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Chem 300 - Ch 29/#2 Today’s To Do List

Chem 300 - Ch 29/#2 Today’s To Do List. Steady-State Approximation Finding a Unique Mechanism? Unimolecular Reactions. The Rate-Determining Step. When 1 step is much slower than the rest ( bottle-neck ). This step will dominate the rate. Example: NO 2 + CO  NO + CO 2

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Chem 300 - Ch 29/#2 Today’s To Do List

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  1. Chem 300 - Ch 29/#2 Today’s To Do List • Steady-State Approximation • Finding a Unique Mechanism? • Unimolecular Reactions

  2. The Rate-Determining Step • When 1 step is much slower than the rest (bottle-neck). • This step will dominate the rate. • Example: NO2 + CO  NO + CO2 • NO2 + NO2 ==> NO3 + NO (k1) (slow) • NO3 + CO ==> NO2 + CO2 (k2) (fast) (k2>>k1) • Rate = k1[NO2]2 (also the experimental law.)

  3. Steady-State Approximation • Consider the consecutive reaction: • A ==> I ==> P • When [A] = [A]0 [I]0 = [P]0 = 0 • Recall: • [I] = {k1[A]0/(k2 – k1)}(e-k(1)t – e-k(2)t) • How [I] varies with t depends on k1/k2

  4. A ==> I ==> P(a) k1=10k2 (b) k2=10k1Steady-State holds in (b)

  5. The SS Approx: d[I]/dt = 0 • Recall the rate laws: • d[A]/dt = -k1[A] • d[I]/dt = k1[A] –k2[I] = 0 • d[P]/dt = k2[I] • [I]ss = k1[A]/k2 • = (k1[A]0/k2)e-k(1)t • d[P]/dt = k2[I]ss = k1[A]0 e-k(1)t • [P] = k1[A]0 e-k(1)tdt = (1 - e-k(1)t )[A]0

  6. Derive Rate Law for 2O33O2 • Possible (more complicated) Mechanism: • O3 ==> O2 + O (k1) • O2 + O ==> O3 (k-1) • O + O3 ==> 2O2 (k2) • Rate Law: • d[O3]/dt = -k1[O3] + k-1[O2][O] – k2[O][O3] • For the SS intermediate (O): • d[O]/dt = k1[O3] – k-1[O2][O] – k2[O][O3] = 0 • [O]ss = k1[O3]/(k-1[O2] + k2[O3])

  7. Continued[O]ss = k1[O3]/(k-1[O2] + k2[O3]) • d[O3]/dt = -k1[O3] + (k-1[O2] – k2[O3])[O] • = -k1[O3] + (k-1[O2] – k2[O3]) k1[O3]/(k-1[O2] + k2[O3]) • = -2k1k2[O3]2/(k-1[O2] + k2[O3]) • SS assumption simplified the math • But still more complex rate law.

  8. Analysis of a “simple” reaction2 NO + O2 2 NO2 • Experim rate law: • Rate = kobs[NO]2[O2] termolecular? • Ea < 0 weird and unrealistic • 2 alternative possible mechanisms. • Fast equilib + slow rate-determ step. • Formation of an intermediate & SS condition valid.

  9. Mechanism 1 • NO + O2<==> NO3 (k1 k-1) fast • NO3 + NO ==> 2 NO2 (k2) rate determ • From 1st (equilibr) • K1 = k1/k-1 = [NO3]/([NO][O2] • [NO3] = K1[NO][O2] • From 2nd (rate law) • (1/2)d[NO2]/dt = k2[NO3][NO] • = k2K1[NO]2[O2] • Consistent with exper rate but kexp not for a single step

  10. Mechanism 2 • NO + NO <==> N2O2 (SS) • N2O2 + O2 ==> 2 NO2 • Rate law: • Rate (NO2) = k2[N2O2][O2] • SS condition: • d[N2O2]/dt = -k-1[N2O2] –k2[N2O2][O2] + k1[NO]2= 0 • [N2O2] = k1[NO]2/(k-1 + k2[O2]) • SS requires back rate >> rate(1) and rate(2) • [N2O2] = k1[NO]2/k-1 • Rate = k2[N2O2][O2] = k2K1[NO]2[O2]

  11. Analysis • Because exper rate suggests termolecular single step doesn’t mean it is. • Both proposed rate laws agree with experimental rate law. • More experiments to detect the intermediates needed to choose between rate laws. • Exper rate law doesn’t imply a unique mechanism.

  12. Unimolecular Reactions • CH3NC ==> CH3CN • Rate = -k[CH3NC] • Valid at high conc • But at low conc Rate = -k[CH3NC]2 • How come?? Is this really an elementary reaction? • Lindemann: Probably not.

  13. Lindemann-HinshelwoodUnimolecular Mechanism

  14. Lindemann Mechanism • A + M <==> A* + M • A* ==> B • Rate (B) = k2[A*] • SS condition: • d[A*]/dt = 0 = k1[A][M] – k-1[A*][M] – k2[A*] • [A*] = k1[M][A]/(k2 + k-1[M]) • Rate = k2 k1[M][A]/(k2 + k-1[M]) = k’[A]

  15. Rate = k2 k1[M][A]/(k2 + k-1[M]) = k’[A] • At high conc: k2 << k-1[M]) • Rate = k ‘ [A] • At low conc: k2 >> k-1[M]) • Rate = k1[M][A]

  16. CH3NC  CH3CN klc =k1[M] khc = k1k2/k-1

  17. Next Time • Chain Reactions • Effect of a Catalyst • Enzyme Catalysis

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