Chem 300 - Ch 29/#3 Today’s To Do List Unimolecular Reactions Chain Reactions Effect of a Catalyst Enzyme Catalysis.

Chem 300 - Ch 29/#3 Today’s To Do List

• Unimolecular Reactions • Chain Reactions• Effect of a Catalyst• Enzyme Catalysis

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.

Lindemann-HinshelwoodUnimolecular Mechanism

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]

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]

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

Chain Reactions

Consider: H2 + Br2 2 HBr Experim. Rate Law:

• ½ d[HBr]/dt = k[H2][Br2]1/2/(1 +k’[HBr]/[Br2])

• How does it do that?? It’s a chain reaction mechanism

A Chain Reaction has Several Unique Steps

Initiation: Br2 + M ==> 2 Br + M k1

• (thermal or photochemical)(193 vs 436 kJ)

Propagation: Br + H2 ==> HBr + H k2

H + Br2 ==> HBr + Br k3

Inhibition: HBr + H ==> Br + H2 k-2

HBr + Br ==> H + Br2 k-3

Termination: 2 Br + M ==> Br2 + M k-1

The Rate Laws

d[HBr]/dt = k2[Br][H2] – k-2[HBr][H] + k3[H][Br2] k-3 0

d[H]/dt = k2[Br][H2] – k-2[HBr][H] - k3[H][Br2]

d[Br]/dt = 2k1[Br2][M] – 2k-1[Br]2[M] – k2[Br][H2] + k-2[HBr][H] +k3[H][Br]

Apply SS condition to: • d[H]/dt = d[Br]/dt = 0 • And solve the 2 simultaneously for [H] & [Br]

Results

[Br] = (k1/k-1)1/2[Br2]1/2

[H] = k2K1/2[H2][Br2]1/2/(k-2[HBr]+k3[Br2]) Substitute into rate law for HBr: ½ d[HBr]/dt = k2K1/2[H2][Br2]1/2/{1+(k-2/k3)[HBr]/[Br2]}

Same functional form as experimental law. At the start of the reaction: [HBr] << [Br2]

• ½ d[HBr]/dt = k2K1/2[H2][Br2]1/2

Catalyst & Kinetics

Catalyst • Increases rate • Provides alternate pathway• Is not consumed • Lowers Ea

Homogeneous Catalyst: Same Phase Heterogeneous Catalyst: Catalyst in

different phase

Effect of Catalyst on Ea

Stratospheric Ozone

2 O3 ==> 3 O2

Mechanism (partial):• O3 O2 + O k1

• O2 + O 2 O3 k-1

• O + O3 2 O2 k2

d[O3]/dt = -k1k2[O3]2/(k-1[O2] + k2[O3])

2 O3 ==> 3 O2

d[O3]/dt = -k1k2[O3]2/(k-1[O2] + k2[O3])

But O + O3 2 O2 Is slow

• k2 small

d[O3]/dt -(k1k2 /k-1 )[O3]2/ [O2]

Ozone Depletion

O3 + O ==> 2 O2 slow Homogeneous Catalysis Proposed by

Rowland & Molina (1974): Chlorofluorocarbons (CFCl3 & CF2Cl2)

CFCl3 + h CFCl2 + Cl

• O3 + Cl ==> ClO + O2

• ClO + O ==> O2 + Cl

• ClO + O3 ==> 2O2 + Cl

Antarctic Ozone Hole

Cl + CH4 CH3 + HCl

ClO + NO2 ClONO2

Heterogeneous Catalysis:• HCl(g) + ClONO2(g) Cl2(g) + HNO3(s)

• Occurs on ice surface in polar strat clouds

• In Spring:

• Cl2(g) + h ==> 2 Cl• Cl is regenerated & reacts with O3 & forms ClO & (ClO)2

Enzyme CatalysisMichaelis-Menten Mechanism

-d[S]/dt =k[S]/(Km + [S])• [S] = Substrate (molecule acted on) conc.

E + S <==> ES <==> E + P• -d[S]/dt = k1[E][S] – k-1[ES]• -d[ES]/dt = (k2 + k-1)[ES] –k1[E][S] –k-2[E][P]• d[P]/dt = k2[ES] – k-2[E][P]

[E]0 = [ES] + [E] = constant Substitute & assume SS for [ES]

SS Solution

[ES] = (k1[S] +k-2[P])[E]0/(k1[S] +k-2[P] + k-1 + k2) Substitute into –d[S]/dt Rate = (k1k2[S] – k-1k-2[P]) [E]0/(k1[S] +k-2[P] +k-1+ k2)

Initially: [S] [S]0 & [P] 0• Initial rate = k2[S]0[E]0/(Km + [S]0)

• Km = (k-1 + k2)/k1 = Michaelis constant

Maximum rate = k2[E]0

Turnover Rate = max rate/[E]0 = k2

• Catalase: 4.0 x 107/s

Lineweaver-Burke Plot1/k = 1/k2 + Km/k2[S]

Next Time

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