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## Chemical kinetics or dynamics

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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 athttp://www.nuigalway.ie/chem/degrees.htm What is kinetics all about?**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 + O21922 TEL; 1930 CFCs • O + O3® 2O2 • ‘Man-made’ CCl2F2 + hu ® Cl + CClF2 • Cl + O3® Cl¾O + O2 • Cl¾O+ 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**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. Chemical kinetics**Thermodynamics vs kinetics**Kinetics determines the rate at which change occurs**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**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**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**Integration of rate laws**• Order of reaction For a reaction aA products, the rate law is: rate of change in the concentration of A**First-order reaction**A plot of ln[A] versus t gives a straight line of slope -kA if r = kA[A]1**Half life: first-order reaction**• 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]0and t = t1/2 in:**When is a reaction over?**• [A] = [A]0 exp{-kt} Technically[A]=0only after infinite time**Second-order reaction**A plot of 1/[A] versus t gives a straight line of slope kA if r = kA[A]2**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) • SlopeÞ a**Example: R0 = k [NO]a[H2]b**2 NO + 2 H2® N2 + 2 H2O • 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 a Now, solve for k from k = R0 / ([NO]a[H2]b)**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.**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**Rate law for elementary reaction**• Law of Mass Action applies: • rate of rxnµproduct 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**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.**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 CO + Cl2´COCl2**- (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: observedk = 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)**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 8.70 6.52 4.89 3.67 2.75 • Show that the reaction is 1st order in azomethane & determine the rate constant at this temperature.**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 90 120 ln [A] 2.16 1.88 1.59 1.30 1.01 • 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**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]