organic reaction mechanisms: Reaction Kinetics

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Reaction kinetics,part-5 end

10.Pre-equilibria

A situation that is only slightly more complicated than the sequential reaction scheme describedin our part-3
A + B ---K1--> C-----k2--> D
<--k1-------

The rate equations for this reaction are:

d[A] / dt = d[B] / dt = -k1[A][B] + k-1[C]

d[C] / dt = k1[A][B] - k2[C]

d[D] / dt = k2[C]

These cannot be solved analytically, and in general would have to be integrated numerically to
obtain an accurate solution. However, the situation simplifies considerably if k-1 >> k2. In this case,
an equilibrium is reached between the reactants A and B and the intermediate C, and the
equlibrium is only perturbed very slightly by C ‘leaking away’ very slowly to form the product D.
If we assume that we can neglect this perturbation of the equilibrium, then once equilibrium is
reached, the rates of the forward and reverse reactions must be equal. i.e.

k1[A][B] = k-1[C]

Rearranging this equation, we find
k1 / k-1 = [C] / [A][B] = K

The equilibrium constant K is therefore given by the ratio of the rate constants k1 and k-1 for the
forward and reverse reactions. The rate of the overall reaction is simply the rate of formation of the
product D, so

ν = d[D] / dt = k2[C] = k2K[A][B]

The reaction therefore follows second order kinetics, with an effective rate constant keff = k2K. Note
that this rate law will not hold until the equilibrium between A, B and C has been established, and
so is unlikely to be accurate in the very early stages of the reaction.

11.The steady state approximation

Apart from the two simple examples described above, the rate equations for virtually all complex
reaction mechanisms generally comprise a complicated system of coupled differential equations
that cannot be solved analytically. In state-of-the-art kinetic modelling studies, fairly sophisticated
software is generally used to obtain numerical solutions to the rate equations in order to determine
the time-varying concentrations of all species involved in a reaction sequence. However, very
good approximate solutions may often be obtained by making simple assumptions about the
nature of reactive intermediates.
Almost by definition, a reactive intermediate R will be used up virtually as soon as it is formed, and
therefore its concentration will remain very low and essentially constant throughout the course of
the reaction. This is true at all times apart from at the very start of the reaction, when [R] must
necessarily build up from zero to some small non-zero value, and at the very end of the reaction in
the case of a reaction that goes to completion, when [R] must return to zero. During the period of
time when [R] is essentially constant, because d[R]/dt is so much less than the rates of change of
the reactant and product concentrations, it is a good approximation to set d[R]/dt = 0. This is
known as the steady state approximation.
Steady state approximation: if a reactive intermediate R is present at low and constant
concentration throughout (most of) the course of the reaction, then we can set d[R]/dt = 0 in the
rate equations.
As we shall see, applying the steady state approximation has the effect of converting a
mathematically intractable set of coupled differential equations into a system of simultaneous
algebraic equations, one for each species involved in the reaction. The algebraic equations may
be solved to find the concentrations of the reactive intermediates, and these may then be
substituted back into the more general equations to obtain an expression for the overall rate law.

12.‘Unimolecular’ reactions – the Lindemann-Hinshelwood mechanism

A number of gas phase reactions follow first order kinetics and apparently only involve one
chemical species. Examples include the structural isomerisation of cyclopropane to propene, and
the decomposition of azomethane (CH2N2CH3 → C2H6 + N2, with experimentally determined rate
law ν = k[CH3N2CH3]) The mechanism by which these molecules acquire enough energy to react
remained a puzzle for some time, particularly since the rate law seemed to rule out a bimolecular
step. The puzzle was solved by Lindemann in 1922, when he proposed the following mechanism
for ‘thermal’ unimolecular reactions..

BJ 13:03 Add Comment

13.Third order reactions

A number of reactions are found to have third order kinetics. An example is the oxidation of NO,
for which the overall reaction equation and rate law are given below.

2NO + O2 →2NO2
d[NO2] / dt = k [NO]2 [O2]

One possibility for the mechanism of this reaction would be a three-body collision (i.e. a true
termolecular reaction).

14.Enzyme reactions – the Michaelis-Menten mechanism

An enzyme is a protein that catalyses a chemical reaction by lowering the activation energy.
Enzymes generally work by having an active sitethat is carefully designed by nature to bind a
particular reactant molecule (known as the substrate). An example of a substrate bound at the
active site of an enzyme is shown on the left.
The activation energy of the reaction for the enzyme-bound substrate is lower than for the
free substrate molecule, often due to the fact that the interactions involved in binding shift the
substrate geometry closer to that of the transition state for the reaction. Once reaction has occurred, the product molecules arereleased from the enzyme.

15.Chain reactions

Chain reactions are complex reactions that involve chain carriers, reactive intermediates which
react to produce further reactive intermediates. The elementary steps in a chain reaction may be
classified into initiation, propagation, inhibition, and termination steps.For more details
16.Explosions and branched chain reactions

An explosion occurs when a reaction rate accelerates out of control. As the reaction speeds up,
gaseous products are formed in larger and larger amounts, and more and more heat is generated.
The rapid liberation of heat causes the gases to expand, generating extremely high pressures, and
it is this sudden formation of a huge volume of expanded gas that constitutes the explosion. The
pressure wave travels at very high speeds, often much faster than the speed of sound, and the
‘bang’ associated with an explosion is the result of a supersonic shock wave.
There are two different mechanisms that may lead to an explosion. These are related to the fact
that the overall reaction rate depends on both the magnitude of the rate constant and the amounts
of reactants present in the reaction mixture.
If the heat generated in a reaction due to the reaction exothermicity cannot be dissipated
sufficiently rapidly, the temperature of the reaction mixture increases. This increases the rate
constant, and therefore the reaction rate, producing more heat and accelerating the reaction rate
still further, and so on until an explosion results. Such explosions are known as thermal
explosions,and in principle may occur whenever the rate of heat production by a reaction mixture
exceeds the rate of heat loss to the surroundings (often the walls of the reaction vessel).
The second category of explosions arise from chain branching within a chain reaction, and are
known as chain branching explosions(or sometimes, somewhat misleadingly, isothermal
explosions). In this case, one or more steps in the reaction mechanism produce two or more chain
carriers from one chain carrier, increasing the number of chain carriers, and therefore the overall
reaction rate.
In practice, both mechanisms often occur simultaneously, since any acceleration in the rate of an
exothermic reaction will eventually lead to an increase in temperature. However, chain branching
is not a requirement for an explosion. As an example, detonation of TNT (2,4,6-trinitrotoluene) is
simply the result of an extremely fast chemical decomposition that generates huge quantities of
gas. The reaction 2H2(g) + O2(g) →2H2O(g)provides an example of a reaction in which both
mechanisms are important.

17.Temperature dependence of reaction rates

It is found experimentally that the rate constants for many chemical reactions follow the Arrhenius
equation..

18.Simple collision theory

As the name suggests, simple collision theory represents one of the most basic attempts to
develop a theory capable of predicting the rate constant for an elementary bimolecular reaction of
the form A + B →P. We begin by considering the factors we might expect a reaction rate to
depend upon. Obviously, the rate of reaction must depend upon the rate of collisions between the
reactants. However, not every collision leads to reaction. Some colliding pairs do not have
enough energy to overcome the activation barrier, and any theory of reaction rates must take this
energy requirement into account. Also, it is highlylikely that reaction will not even take place on
every collision for which the energy requirement is met, since the reactants may need to collide in
a particular orientation (e.g. SN2 reactions) or some of the energy may need to be present in a
particular form (e.g. vibration in a bond coupled to the reaction coordinate).

BJ 13:04 Add Comment

5. Integrated rate laws

A rate law is a differential equation that describes the rate of change of a reactant (or product)
concentration with time. If we integrate the rate law then we obtain an expression for the
concentration as a function of time, which is generally the type of data obtained in an experiment.
In many simple cases, the rate law may be integrated analytically....

6.Half lives

The half life, t1/2, of a substance is defined as the time it takes for the concentration of the
substance to fall to half of its initial value...

7.Determining the rate law from experimental data

A kinetics experiment consists of measuring the concentrations of one or more reactants or
products at a number of different times during the reaction

8.Complex reactions

In kinetics, a ‘complex reaction’ simply means a reaction whose mechanism comprises more than
one elementary step. In the previous sections we have looked at experimental methods for
measuring reaction rates to provide kinetic data that may be compared with the predictions of
theory. In the following sections, we will look at a range of different types of complex reactions and
the rate laws that may be predicted from their kinetic mechanisms. Disagreement of a predicted
rate law with the experimental data is enough to rule out the corresponding proposed mechanism,
while agreement inspires some confidence that the proposed mechanism is the correct one. It
should be noted though that agreement between the predicted and measured kinetics is not
always enough to assign a mechanism. The proposed mechanism must be able to account for all
other properties of the reaction, which may include quantities such as the product distribution,
product stereochemistry, kinetic isotope effects, temperature dependence, and so on.
The types of complex mechanisms that we will cover are: consecutive (or sequential) reactions;
competing reactions; pre-equlibria; unimolecular reactions; third order reactions; enzyme reactions;
chain reactions; and explosions.

9.Consecutive reactions

If the rate constants for the following reaction are $k_1$ and $k_2$ ; $A \rightarrow \; B \rightarrow \; C$ , then the rate equation is:

For reactant A:
$\frac{d[A]}{dt} = -k_1 [A]$

For reactant B:
$\frac{d[B]}{dt} = k_1 [A] - k_2 [B]$

For product C:

$\frac{d[C]}{dt} = k_2 [B]$

BJ 06:31 Add Comment

3.Rate laws

The rate law is an expression relating the rate of a reaction to the concentrations of the chemical
species present, which may include reactants, products, and catalysts. Many reactions follow a
simple rate law, which takes the form

ν = k [A]a[B]b[C]c

i.e. the rate is proportional to the concentrations of the reactants each raised to some power. The
constant of proportionality, k, is called the rate constant. The power a particular concentration is
raised to is the order of the reaction with respect to that reactant. Note that the orders do not have
to be integers. The sum of the powers is called the overall order. Even reactions that involve
multiple elementary steps often obey rate laws of this kind, though in these cases the orders will
not necessarily reflect the stoichiometry of the reaction equation. For example,

H2 + I2 → 2HI ν = k [H2][I2]
3ClO− → ClO3 + 2Cl− ν = k [ClO−]2

Other reactions follow complex rate laws. These often have a much more complicated
dependence on the chemical species present, and may also contain more than one rate constant.
Complex rate laws always imply a multi-step reaction mechanism. An example of a reaction with a
complex rate law is

H2 + Br2 → 2HBr ν =[H2][Br2]1/2 / [1 + k'[HBr]/[Br2] ]

In the above example, the reaction has order 1 with respect to [H2], but it is impossible to define
orders with respect to Br2 and HBr since there is no direct proportionality between their
concentrations and the reaction rate. Consequently, it is also impossible to define an overall order
for this reaction.
To give you some idea of the complexity that may underlie an overall reaction equation, a
slightly simplified version of the sequence of elementary steps involved in the above reaction is
shown below.

Br2 → Br + Br
Br + H2 → H + HBr
H + Br2 → Br + HBr
Br + Br → Br2

As well as having rate laws for overall reactions, we can of course also write down individual rate
laws for elementary steps. Elementary processes always follow simple rate laws, in which the
order with respect to each reactant reflects the molecularity of the process (how many molecules
are involved). For example,

Unimolecular decomposition A → B ν = k [A]

Bimolecular reaction A + B → P ν = k [A][B]
                                 A + A → P ν = k [A][A] = k [A]2

Multi-step processes may follow simple or complex rate laws, and as the above examples have
hopefully illustrated, the rate law generally does not follow from the overall reaction equation. This
makes perfect sense, since the overall reaction equation for a multi-step process is simply the net
result of all of the elementary reactions in the mechanism. The ‘reaction’ given in the overall
reaction equation never actually takes place! However, even though the rate law for a multi-step
reaction cannot immediately be written down from the reaction equation as it can in the case of an
elementary reaction, the rate law is a direct result of the sequence of elementary steps that
constitute the reaction mechanism. As such, it provides our best tool for determining an unknown
mechanism. As we will find out later in the course, once we know the sequence of elementary
steps that constitute the reaction mechanism, we can quite quickly deduce the rate law.
Conversely, if we do not know the reaction mechanism, we can carry out experiments to determine
the orders with respect to each reactant (see Sections 7 and 8) and then try out various ‘trial’
reaction mechanisms to see which one fits best with the experimental data. At this point it should
be emphasized again that for multi-step reactions, the rate law, rate constant, and order are
determined by experiment, and the orders are not generally the same as the stoichiometric
coefficients in the reaction equation.
A final important point about rate laws is that overall rate laws for a reaction may contain reactant,
product and catalyst concentrations, but must not contain concentrations of reactive intermediates
(these will of course appear in rate laws for individual elementary steps).

4. The units of the rate constant

A point which often seems to cause endless confusion is the fact that the units of the rate constant
depend on the form of the rate law in which it appears i.e. a rate constant appearing in a first order
rate law will have different units from a rate constant appearing in a second order or third order rate
law. This follows immediately from the fact that the reaction rate always has the same units of
concentration per unit time, which must match the overall units of a rate law in which
concentrations raised to varying powers may appear. The good news is that it is very
straightforward to determine the units of a rate constant in any given rate law.
Below are a few examples.

(i) Consider the rate law

ν = k[H2][I2].

If we substitute units into the equation,
we obtain

(mol dm-3 s-1) = [k] (mol dm-3) (mol dm-3)

where the notation [k] means ‘the units of k’. We can rearrange this expression to
find the units of the rate constant, k.

[k] =(mol dm-3 s-1) / (mol dm-3) (mol dm-3) = mol-1 dm3 s-1

(ii) We can apply the same treatment to a first order rate law,
for example

ν = k [CH3N2CH3].

(mol dm-3 s-1) = [k] (mol dm-3)

[k] = (mol dm-3 s-1) / (mol dm-3) = s-1

(iii) As a final example, consider the rate law

ν = k [CH3CHO]3/2.

(mol dm-3 s-1) = [k] (mol dm-3)3/2

[k] = (mol dm-3 s-1) / (mol dm-3)3/2 = mol-1/2 dm3/2 s-1

An important point to note is that it is meaningless to try and compare two rate constants unless
they have the same units.

BJ 00:39 Add Comment

1. Introduction

Chemical reaction kinetics deals with the rates of chemical processes. Any chemical process maybe broken down into a sequence of one or more single-step processes known either as elementaryprocesses, elementary reactions, or elementary steps.
Elementary reactions usually involve eithera single reactive collision between two molecules, which we refer to as a a bimolecular step, ordissociation/isomerisation of a single reactant molecule, which we refer to as a unimolecular step.
Very rarely, under conditions of extremely high pressure, a termolecular step may occur, which involves simultaneous collision of three reactant molecules. An important point to recognise is that many reactions that are written as a single reaction equation in actual fact consist of a series of elementary steps. This will become extremely important as we learn more about the theory of chemical reaction rates.

As a general rule, elementary processes involve a transition between two atomic or molecular states separated by a potential barrier. The potential barrier constitutes the activation energy of the process, and determines the rate at which it occurs. When the barrier is low, the thermal energy of the reactants will generally be high enough to surmount the barrier and move over to products, and the reaction will be fast. However, when the barrier is high, only a few reactants will have sufficient energy, and the reaction will be much slower. The presence of a potential barrier to reaction is also the source of the temperature dependence of reaction rates.
The huge variety of chemical species, types of reaction, and the accompanying potential energy surfaces involved means that the timescale over which chemical reactions occur covers many orders of magnitude, from very slow reactions, such as iron rusting, to extremely fast reactions,such as the electron transfer processes involved in many biological systems or the combustion reactions occurring in flames.
A study into the kinetics of a chemical reaction is usually carried out with one or both of two main
goals in mind:

1. Analysis of the sequence of elementary steps giving rise to the overall reaction. i.e.the reaction mechanism.

2. Determination of the absolute rate of the reaction and/or its individual elementary steps.The aim of this course is to show you how these two goals may be achieved.

2. Rate of reaction When we talk about the rate of a chemical reaction, what we mean is the rate at which reactants are used up, or equivalently the rate at which products are formed. The rate therefore has units of
concentration per unit time, mol dm-3 s-1 (for gas phase reactions, alternative units of concentration
are often used, usually units of pressure – Torr, mbar or Pa).
To measure a reaction rate, we simply need to monitor the concentration of one of the reactants or products as a function of time.
There is one slight complication to our definition of the reaction rate so far, which is to do with the stochiometry of the reaction. The stoichiometry simply refers to the number of moles of each reactant and product appearing in the reaction equation.

For example, the reaction equation for the well-known Haber process, used industrially to produce ammonia, is:

N2 + 3H2 = 2NH3

N2 has a stochiometric coefficient of 1, H2 has a coefficient of 3, and NH3 has a coefficient of 2.We could determine the rate of this reaction in any one of three ways, by monitoring the changing concentration of N2, H2, or NH3. Say we monitor N2, and obtain a rate of

-d[N2]/dt = x mol dm-3 s-1.

Since for every mole of N2 that reacts, we lose three moles of H2, if we had monitored H2 instead of
N2 we would have obtained a rate

-d[H2]/dt = 3x mol dm-3 s-1.

Similarly, monitoring the concentration of NH3 would yield a rate of 2x mol dm-3 s-1.

Clearly, the same reaction cannot have three different rates, so we appear to have a problem. The solution is actually very simple: the reaction rate is defined as the rate of change of the concentration of a reactant or product divided by its stochiometric coefficient. For the above reaction, the rate (usually given the symbol ν) is therefore

ν = -d[N2]/dt = -1/3{d[H2]/dt} =1/2d{[NH3]/dt}

Note that a negative sign appears when we define the rate using the concentration of one of the reactants. This is because the rate of change of a reactant is negative (since it is being used up in the reaction), but the reaction rate needs to be a positive quantity.

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