1 Chemical kinetics or dynamics 3 lectures leading to one exam question l Texts: “Elements of Physical Chemistry” Atkins & de Paula l Specialist text.

1

Chemical kinetics or dynamics3 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?

2

Academic? Ozone chemistry Ozone; natural formation ( 185240 nm)– O2 + h2O– O + O2 O3

Ozone; natural destruction ( 280320 nm) Thomas Midgely

– O3 + hO + O2 1922 TEL; 1930 CFCs

– O + O3 2O2

‘Man-made’ CCl2F2 + hCl + CClF2

– Cl + O3 ClO + O2

– ClO + 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 1987

3

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.

4

Thermodynamics vs kinetics

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

5

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

6

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

7

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

8

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

9

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

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First-order reaction

)()][]ln([

][

][

][

][

][][

00

][

][ 0

1

0

ttkAA

dtkA

Ad

dtkA

Ad

Akdt

Adr

At

A

A

t

A

A

A

t

11

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

12

First-order reaction

tkt

tkt

At

At

A

A

eAA

eA

A

tkA

A

ttkAA

0

0

0

00

][][

][

][

][

][ln

)(]ln[]ln[

13

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

14

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

15

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:

16

Half life: first-order reaction

2/12/1

2/1

693.0693.0

693.02

1ln

tkor

kt

tk

AA

A

17

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

Technically [A]=0 only after infinite time

18

Second-order reaction

tA

A

t

A

A

A

dtkA

Ad

dtkA

Ad

Akdt

Adr

][

][ 02

2

2

0][

][

][

][

][][

19

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

20

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

21

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

22

Initial Rate Method5 Br- + BrO3

- + 6 H+ 3 Br2 + 3 H2O General example: A + B +… P + Q + … Rate law: rate = k [A] [B] …??log R0 = log[A]0 + (log k+ 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

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Example: R0 = k [NO][H2]

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

– 1 25 10 2.410-3

– 2 25 5 1.210-3

– 3 12.5 10 0.610-3

Deduce orders wrt NO and H2 and calculate k. Compare experiments #1 and #2 Compare experiments #1 and #3 Now, solve for k from k = R0 / ([NO][H2])

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How to measure initial rate? Key: - (d[A]/dt) -([A]/t) ([P]/dt)

A + B + … P + Q + … t=0 100 100 0 0 mol m3

s 99 99 1 1 ditto Rate? (100-99)/10 = -0.10 mol m3 s1

+(0-1)/10 = -0.10 mol m3 s1

Conclusion? Use product analysis for best accuracy.

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Isolation / floodingIO3

+ 8 I + 6 H 3 I3 + 3 H2O Rate = k [IO3

-] [I-] [H+]…– Add excess iodate to reaction mix– Hence [IO3

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

Rate k [I-]

Use integral or initial rate methods as desired

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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

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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.

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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 (?)

30

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)

31

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.

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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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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]

34

Rate constant expression

RT

EAk aexp

RT

EE

k

k

RT

ERT

E

A

A

k

k

RT

EAk

aa

a

a

a

)(exp

)(

)(

exp

exp

21

2

1

2

1

2

1

11