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  • Atkins & de Paula: Elements of Physical Chemistry: 5eChapter 11: Chemical Kinetics: Accounting for the Rate Laws

  • End of chapter 11 assignmentsDiscussion questions:2, 5

    Exercises:5, 13, 14, 18, 25

    Use Excel if data needs to be graphed

  • Time for me to nag you!Did you:Read the chapter?Work through the example problems?Connect to the publishers website & access the Living Graphs?Examine the Checklist of Key Ideas?Work assigned end-of-chapter exercises?Review terms & concepts that you should remember from previous chapters or courses

  • Reaction schemesAll chemical rxns proceed toward a state of equilibriumThe reverse rxn becomes increasingly important as the rxn proceedsMany rxns proceed through a series of intermediatesThese intermediates may be important

  • Rate LawsRate laws are always determined experimentally.Reaction order is always defined in terms of reactant (not product) concentrations.

    14.2

  • Rate LawsThe order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation.

    The stoichiometric coefficient of the reactant in the balanced chemical equation is the exponent in the K expression.14.2rate = k [F2][ClO2]1

  • Digression: Rate law and KeqKeq is concerned with the initial and final conditions of the rxn (overall)Rate law, order, kinetics are concerned with everything in betweenLet me show you what IUPAC says about the rate law and Keq

  • Rate law and KeqThe equilibrium law or the equilibrium constant are in the first instance derived from the thermodynamics of the reaction, the value of the equilibrium constant Kc depending entirely upon the characteristics of the reactants and these of the reaction products.

    --continued

  • Rate law and KeqRate of reaction and reaction mechanism are kinetic aspects of the reaction. They have nothing to do with the expression for Kc or with the value of Kc. It would be simplistic and imprudent to assume all sorts of unrealistic reaction equations to obtain an expression for Kc.

  • Rate law and KeqQuoted from IUPACs web sitehttp://www.iupac.org/Specifically: http://www.iupac.org/didac/Didac%20Eng/ Didac02/Content /E06%20-%20E07%20-%20E08.htm

  • Order & molecularityThe overall reaction order is the sum of the powers of the concentrations of all of the substances appearing in the experimental rate law for the reaction; hence, it is the sum of the individual orders (exponents) associated with a given reactant (or product). Reaction order is an experimentally determined, not theoretical, quantity, although theory may attempt to predict it.

    Quoted from another book by Atkins

  • Order & molecularityMolecularity is the number of reactant molecules partici-pating in an elementary reaction. This concept has meaning only for an elementary reaction, but reaction order applies to any reaction. In general, reaction order bears no necessary relation to the stoichiometry of the reaction, with the exception of elementary reactions, where the order of the reaction corresponds to the number of molecules participating in the reaction; that is, to its molecularity. Thus for an elementary reaction, overall order and molecularity are the same and are determined by the stoichiometry. Quoted from another book by Atkins

  • Profile of an exothermic rxnFig 11.1 (258)Ea is greater for the reverse rxn than for the forward rxnSo the Rf increases less sharply (than the Rr) as T increasesAs the T is raised, the products are favored

  • The approach to equilibriumDerivation 11.1 (259)You should work through this derivation

  • The approach to equilibriumFig 11.2 (259)The approach to equilibrium of a rxn that is 1st order in both directionsIn this example, k = 2k, so at equilibrium the ratio of b:a is 2:1

    A B

  • Relaxation methodsSome externally applied influence shifts the rxn equilibrium abruptlyTemperature jumpPressure jumpThe adjustment back to equilibrium is called relaxation

  • Relaxation methodsFig 11.3 (259)Relaxation to a new equilibrium following a sud-den change of temperature from T1 to T2

  • Relaxation methodsWhen a sudden temperature jump shifts the equilibrium of a 1st order rxn:

    x = x0et/ and = ka + kb

    x is the shift in equilibrium at the new tx0 is the shift in equilibrium immediately after the temperature change is the relaxation timet is the time (s)

  • Consecutive reactionsCommon examples are radioactive decay chains and biochemical rxn sequences (e.g., gylcoysis, Krebs cycle, etc)

  • The Uranium Decay Series

  • Glycolysis

  • Consecutive reactionsFig 11.4 (263)Concentrations in consecutive rxns, A I PA = reactantI = intermediateP = productAt any time the sum [A] + [I] + [P] is a constant

  • Elementary reactionsMany reactions occur as a series of steps called elementary reactionsMolecularity is the number of molecules coming together to reactUnimolecularBimolecular Whats next?

  • Elementary reactionsFig 11.5 (263)A unimolecular elementary rxn

  • Elementary reactionsFig 11.6 (264)A bimolecular elementary rxn

  • The formulation of rate lawsAssuming that a chemical rxn is a series of elementary rxnsOne of these elementary rxns is the rate determining stepAn acceptable rate law for an overall rxn is expressed only in terms of the concentrations of the species in the overall rxn, not in terms of an intermediate(264)

  • The formulation of rate lawsConsider this rxn:2 NO(g) + O2(g) d 2 NO2(g)Experimentally: rate = k[NO]2[O2] (which is 3rd order overall)How can we explain this? Is it a termolecular rxn (2 NOs and 1 O2)?Lets see

    (264)

  • Proposed mechanism: 2 NO(g) + O2(g) d 2 NO2Step 1. NO + NO d N2O2Rate of formation of N2O2 = ka[NO]2Step 2. N2O2 d NO + NORate of decomposition of N2O2 = ka[N2O2]Step 3. N2O2 + O2 d NO2 + NO2Rate of consumption of N2O2 = 2kb[N2O2][O2]

    Thus the net rate of formation of N2O2 is = ka[NO]2 ka[N2O2] 2kb[N2O2][O2]Does text have an error in this last eqn?

  • Proposed mechanism: 2 NO(g) + O2(g) d 2 NO2Net rate of formation of N2O2 = ka[NO]2 ka[N2O2] 2kb[N2O2][O2]Solving this eqn involves solving a very difficult differential eqn and gives a complex expressionThis drives us to consider The Steady-State Approximation(265)

  • The steady-state approximationOur assumption: the concentrations of all intermediates remain constant and small throughout the rxn (except at the very beginning and at the very end)So, the net rate of formation of N2O2 = 0So, ka[NO]2 ka[N2O2] 2kb[N2O2][O2] = 0Solving for [N2O2]Solving for [NO2] gives 11.10, which accounts for overall 3rd order rxn (265)

  • The rate-determining stepIf step 3 is much faster than steps 1 and 2, (this is the case if the [O2] is high) then ka can be ignoredNow the rate law simplifies to 11.12

    [NO2] = = 2ka[NO]2

    Now the rxn is 2nd order and [O2] does not appear in the rate lawkb [NO]2[O2]2kakb[NO]2[O2]

  • The rate-determining stepThe rate determining step (RDS) must have two characteristics:It must be the slowest step, andIt must be a crucial part of the path from reactants to products

  • The rate-determining stepFig 11.7 (266)Diagrammatic illustration of the RDSHeavy lines = fast stepsThin lines = slow steps

  • The rate-determining stepFig 11.7 (266)Heavy lines = fast stepsThin lines = slow stepsIn (b), the thin line is not a bottleneck because there is an alternate route

  • Kinetic controlConsider these competing rxns:A + B d P1 Rate of formation of P1 = k1[A][B]A + B d P2 Rate of formation of P2 = k2[A][B]

    Represents kinetic control

  • Kinetic controlIf a rxn is allowed to reach equilibrium, then thermodynamics (rather than kinetic considerations) determines the proportions of products, and the ratio of concentrations is controlled by standard Gibbs energies of reactants and products

    (267)

  • Reaction profileFig 11.7 (267)Multistep mechanism in which the first step is the RDS (highest Ea)

  • Unimolecular reactionsMany gas-phase rxns proceed by two gas molecules colliding but are 1st orderThese collisions are bimolecular events and the rxn should be 2nd orderExplain! The Lindemann mechanism posits that the RDS is unimolecular

    (267f)

  • Activation control & diffusion controlNow we consider reactions in solution

  • Activation control & diffusion controlkd is the rate constant (of diffusion)Diffusion-controlled limitthe rate of the rxn is controlled by the rate at which reactants diffuse (kd) Activation-controlled limitthe rate of the rxn depends on the rate at which energy accumulates in the encounter pair (ka)

  • Activation control & diffusion controlNow read next-to-last paragraph, p.269A lesson to learn from this analysis is the concept of the rate-determining step is rather subtle. Thus, in the

  • Activation control & diffusion controlVis-a-vis the rate of diffusion in a liquid, the rate constant (kd) is related to the coefficient of viscosity, How are kd and related?

    kd =

    is the Greek lowercase letter eta

  • DiffusionThis section relates directly to the lab experiment youve worked on this semester

    (270)

  • DiffusionTerms: diffusion, random walk, flux (J)Rate of diffusion Concentration gradient

    J =

    J = D Concentration gradient (11.18b)D is the diffusion coefficient (area/time)This is Ficks first law of diffusion

  • DiffusionFig 11.9 (271)The flux of solute particles is proportional to