Consider a mechanism with first step as rate

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Consider a mechanism with first step as rate-determining step.Example: 2 NO2+ F22 NO2FRate law found experimentally to berate=kobs[NO2][F2]Possible mechanism:NO2+F2k1!NO2F+F (slow)NO2+Fk2!NO2F (fast)Overall ratedetermined byfirst stepRate Determining StepReaction of alkyl halides with water. Overall reaction fort-butyl bromide::(CH3)3CBr + H2O(CH3)3COH + H++ BrExperimental rate law: rate =k[(CH3)3CBr]Accepted mechanism:(1) (CH3)3C–Br(CH3)3C++ Br(slow)(2) (CH3)3C++ H2O(CH3)3C–OH2+(fast)(3) (CH3)3C–OH2+H++ (CH3)3C–OH (fast)Show that reaction is consistent with experimental rate law.Why does H2O not appear in the rate law?What are the reaction intermediates?
Intro Chemistry II – 030.102 - Chapter 13Spring, 20084Reaction Mechanisms and RateMechanisms in which rate-determining step is after one or more faststeps often revealed by reaction order > 2, non-integral order, orinverse concentration dependence of a species.Example: 2 NO + O22 NO2Rate law found experimentally to berate=kobs[NO]2[O2]Alternative mechanism:NO+NOk–1k1!N2O2(fast equilibrium)N2O2+O2k2!2 NO2(slow)Rate law consistent with single termolecular step, but these are rare.Reaction Mechanisms and RateSince 2nd step determines overall rate, we can writeHowever, concentration of N2O2intermediate cannot be controlled.since 2nd step is slow, we can assume elementary reactionsbeforerate-determining step are in equilibrium. Hence, for 1st step,rate=k2[N2O2][O2][N2O2][NO]2=k1k–1=K1OR[N2O2]=K1[NO]2Substituting, we have rate=k2K1[NO]2[O2]This is consistent with observed reaction order, withkobs=k2K1.
Intro Chemistry II – 030.102 - Chapter 13Spring, 20085Determining Reaction MechanismsAnother mechanism has been proposed for oxidation of NO:NO+O2k–3k3!NO3(fast equilibrium)NO3+NOk4!2 NO2(slow)Since 2nd step again determines overall rate, we can writerate=k4[NO3][NO]Assume equilibrium in the first step:[NO3][NO][O2]=k3k–3=K3OR[NO3]=K3[NO][O2]Substituting, we have rate=k4K3[NO]2[O2]

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