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 l When 1 step is much slower than the rest (bottle-neck). l This step will dominate the rate. l Example: NO 2 + CO  NO + CO 2 NO 2 + NO 2 ==> NO 3 + NO (k 1 ) (slow) NO 3 + CO ==> NO 2 + CO 2 (k 2 ) (fast) (k 2 >>k 1 ) Rate = k 1 [NO 2 ] 2 (also the experimental law.)

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

4. A ==> I ==> P (a) k 1 =10k 2 (b) k 2 =10k 1 Steady-State holds in (b)

5. The SS Approx: d[I]/dt = 0 l Recall the rate laws: d[A]/dt = -k 1 [A] d[I]/dt = k 1 [A] –k 2 [I] = 0 d[P]/dt = k 2 [I] l [I] ss = k 1 [A]/k 2 l = (k 1 [A] 0 /k 2 ) e -k(1)t l d[P]/dt = k 2 [I] ss = k 1 [A] 0 e -k(1)t l [P] = k 1 [A] 0  e -k(1)t dt = (1 - e -k(1)t )[A] 0

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

7. Continued [O] ss = k 1 [O 3 ]/(k -1 [O 2 ] + k 2 [O 3 ]) l d[O 3 ]/dt = -k 1 [O 3 ] + (k -1 [O 2 ] – k 2 [O 3 ])[O] = -k 1 [O 3 ] + (k -1 [O 2 ] – k 2 [O 3 ]) k 1 [O 3 ]/(k -1 [O 2 ] + k 2 [O 3 ]) = -2k 1 k 2 [O 3 ] 2 /(k -1 [O 2 ] + k 2 [O 3 ]) l SS assumption simplified the math l But still more complex rate law.

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

9. Mechanism 1 l NO + O 2 NO 3 (k 1 k -1 ) fast l NO 3 + NO ==> 2 NO 2 (k 2 ) rate determ l From 1 st (equilibr) K 1 = k 1 /k -1 = [NO 3 ]/([NO][O 2 ] [NO 3 ] = K 1 [NO][O 2 ] l From 2 nd (rate law) (1/2)d[NO 2 ]/dt = k 2 [NO 3 ][NO] = k 2 K 1 [NO] 2 [O 2 ] Consistent with exper rate but k exp not for a single step

10. Mechanism 2 l NO + NO N 2 O 2 (SS) l N 2 O 2 + O 2 ==> 2 NO 2 l Rate law: Rate (NO 2 ) = k 2 [N 2 O 2 ][O 2 ] l SS condition: d[N 2 O 2 ]/dt = -k -1 [N 2 O 2 ] –k 2 [N 2 O 2 ][O 2 ] + k 1 [NO] 2 = 0 [N 2 O 2 ] = k 1 [NO] 2 /(k -1 + k 2 [O 2 ]) SS requires back rate >> rate(1) and rate(2) [N 2 O 2 ] = k 1 [NO] 2 /k -1 Rate = k 2 [N 2 O 2 ][O 2 ] = k 2 K 1 [NO] 2 [O 2 ]

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

12. Unimolecular Reactions l CH 3 NC ==> CH 3 CN Rate = -k[CH 3 NC] Valid at high conc But at low conc Rate = -k[CH 3 NC] 2 How come?? Is this really an elementary reaction? l Lindemann: Probably not.

13. Lindemann-Hinshelwood Unimolecular Mechanism

14. Lindemann Mechanism l A + M A * + M l A * ==> B l Rate (B) = k 2 [A * ] l SS condition: d[A * ]/dt = 0 = k 1 [A][M] – k -1 [A * ][M] – k 2 [A * ] [A * ] = k 1 [M][A]/(k 2 + k -1 [M]) Rate = k 2 k 1 [M][A]/(k 2 + k -1 [M]) = k’[A]

15. l At high conc: k 2 << k -1 [M]) Rate = k ‘ [A] l At low conc: k 2 >> k -1 [M]) Rate = k 1 [M][A]

16. CH 3 NC  CH 3 CN k lc =k 1 [M] k hc = k 1 k 2 /k -1

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