Rates of Chemical Reactions

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Chapter 25 Rates of Chemical Reactions
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OUTLINE SECTION 25.1 - Experimental
techniques SECTION 25.2 - Rates of
reactions SECTION 25.3 - Integrated rate
laws SECTION 25.4 - Reactions approaching
equilibrium SECTION 25.5 - Temperature dependence
of reaction rates SECTION 25.6 - Elementary
reactions SECTION 25.7 - Consecutive elementary
reactions SECTION 25.8 - Unimolecular reactions
HOMEWORK EXERCISES 25.5 - 25.18 Parts (a)
only. PROBLEM 25.21(a)
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Introductory Comments
Reactions occur at different speeds (rates)

• An explosion is usually a very rapid oxidation
  reaction
• Rusting is a very slow oxidation reaction

The study of reaction rates is called chemical
kinetics. It is an empirical science.
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Introductory Comments

• We will show that the rate of any given reaction
  has a dependence upon the concentrations of the
  reactants (and products).
• Dependence expressed mathematically using a diff.
  eqn. which we call a rate law.
• Solving a rate law allows us to determine the
  concentration of products in a reaction at any
  given time after initiation.
• We can also use the form of the rate law to tell
  us a something about the mechanism of the
  reaction.

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Experimental techniques - General

• If
• One or more component of the reaction mixture is
  a gas ? can use a barometer
• One or more component is colored ? can use a
  spectrophotometer
• The reaction changes the number of ions ?
  conductivity of the mixture
• The reaction changes the number of H ions ? pH
  meter

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Experimental techniques - Application

• Real-time analysis
• Flow method
• Stopped flow method
• Flash photolysis

• Quenching
• Chemical quench flow method
• Freeze quench method

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Rates of reactions
Consider A 2B ? 3C D
Rate of consumption of the reactants is dR/dt
Rate of formation of the products is dP/dt
We have four rates describing the same
reaction. Instead we use the unique rate of
reaction, v
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Rates of reactions
Example Assuming constant volume write the
reaction 2N2O5 (g) ? 4NO2 (g) O2 (g) in terms
of the concentration changes for each species in
the reaction.
Self-test 25.2 The rate of change of molar
concentration of CH3 radicals in the reaction
2CH3 (g) ? C2H6 (g) was reported as dCH3/dt
-1.2 mol L-1 s-1 under particular conditions.
What is (a) the rate of reaction and (b) the rate
of formation of C2H6?
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Rates of reactions rate laws and rate constants
The rate of a reaction is often found to be
proportional to the concentrations of the
reactants raised to a power. For example
v k AB
Rate of reaction
Rate Constant
Rate laws are determined experimentally and
cannot in general be inferred from the chemical
equation for the reaction. e.g. H2 (g) Br2 (g)
? 2HBr (g)
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Rates of reactions Reaction order
Many reactions are found to have rate laws of the
form
v k AaBb
The power to which the conc. of a species is
raised gives the order of the reaction with
respect to that species.
The overall order of a reaction like that given
above is the sum of the individual orders.
v k AB

• Three problems
• What is the rate law and rate constant?
• How do we construct a reaction mechanism
  consistent with the rate law?
• Must account for the values of the rate constants
  and their dependence on T.

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Rates of reactions The determination of the
rate law
The isolation method Here the concentrations of
all the reactants except one are in large excess.
E. g. if B is in large excess to a good
approximation its concentration is constant
throughout the reaction. So although the true
rate law might be
v k AB
We can approximate B by B0 and write
v k A k k B0
Pseudo first-order rate law
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Rates of reactions The determination of the
rate law
The method of initial rates
If from the isolation method we have found
v k Aa
Then its initial rate, v0, is given by the
initial values of the concentration of A
v0 k A0a
Taking logarithms gives
log v0 a log A0 log k
y mx c
c.f.
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Rates of reactions The determination of the
rate law
v0 k I02Ar0
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Rates of reactions The determination of the
rate law
The method of initial rates
Disadvantage The initial rate law might not
be the full, real rate law.
The products may participate in the
reaction.
In practice simple rate laws are often
determined by making guesses as to the form of
the equation and then fitting this equation to
the experimental data. We should also test our
law by altering the concentrations of components
in the reaction mixture, adding product etc.
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Integrated Rate Laws
v k AB

Rate laws are differential equations.
If we want to determine concentration as a
function of time we must integrate.
First Order Reactions
Lets imagine we have a reaction A ? B C that
has the first-order rate law
v k A
We have the solution
or
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Integrated Rate Laws - First Order Reactions
or
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Integrated Rate Laws - First Order Reactions
The half-life, t1/2, of a substance is the time
taken for the concentration of a reactant to fall
to half its initial value. So we calculate
the time, t, taken for A to decrease from A0
to ½ A0.
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Integrated Rate Laws - First Order Reactions
Nucleotide Half-life 18F
1.8 hrs 238U 4.51 109
yrs 280Rg 3.6 s
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Integrated Rate Laws - Second Order Reactions
Lets imagine we have a reaction A ? B C that
has the second-order rate law
v k A2
We have the solution
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Integrated Rate Laws - Second Order Reactions
Lets imagine we have a reaction A B ? C D
that has the second-order rate law
v k AB
We have the solution
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Integrated Rate Laws
Recap For the reaction A ? Products A
versus time. If a straight line, the reaction is
zero order. ln A versus time. If a straight
line, the reaction is first-order. 1/A versus
time. If a straight line, the reaction is
second-order.
For the reaction A B ? Products
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Example The reaction 2 A ? Products has a
second order rate law with k 3.5 x 10-4 L
mol-1 s-1. Calculate the time required for the
concentration of A to change from 0.26 mol L-1 to
0.011 mol L-1.
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Reactions Approaching Equilibrium
So far we have only considered reactions far from
equilibrium (i.e. reactions at the start).
When a reaction is near equilibrium then the rate
of the reverse reaction should be considered.
Lets take a simple reaction (perhaps an
isomerization) A ? B v k A B ? A v k
B
The concentration of A is reduced by the forward
reaction and is increased by the reverse reaction.
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Reactions Approaching Equilibrium
The change in the concentration of A is therefore
If the initial concentration is A0 and B 0
then at all times A B A0
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Reactions Approaching Equilibrium
The solution of this first order differential
equation is
As t ? 8, the concentrations reach their
equilibrium values
Thus
K ?
k Aeq k Beq
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The Temperature Dependence of Reaction Rates
The rates of most reactions increase by a factor
between 2 and 4 by increasing the temperature
from 25 oC to 35 oC
It is found experimentally that for many
reactions a plot of ln k versus 1/T is a straight
line.
We can fit a straight line to the experimental
data. It will have the form y mx c
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The Temperature Dependence of Reaction Rates
It has been discovered that the two fitted
constants (m and c in y mx c) may be
interpreted to have some physical meaning.
The Arrhenius Equation
The parameter A corresponds to the intercept of
the line at 1/T 0 (infinite temperature). It
is called the pre-exponential factor.
The parameter Ea is obtained from the slope of
the line. It is called the activation energy.
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The Temperature Dependence of Reaction Rates
The Interpretation of the Arrhenius Parameters
It is useful to consider the potential energy
changes during a reaction A B ? Products where
the chemical reaction proceeds via a collision
between A and B.
A
B
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The Temperature Dependence of Reaction Rates
The Interpretation of the Arrhenius Parameters
Ea, the activation energy, is the minimum kinetic
energy that the reactants must have in order to
form products.
A, the pre-exponential factor, is a measure of
the rate at which collisions occur irrespective
of their energy.
Therefore the product of A and e-(Ea/RT) gives
the rate of successful collisions.
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Example The rate constant for the decomposition
of a certain substance is 1.7 x 10-2 L mol-1 s-1
at 24 oC and 2.01 x 10-2 L mol-1 s-1 at 37 oC.
Evaluate the Arrhenius parameters of the reaction.
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Reaction Mechanisms
In any chemical reaction we start with reactants
and end up with products. We have demonstrated
a number of ways in which the rate (speed) of the
reaction may be measured and the form of the rate
law discovered. Clearly, as we progress from
reactants to products the route will effect the
rate. We may attempt to discover the route, or
mechanism, using the form of the rate law and
some clever detective work.
N.B. Just like the rate this route, or mechanism,
cannot be obtained from the stoichiometry of the
reaction.
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Reaction Mechanisms
Most reactions occur in a sequence of steps
called elementary reactions.
Molecularity - of molecules coming together to
react in an elementary reaction. We can classify
elementary reactions according to their
molecularity. unimolecular elementary reaction
single molecule breaks apart. bimolecular
elementary reaction a pair of molecules collide
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Reaction Mechanisms
Some reactions have a mechanism (route) that
involves the formation of an intermediate, I
E. g. two consecutive unimolecular elementary
reactions
We may discover whether a reaction involves the
formation of an intermediate by looking at the
rate of change of the concentration of each
substance.
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Reaction Mechanisms
The variation of concentration with time.
Three rates to consider
As we go on we consider that at the start of the
reaction only A is present and its concentration
is A0. This means that at all times A I
P A0.
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Reaction Mechanisms
The variation of concentration with time.
(Setting I0 0)
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Reaction Mechanisms
The variation of concentration with time.
At all times A I P A0. Hence
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Reaction Mechanisms
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Example 25.6 Analysing consecutive
reactions Suppose that for A ? I ? P we
actually want I not P. When is I at its greatest
concentration?
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Reaction Mechanisms
The Rate-Determining Step
Suppose now that kb gtgt ka then
In this case A ? I is called the rate-determining
step
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Reaction Mechanisms
We have considered the case of a reaction that
proceeds via two unimolecular elementary
reactions with one intermediate. The level of
mathematics required for even this simple
mechanism suggests that if we consider a reaction
with two intermediates and/or consider higher
molecularity elementary steps then the math will
be very complicated! As in may other areas
of science understanding may still be obtained by
making approximations
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Reaction Mechanisms
The Steady State Approximation
We may illustrate this approximation by once
again using the simple reaction
In this approximation, after an induction period,
the concentration of I is assumed to be small and
steady i.e.
Before we had
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Reaction Mechanisms
The Steady State Approximation
Same result obtained with the rate-determing step
mechanism!
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Example 25.7 Using the steady-state
approximation.
Devise the rate law for the decomposition of
N2O5, 2N2O5(g) ? 4NO2(g) O2(g) On the basis
of the following mechanism N2O5 ? NO2
NO3 ka NO2 NO3 ? N2O5 ka NO2 NO3 ? NO2
O2 NO kb NO N2O5 ? 3 NO2 kc