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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, 1995 These notes available On NUI Galway web pages at What is kinetics all about? 1

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Academic? Ozone chemistry Ozone; natural formation ( 185 240 nm) –O 2 + h 2O –O + O 2 O 3 Ozone; natural destruction ( 280 320 nm) Thomas Midgely –O 3 + h O + O TEL; 1930 CFCs –O + O 3 2O 2 ‘Man-made’ CCl 2 F 2 + h Cl + CClF 2 –Cl + O 3 Cl ̶ O + O 2 –Cl ̶ O + O Cl + O 2 – –Net result is: O + O 3 2 O 2 1995 Nobel for chemistry: Crutzen, Molina & Rowland 1996 CFCs phased out by Montreal protocol of

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Chemical kinetics Thermodynamics –Direction of change Kinetics –Rate of change –Key variable: time What times? –10 18 s age of universe – s atomic nuclei –10 8 to 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. 3

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4 Kinetics determines the rate at which change occurs Thermodynamics vs kinetics

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Pressure vs CAD in an engine 5

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⇌ Kinetics and equilibrium kinetics equilibrium 6

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Hierarchical structure C.K. Westbrook and F.L. Dryer Prog. Energy Combust. Sci., 10 (1984) 1–57. 7

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Reaction A f n f Ea f A r n r Ea r k = A T n exp(-E a /RT) Reaction mechanism c2h5oh=c2h4+h2o1.25E E E E+03 c2h5oh=ch2oh+ch32.00E E E E+03 c2h5oh=c2h5+oh2.40E E E E+03 c2h5oh=ch3cho+h27.24E E E E+04 c2h5oh+o2=pc2h4oh+ho22.00E E E E+02 c2h5oh+o2=sc2h4oh+ho21.50E E E E+03 c2h5oh+oh=pc2h4oh+h2o1.81E E E E+04 c2h5oh+oh=sc2h4oh+h2o6.18E E E E+04 c2h5oh+oh=c2h5o+h2o1.50E E E E+04 c2h5oh+h=pc2h4oh+h21.88E E E E+03 c2h5oh+h=sc2h4oh+h28.95E E E E+04 c2h5oh+h=c2h5o+h25.36E E E E+03 c2h5oh+ho2=pc2h4oh+h2o22.38E E E E+03 c2h5oh+ho2=sc2h4oh+h2o26.00E E E E+03 c2h5oh+ho2=c2h5o+h2o22.50E E E E+03 8

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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/m 3 /s, more practically mol dm –3 s –1 9

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10 5Br - + BrO H + = 3Br 2 + 3H 2 O Rate?conc/time or in SI mol dm -3 s -1 – (1/5)(d[Br - ]/dt) = – (1/6) (d[H + ]/dt) = (1/3)(d[Br 2 ]/dt) = (1/3)(d[H 2 O]/dt) Rate law? Comes from experiment Rate = k [Br - ][BrO 3 - ][H + ] 2 where k is the rate constant (variable units) Rate of reaction {symbol: R,n,..}

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11 Rate Law How does the rate depend upon [ ]s? Find out by experiment The Rate Law equation R = k n [A] [B] … (for many reactions) –order, n = + + … (dimensionless) –rate constant, k n (units depend on n) –Rate = k n when each [conc] = unity

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12 Experimental rate laws? CO + Cl 2 COCl 2 Rate = k [CO][Cl 2 ] 1/2 –Order = 1.5 or one-and-a-half order H 2 + I 2 2HI Rate = k [H 2 ][I 2 ] –Order = 2 or second order H 2 + Br 2 2HBr Rate = k [H 2 ][Br 2 ] / (1 + k’ {[HBr]/[Br 2 ]} ) –Order = undefined or none

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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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14 Integration of rate laws Order of reaction For a reaction aA products, the rate law is: rate of change in the concentration of A

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15 First-order reaction

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16 First-order reaction A plot of ln[A] versus t gives a straight line of slope ̶ k A if r = k A [A] 1

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17 First-order reaction

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18 A P assume that -(d[A]/dt) = k [A] 1

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19 Integrated rate equation ln [A] = -k t + ln [A] 0 Slope = -k 1 st Order reaction

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20 Half life: first-order reaction The time taken for [A] to drop to half its original value is called the reaction’s half-life, t 1/2. Setting [A] = ½[A] 0 and t = t 1/2 in:

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21 Half life: first-order reaction

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22 When is a reaction over? [A] = [A] 0 exp{-kt} Technically [A]=0 only after infinite time

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23 Second-order reaction

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24 Second-order reaction A plot of 1/[A] versus t gives a straight line of slope k A if r = k A [A] 2

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25 Second order test: A + A P Slope = k 2 nd Order reaction

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26 Half-life: second-order reaction

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Kinetics and equilibrium kinetics equilibrium 27

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Initial Rate Method 5 Br - + BrO H + 3 Br H 2 O General example: A + B +… P + Q + … Rate law: rate = k [A] [B] …?? log R 0 = 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 R 0 Plot log R 0 (Y-axis) versus log [A] 0 (X-axis) Slope 28

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Example: R 0 = k [NO] [H 2 ] 2 NO + 2H 2 N 2 + 2H 2 O Expt.[NO] 0 [H 2 ] 0 R 0 – × – × – × Deduce orders wrt NO and H 2 and calculate k. Compare experiments #1 and #2 Compare experiments #1 and #3 Now, solve for k from k = R 0 / ([NO] [H 2 ] ) 29

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How to measure initial rate? Key: - (d[A]/dt) - ( [A]/ t) ( [P]/dt) A + B + … P + Q + … t= 0 0 mol m 3 10 s 1 1 ditto Rate? (100-99)/10 = mol m 3 s 1 +(0-1)/10 = mol m 3 s 1 Conclusion? Use product analysis for best accuracy. 30

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Isolation / flooding IO I H + 3 I H 2 O Rate = k [IO 3 - ] [I - ] [H + ] … –Add excess iodate to reaction mix –Hence [IO 3 - ] 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 31

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32 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 + Qrate = k 1 [A] 1 – A + B C + Drate = k 2 [A] 1 [B] 1 –2A + B E + F + Grate = k 3 [A] 2 [B] 1

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33 Molecularity of elementary reactions? Unimolecular (decay) A P –(d[A]/dt) = k 1 [A] Bimolecular (collision) A + B P –(d[A]/dt) = k 2 [A] [B] Termolecular (collision) A + B + C P –(d[A]/dt) = k 3 [A] [B] [C] No other are feasible! Statistically highly unlikely.

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CO + Cl 2 X COCl 2 34 Experimental rate law: –(d[CO]/dt) = k [CO] [Cl 2 ] 1/2 –Conclusion?: reaction does not proceed as written –“Elementary” reactions; rxns. that proceed as written at the molecular level. Cl 2 Cl + Cl(1) Cl + CO COCl(2) COCl + Cl 2 COCl 2 + Cl(3) Cl + Cl Cl 2 (4) –Steps 1 thru 4 comprise the “mechanism” of the reaction. decay collisional

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35 - (d[CO]/dt) = k 2 [Cl] [CO] If steps 2 & 3 are slow in comparison to 1 & 4 then, Cl 2 ⇌ 2Cl or K = [Cl] 2 / [Cl 2 ] So [Cl] = K × [Cl 2 ] 1/2 Hence: - (d[CO] / dt) = k 2 × K × [CO][Cl 2 ] 1/2 Predict that: observed k = k 2 × K Therefore mechanism confirmed (?)

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H 2 + I 2 2 HI Predict: + (1/2) (d[HI]/dt) = k [H 2 ] [I 2 ] But if via: – I 2 2 I –I + I + H 2 2 HI rate = k 2 [I] 2 [H 2 ] – I + I I 2 Assume, as before, that 1 & 3 are fast cf. to 2 Then: I 2 ⇌ 2 I or K = [I] 2 / [I 2 ] Rate = k 2 [I] 2 [H 2 ] = k 2 K [I 2 ] [H 2 ] (identical) Check? I 2 + h 2 I (light of 578 nm) 36

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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 [A] / mmol dm Show that the reaction is 1 st order in azomethane & determine the rate constant at this temperature. 37

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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 ln [A] Plot ln [A] along y-axis; t along x-axis Is it linear? Yes. Conclusion follows. Calc. slope as: so k = × min -1 38

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More recent questions … Write down the rate of rxn for the rxn: C 3 H O 2 = 3 CO H 2 O for both products & reactants[8 marks] For a 2 nd 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 NO 2 + NO 2 N 2 O 4 referred to as a bimolecular rxn?[3 marks] 39

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