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Mechanisms!

•Exam 1 today, 5-6pm, Old Chem 111

•Bring:–ID (catcard)!–calculator

•Know your section!

General Chemistry IIGeneral Chemistry IICHEM 152CHEM 152

Week 4

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Week 4 Reading Assignment

Chapter 13 – Section 13.6 (mechanisms)

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Reaction Reaction MechanismsA Microscopic View of ReactionsA Microscopic View of ReactionsReaction Reaction Mechanisms

A Microscopic View of ReactionsA Microscopic View of Reactions

Mechanism: how reactants are converted Mechanism: how reactants are converted to products at the molecular level.to products at the molecular level.

RATE LAW RATE LAW MECHANISMMECHANISMexperiment experiment theorytheory

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Isomerization Simulation

Mechanism and Order A(blue) → B(red)

An isomerization reaction occurs in

a single step (mechanism).

This is a “unimolecular

” process.

What is the order of this

reaction?

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Reaction MechanismsReaction Mechanisms

UNIMOLECULARUNIMOLECULAR: the structure of a : the structure of a single particle rearranges to produce single particle rearranges to produce a different particle or particles. a different particle or particles.

Isomerization Isomerization

Reaction Order?

Rate= k[A] 1st-Order

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UNIMOLECULAR processUNIMOLECULAR processConversion of trans to cis butene Conversion of trans to cis butene

Activation energy barrier

ONE moleculechanging

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Dimerization simulation

This is a “bimolecul

ar” process.

Mechanism and Order 2A(blue) → A2(red)

What is the order of the

reaction?

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What is the order of the step?

1.12.23.04.what the....?

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Reaction Mechanisms

BIMOLECULAR: Two particles collide and rearrange into products

Reaction Order?

RATE= k[A][B]2nd Order

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Elementary Step

Molecularity

Rate Law

A product

2A product

A + B product

2A + B product

Unimolecular

Bimolecular

Bimolecular

Termolecular

Rate = k [A]

Rate = k [A]2

Rate = k [A][B]

Rate = k [A]2[B]

Rate Laws for General Elementary Steps

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PLAN:

Determining Molecularity and Rate Laws for Elementary Steps

The following two reactions are proposed as elementary steps in the mechanism of an overall reaction:

(1) NO2Cl(g) NO2(g) + Cl (g)

(2) NO2Cl(g) + Cl (g) NO2(g) + Cl2(g)

(a) Write the overall balanced equation.

(b) Determine the molecularity of each step.

(a) The overall equation is the sum of the steps.(b) The molecularity is the sum of the reactant particles in the step.

(c) Write the rate law for each step.

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SOLUTION:

2NO2Cl(g) 2NO2(g) + Cl2(g)

rate2 = k2 [NO2Cl][Cl]

(1) NO2Cl(g) NO2(g) + Cl (g)(2) NO2Cl(g) + Cl (g) NO2(g) + Cl2(g)

(a)

Step(1) is unimolecular.Step(2) is bimolecular.

(b)

rate1 = k1 [NO2Cl](c)

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The Rate-Determining Step of a Reaction Mechanism

The overall rate of a reaction is related to the rate of the slowest, or rate-determining step.

Correlating the Mechanism with the Rate Law

The elementary steps must add up to the overall equation.

The elementary steps must be physically reasonable.The mechanism must correlated with the rate law.

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The mechanism for the gas-phase reaction 2NO + Cl2 → 2NOCl is suggested to be:

(1) NO + NO → N2O2   (slow)

(2) N2O2 + Cl2 → 2NOCl    (fast)

Derive the rate law from this mechanism

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Consider the following mechanism for the oxidation of bromide ions by hydrogen peroxide in aqueous acid solution.

(1) H+ + H2O2 H3O2+     (fast)

(2) H3O2+ + Br¯ → HOBr + H2O    (slow)

(3) HOBr + H+ + Br¯ → Br2 + H2O   (fast)

What is the overall reaction equation and rate law?Why do we not want intermediates in the rate law if it can be avoided?

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What would be a reasonable rate law for a reaction with the following mechanism?

(1) NO(g) + Br2(g) NOBr2(g) fast

(2) NOBr2(g) + NO(g) 2NOBr(g) slow

What is the overall reaction equation and rate law?

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Reaction energy diagram for the two-step NO2-F2

reaction

NO2+F2 NO2F + F

F + NO2 NO2F

RATE LAW FOR THE

REACTION?

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Reaction energy diagram for the two-step NO2-F2

reaction

NO2+F2 NO2F + F

F + NO2 NO2F

2NO2 + F2 → 2NO2F

Slow Rate = k[NO2][F2] = Rate overall

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Consider the reaction O3 + NO O2 + NO2

where the rate law was found experimentally to be

rate = k[O3][NO]

Which of the following mechanisms is consistent with the rate law?

(1) O3 + NO O + NO3 (slow) O + O3 2O2 (fast) NO3 + NO 2NO2 (fast)

(2) O3 O2 + O (fast) NO + O NO2 (slow)

(3) O3 + NO O2 + NO2

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SN2 MechanismKinetics

CH3I + OH– CH3OH + I–

find: Rate = k[CH3I][OH–], bimolecular

Both CH3I and OH– involved in rate-limiting step

reactivity: R-I > R-Br > R-Cl >> R-F

C-X bond breaking involved in rate-limiting step

single-step mechanism:

CH3I + OH–

CH3OH + I–

[HO---CH3---I]–

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SN2 Mechanism

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SN2 Mechanism

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Group Activity:Molecules can be designed so that they undergo this SN2 faster or slower –

If you want the SN2 reaction to happen – you choose a molecule which can easily undergo the reaction.If you want another reaction to occur other than the SN2 reaction – you choose a molecule which is harder to have the SN2 reaction occur.

How would you alter the following molecule so that the SN2 reaction is LESS LIKELY to occur? (same conc./T)

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Group Activity:Draw 2 reaction coordinate diagrams to express the steric effect difference.

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SN2 MechanismSteric effects

Compound Rel. Rate

Methyl CH3Br 150

1º RX CH3CH2Br 1

2º RX (CH3)2CHBr 0.008

3º RX (CH3)3CBr ~0

increasingsteric hindrance

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SN2 MechanismSteric effects

C Br

H

HH

I

H C

H

C Br

CHH

H

H

H

H

CHI

minimal steric hindrance

maximum steric hindrance

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Summary Activity:An important reaction in the formation of smog is the reaction between ozone and NO

NO(g) + O3(g) NO2(g) + O2(g)Experimentally, the [NO] was monitored and the ln[NO] vs time was fairly linear. In another experiment, the O3 concentration was tripled – and the rate also went up by a factor of 3. The rate constant for the reaction is 80. L/(mol s) at 25 C.If this reaction were to occur in a single step, would the rate law be consistent with the observed order of the reaction?

What is the rate of the reaction at 25 C when [NO] = 3.0 x 10-6 M and [O3] = 5.0 x 10-9 M?

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Is the rate law consistent with a single step?

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What is the rate of the reaction at 25 ºC when [NO] = 3.0 x 10-6 M

and [O3] = 5.0 x 10-9 M?

Enter scientific notation in the format: #.#E#

rate = ___________ M/s