CHEM 106 1nd Edition Lecture 15Outline of Last Lecture I. Arrhenius EquationII. Collision TheoryIII. Potential Energy ProfileIV. Reaction MechanismsOutline of Current Lecture II. Elementary Stepsa. ExamplesIII. CatalysisCurrent LectureChapter 15: Chemical Kinetics (continued)Elementary Steps:1: 2NO2 → NO + O32: NO3 → NO + O2Unimolecular: involves a single moleculeBimolecular: 2 molecules reactingTermolecular: 3 molecules reactingStep 1: (first transition state) formation of activated complex Fig. 15.212NO2 → NO + NO3NO3 is intermediateStep 2: (second transition state) a different activated complexNO3 → NO + O2Fig. 15.22 pg 735Activation energies for each of the two elementary steps.Intermediate NO3 is a local energy minimum. Each step has its own Ea for each direction.Small Ea → large kLarge Ea → small kRemember: proposed mechanism must be consistent with rate law derived from experimental data. Recall tbl. 15.9 pg 721 from last weekRate = k [NO2]2 Is proposed mechanism consistent with this rate law?Note that the reaction rates for each of the two elementary steps not necessarily equalRate determining state: slowest of the two elementary steps, controls overall reaction rate. For any mechanism, if 1st step is slow, all others fast, then 1st step controls the overall rate. Ex: step 1: A → B slowStep 2: B + A → C fastOverall: 2A → CPredicted rate law would be: rate = k1[A]; k for step 1Use this approach for NO2 decompositionStep 1: 2NO2 → NO + O3 slowStep 2: NO3 → NO + O2 fastSince Ea (step 1) >> Ea (step 2), the rate law is: rate = k [NO2]2Is consistent with experiment! Now let’s consider the reverse reaction:2 NO + O2 → 2NO2Step 1: NO + O2 (k1) → NO3Step 2: NO3 + NO (k2) → 2NO2 (proposed mechanism)Last week, we looked at experimental data (Tbl 15.3, pg 712) to determine the following:Rate = k [NO]2 [O2]Based on our elementary steps above, we can write the following rate laws:Step 1: rate = k1 [NO] [O2]Step 2: rate = k2 [NO] [NO3]But we know from Ea’s in Fig. 15.22 that the 2nd step is slowest, and its rate law is not consistent with experimental rate lawNote that NO3 is an unstable intermediate, so difficult to know its concentration.If step 1 is reversible, that simplifies things:NO + O2 ↔ (kf,kf) NO3Rate forward = kf [NO] [O2]Rate reverse= kr [NO3]If step 1 is reversible and at equilibrium:Rate forward = rate reverseSo, kf [NO] [O2] = kr [NO3] (solve for [NO3])[NO3] = kfkr [NO] [O2] (can now substitute for [NO3])Consider step 2 is irreversible:Rate step 2 = k2 [NO] [NO3] (substitute)Rate 2 = k2 [NO] (kfkr [NO] [O2] )Rate 2 = kfk2kr [NO]2 [O2]Rate = koverall [NO]2 [O2] (3rd order overall)This rate expression is consistent with the rate law obtained from experimental data!This is a possible mechanism.Note that the observed rate constant is not the rate constant for any single step in the mechanisms.The proposed mechanism is also consistent with energy profile (Fig. 15.22) in reverse!Another example: Zeroth orderNO2 + CO → CO2 + NORate = k [NO2]2so zeroth order in [CO]Possible mechanism:Step 1: 2NO2 → NO + NO3Step 2: NO3 + CO → CO2 + NOIf step 1 is the rate determining step, thenRate = k [NO2]2Consistent with observed rate law 15.6 CatalysisCatalyst: substance added to reaction to increase the reaction rate, but is not consumedEnzymes as catalysts (Fig. 15.25, pg 742)Reactants same, products same, lower energy pathwayHomogenous catalyst: same phase gas or solutionHeterogeneous catalyst: solid phaseConsider the following:2O3 → 3O2One mechanism: O3 → O2 + OO + O3 → 2O2But there is another reaction:CCl3F → CCl2F + ClThen:Cl + O3 → ClO + O2ClO + O3 → Cl + 2O2Overall: 2O3 → 3O2This reaction is catalyzed by ClIntermediate is ClOCatalytic converter: heterogeneous catalysis Increases rate of conversion of NO to N2 & O2 by lowering