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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 http://www.nuigalway.ie/chem/degrees.htm 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 2 1922 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 1987 2

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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 –10 -15 s atomic nuclei –10 8 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. 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+140.12.00E+041.11E+071.778.08E+03 c2h5oh=ch2oh+ch32.00E+23-1.689.64E+048.38E+14-0.227.02E+03 c2h5oh=c2h5+oh2.40E+23-1.629.95E+049.00E+15-0.244.65E+03 c2h5oh=ch3cho+h27.24E+110.19.10E+044.91E+070.997.50E+04 c2h5oh+o2=pc2h4oh+ho22.00E+1305.28E+042.19E+100.284.43E+02 c2h5oh+o2=sc2h4oh+ho21.50E+1305.02E+041.95E+110.094.88E+03 c2h5oh+oh=pc2h4oh+h2o1.81E+110.47.17E+024.02E+080.921.79E+04 c2h5oh+oh=sc2h4oh+h2o6.18E+100.5-3.80E+021.63E+090.832.39E+04 c2h5oh+oh=c2h5o+h2o1.50E+100.82.53E+037.34E+090.911.72E+04 c2h5oh+h=pc2h4oh+h21.88E+033.27.15E+033.93E-013.839.48E+03 c2h5oh+h=sc2h4oh+h28.95E+042.533.42E+032.21E+022.971.28E+04 c2h5oh+h=c2h5o+h25.36E+042.534.41E+032.47E+032.744.19E+03 c2h5oh+ho2=pc2h4oh+h2o22.38E+042.551.65E+042.88E+032.482.83E+03 c2h5oh+ho2=sc2h4oh+h2o26.00E+1201.60E+048.59E+12-0.269.42E+03 c2h5oh+ho2=c2h5o+h2o22.50E+1202.40E+046.66E+13-0.487.78E+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 3 - + 6H + = 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 3 - + 6 H + 3 Br 2 + 3 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 –1 25 102.4 × 10 -3 –2 25 51.2 × 10 -3 –3 12.5 100.6 × 10 -3 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=0 100 100 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. 30

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Isolation / flooding IO 3 - + 8 I - + 6 H + 3 I 3 - + 3 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 0 30 60 90120 [A] / mmol dm 3 8.706.524.893.672.75 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 0 30 60 90120 ln [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 -1 38

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More recent questions … Write down the rate of rxn for the rxn: C 3 H 8 + 5 O 2 = 3 CO 2 + 4 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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Chapter 14 Kinetics Chapter 10 provided an introduction to kinetics and equilibrium. In this chapter we expand the quantitative treatment of chemical kinetics.

Chapter 14 Kinetics Chapter 10 provided an introduction to kinetics and equilibrium. In this chapter we expand the quantitative treatment of chemical kinetics.

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