: A + B +… ® P + Q + …

Rate law: rate = k [A]a [B]b …??log R0 = a log[A]0 + (log k+ b log[B]0 +…) y = mx + c Do series of expts. in which all [B]0, etc

are constant and only [A]0 is varied; measure R0

Plot log R0 (Y-axis) versus log [A]0 (X-axis) Slope Þ a

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Example: R0 = k [NO]a[H2]b

2 NO + 2H2 ® N2 + 2H2O Expt. [NO]0 [H2]0 R0

– 1 25 10 2.4×10-3

– 2 25 5 1.2×10-3

– 3 12.5 10 0.6×10-3

Deduce orders wrt NO and H2 and calculate k.

Compare experiments #1 and #2 Þ b Compare experiments #1 and #3 Þ aNow, solve for k from k = R0 / ([NO]a[H2]b)

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How to measure initial rate? Key: - (d[A]/dt) » - (d[A]/dt) » (d[P]/dt) A + B + … ® P + Q + … t=0 100 100 ® 0 0 mol m-3

10 s 99 99 ® 1 1 ditto Rate?- (100-99)/10 = -0.10 mol m-3 s-1

+(0-1)/10 = -0.10 mol m-3 s-1

Conclusion? Use product analysis for best accuracy.

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Isolation / flooding

IO3- + 8 I- + 6 H+ ® 3 I3- + 3 H2O

Rate = k [IO3-]a [I-]b [H+]g …

– Add excess iodate to reaction mix– Hence [IO3

-] is effectively constant– Rate = k¢ [I-]b [H+]g …– Add excess acid– Therefore [H+] is effectively constant

Rate » k² [I-]a

Use integral or initial rate methods as desired

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32

Rate law for elementary reaction Law of Mass Action applies:

– rate of rxn µ 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

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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.

CO + Cl2 X COCl2

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Experimental 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 collision

al collision

al collision

al

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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] = ÖK × [Cl2]1/2

Hence:

- (d[CO] / dt) = k2 × ÖK × [CO][Cl2]1/2

Predict that: observed k = k2 × ÖK 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 rate = k2 [I]2 [H2]– I + I ® I2

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)36

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 dm-3 8.706.524.893.67

2.75 Show that the reaction is 1st order in

azomethane & determine the rate constant at this temperature.

37

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 = + 9.6×10-3

min-138

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]

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