chem 2880 - kinetics kinetics - mechanisms iii - mechanis… · chem 2880 - kinetics 104 for the...

CHEM 2880 - Kinetics

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Kinetics - Mechanisms

Tinoco Chapter 7


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Reaction Mechanisms

• many reactions occur in a series of steps calledelementary reactions

• elementary reactions involve only one or twomolecules and occur in one step

• they are denoted by writing the chemical equationwithout physical states

• for example the elementary reaction:

2H + Br 6 HBr + Brspecifies that a hydrogen atom reacts with a brominemolecule to produce a hydrogen bromide moleculeand a bromine atom

• the rate laws for elementary reactions can bedetermined by inspection. For the reaction above:

2v = k[H][Br ]

• a reaction mechanism is a series of elementaryreactions the net result of which is the stoichiometricreaction (for stoichiometric reactions the physicalstates are included)

• stoichiometric reactions do not imply any particularmechanism nor rate law

• kinetic data cannot tell us what the mechanism of areaction is, but it can provide useful information fordeducing a possible mechanism


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Molecularity

The molecularity of an elementary reaction indicates thenumber of particles reacting. Molecularity applies only toelementary reactions and not to stoichiometric reactions.

Unimolecular reactions • one particle rearranges (isomerization) or falls apart

(radioactive decay)A 6 products v = k[A]

• uncommon because most reactions require anactivation energy provided by the collision ofparticles

Bimolecular• two particles collide and exchange energy, atoms, or

groups of atoms or undergo some other change2A 6 products v = k[A]2

A + B 6 products v = k[A][B]• examples: nucleophilic substitutions in organic

chemistry, enzyme-catalyzed reactions (enzyme +substrate)

Termolecular• three particles collide• more likely to occur at high pressures or in solution

2A + B 6 products v = k[A] [B]2


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Very important to note:

The molecularity of an elementary reaction implies therate law for that reaction i.e. a bimolecular reaction hassecond order kinetics. However the reverse is not true. Second order kinetics could indicate a bimolecularreaction or the reaction may be more complex.

In other words:

The mechanism does provide the rate law, but the ratelaw does not provide the mechanism.

Rate-Determining Step

Some reactions consist of only one elementary step andthe rate law can be determined by inspection.

However most reactions and processes in nature are muchmore complex, consisting of a series of elementaryreactions. In this case there may be one step (usually theslowest) which determines the overall kinetics of thereaction, this is the rate-determining step.


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For a complicated mechanism we would be interested in:

• the sequence and type of elementary steps• reaction intermediates (energy minima)• structures and energies of transition states (energy

maxima)• energy barriers involved


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Types of Reactions

Parallel

Parallel reactions are of the following type:

1 2Species A can undergo reaction to B via k or to C via k . The rate expression for the reaction of A is:

The integrated form of this equation is:

A decomposes exponentially at a rate determined by thesum of the two individual rate constants.


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The rate expressions for the production of B and C are:

Substituting for [A] from the previous page:

0 0Integrating and setting [B] = [C] = 0

1 2The ratio of [B]/[C] is equal to k /k throughout thereaction.


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Graphically, the progress of the reactions looks like:


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Series

Series reactions are of the following type:

Species A reacts to form B which in turn reacts to form C.

The rate expressions are:

0If [B] = 0 the integrated rate expression is:


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Similarly, for C:

0and with [C] = 0:

The best way to understand these equations is graphically:


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[A] follows a straight forward exponential decay asexpected for a 1 -order reaction.st

[B] starts at 0 and then climbs fairly rapidly as A reacts,eventually reaching a maximum. As [B] increases, therate of formation of C increases and eventually B reacts toC faster than it is formed from A and [B] declines until itreaches 0.

[C] starts at 0 and rises more slowly than [B]. The curvefor C reaches an inflection point when [B] reaches amaximum and then continues to rise to its maximum asthe reaction reaches completion.


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For a series reaction such as this, the slower step will berate-determining. There are two possibilities :

1 2 1 2k >>k or k <<k

1 2If k >>k , then the second step is rate-determining. Areacts to B rapidly, and then for most of the reaction will

2undergo 1 -order reaction to C, controlled by k . Thest

integrated rate expression reduces to:

Graphically, [A] drops quickly and [B] rises quickly to arelatively high value. [C] rises more slowly than B.


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1 2If k <<k , then the first step is rate-determining. A slowly

1reacts to B controlled by k , and then B quickly convertsto C. [B] remains small throughout the reaction, andessentially C appears as A disappears. The integrated rateexpression reduces to:


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Both cases have rate expressions for the formation ofproducts that are exponential growth curves, the first

1 2 2where k >>k is dependant on k and the second where

1 2 1k <<k is dependant on k . i.e. 1 -order kinetics withst

rates dependant on the rate constant for the rate-determining step.


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Equilibrium

All reactions approach equilibrium. Some equilibriaheavily favour products, so we consider them to go tocompletion. Others heavily favour reagents and weconsider them to not go at all or in other words thereverse reaction goes to completion.

If we consider all reactions to approach equilibrium, thenwe must consider the reverse reaction:

-1If k is very small, or [B] is kept small, the reversereaction can be ignored. Otherwise the rate expression is:

At equilibrium, [A] is constant, -d[A]/dt = 0 and:

The equilibrium constant is equal to the ratio of theforward and reverse rate constants.


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Complex reactions

Many reactions involve a series of steps with a number ofintermediates between the reagents and products of thestoichiometric reaction. Solving the rate expressions forthese reactions becomes more complex as more steps areadded because of the number of species (reagents,products and intermediates) involved.

One method for simplifying complex reactions is theSteady-State Approximation.

The Steady-State Approximation

The steady-state approximation considers that anyintermediate formed in a series of reactions is veryreactive. They tend to react as quickly as they areformed. Thus after an initial induction period duringwhich the concentrations of the intermediates rise fromzero, the concentrations of all intermediates areunchanging and negligibly small.


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For the reaction:

the intermediate X reacts as quickly as it forms.

A more general form of the steady state approximation isto set the rate of formation of the intermediate to 0:

Thus [X] is not zero, but d[X]/dt is at or close to zero. The plot of [X] over time remains relatively flat and [X]remains relatively unchanged throughout most of thereaction.

This method for simplifying complex reactions does notwork in every case. For the example given under Series

1 2Reactions where k >>k , the steady-state approximationwould not work because [B] changes throughout the


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1 2reaction. However for the case where k <<k , [B] is lowand changes little throughout the reaction and the steady-state approximation would work well.

Using the steady-state approximation on our example

1 2with k <<k also enables us to reach the same rateexpression with fewer steps:

The result is simple first order kinetics wrt A.


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0Assuming [C] = 0, the integrated rate expression isderived:


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Deducing a Mechanism from Kinetic Data

There is no single way to deduce a mechanism fromkinetic data. There will often be more than one possiblemechanism that fits the experimental data, and moreexperiments may be required to support one over theother.

In general, it is best to start with a simple mechanism thatfits all the data available and calculate its rate expression. Any proposed mechanism will have a unique rateexpression which may or may not match the empiricalrate expression.


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An example from Tinoco (p. 353)

For the stoichiometric reaction:A + B 6 C

the rate expression is found to be:

A proposed mechanism is:

The first step has A reacting with OH which matches-

with the observed rate expression. The second stepaccounts for the observed stoichiometry. Rateexpressions for B and C can be calculated and correlatedwith experimental data.


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A second proposed mechanism is:

The rate-determining step is:

but

Therefore

The first and third steps are fast, so every time N reacts toform P, A reacts to form N and B reacts to form C. Therate expressions are therefore:

chem 2880 - kinetics kinetics - mechanisms iii - mechanis… · chem 2880 - kinetics 104 for the...

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