Lecture7.ppt

IO- + ClTS = {IOClH-}
d[IO# ] [I # ][OCl# ] = k' = k''[I # ][HOCl] dt [OH # ]
OCl- + H2O
!
HOCl + OHHOI + ClH2O + IOIO- + ClSlow (rds)
I- + HOCl OH- + HOl IO- + OCl-
Rate decreases with [H+] !
H: Similar to A: Again Preq of the form (but now X- is low [ ] reactive species): All of these ideas can also HX = X- + H+ I: Similar to B: has [HX] and [X-] in separate rds
10. A sum of n terms in the denominator indicates a succession of n steps with the first n-1 reversible Non integral order REQUIRES intermediates
VERY USEFUL PAPER!
Rate law
Mechanism
You have collected and plotted the data and have the rate law: Now what?
rate = k[A]x [B]y + ......
Some “rules” for interpreting rate laws (i.e. converting the rate ! law to a mechanism) If it is of simple form (x = 1 or 2) you are essentially done:
Some rate law interpretation “rules” Chapter 6 of Espenson and selected papers
1a. The rate law usually gives the composition of the transition state. (Especially for preequilibrium cases; Exception: for chain mechanisms) Tl3+ + Hg22+
5. The first intermediate after the TS (of the rds) contains no atoms not present in the TS (but can contain less)
!
!
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6. Alternative mechanisms that lead to the same formulation of the TS are not kinetically distinguishable
Cl2 + H2C2O4 Observed rate law " = Cl2 + H2O H2C2O4 !
2 CO2 + 2 Cl- + 2 H+
d[CO2 ] [Cl ][H C O ] = k' 2 + 2 2 2 4 dt [H ] [Cl# ]
TS = {ClC2O4-}
HOCl + Cl- + H+ H+ + HC2O4TS = {(H2O)ClC2O4-}
4. Inverse orders are due to rapid preequilibria (usually the reagent of
inverse order complexes the reactive species). The species in the denominator are produced before the rds.
k1[H + ] K + [H + ]
k 2K + k1[H + ] K + [H + ]
both forms reactive (separate rds; k1 for HX and k2 for X-), protonated more reactive. Collapses to A, B, C in various limits! A: K>> [H+], only HX reactive kobs = k1[H+]/K B: K>> [H+], both reactive kobs = k1[H+]/K + k2 C: K≈ [H+], only HX reactive kobs = k1[H+]/(K &[H + ][Cl" ] [Cl2 ]
" [H + ][HC2O4 ] [H 2C2O4 ]
d[CO2 ] " = k[HOCl][HC2O4 ] dt
[HOCl] = K1 [Cl2 ] [H + ][Cl" ]
K a1 = !
" [HC2O4 ] =
!
K a1[H 2C2O4 ] ! + [H ]
kobs = k a [H + ] + kb [H + ]2 kobs = k + k a [H + ] + kb [H + ]2
G
OMG Study something simpler!
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VERY USEFUL PAPER!
H kobs = k1 /[H + ]
I
!
kobs = k1 + k 2 /[H + ]
Also: Read 6.5 and 6.6 of Espenson for different version of same information be extended to other types of rapid preequilibria: See section 6.7 for examples
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9. Mechanisms that have all species at equilibrium prior to the rds cannot be catalyzed or inhibited. (TS is at equilibrium with the all the reactants: can’t catalyze such a reaction)
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2. The number of positive terms in the rate law is the number of independent, parallel pathways to product. (microscopic reversibility). Negative terms represent the reverse reaction. 3 I - + H 2 O2 + 2 H + I3- + 2 H2O
" = k1[I # ][H 2O2 ] + k2 [I # ][H 2O2 ][H + ]
Reaction occurs by 2 separate pathways
!
TS1 is {H2O2I-} and TS2 is {H3O2I}
3. A reaction order greater than 3 means the mechanism has one or more rapid preequilibria (and intermediates) before the rds Elementary steps are either unimolecular or bimolecular
!
8. The sum of all the mechanistic steps must add up to the net reaction (If a reactant stoichiometry coefficient is larger than the order of that reactant, a fast reaction that uses that reactant occurs after the rds) I- + OCl"=
Cl2 + H2C2O4 Observed rate law " = Cl2 Cl+ + Cl-
2 CO2 + 2 Cl- + 2 H+
d[CO2 ] [Cl ][H C O ] = k' 2 + 2 2 2 4 dt [H ] [Cl# ]
TS = {ClC2O4-}
K1 = K a1K a 2 =
"=k
Tl+ + 2 Hg2+ TS is {TlHg3+}
2+ [Tl 3+ ][Hg2 ] [Hg 2+ ]
1b. Remember: If you didn’t vary the concentration of something that actually mattered (the solvent, oxygen, ! water in the solvent, etc.) AND IT DIDN’T CHANGE! then its concentration is included in the rate constant.