Transcript Chapter 13 THE RATES OF REACTIONS Reaction Rate  The reaction rate is defined as the change in concentration of a species with.

Chapter 13
THE RATES OF REACTIONS
Reaction Rate
 The reaction rate is defined as the change in
concentration of a species with time.
 Consider the reaction 2HI(g)H2(g) + I2(g)
 As the HI decomposes its concentration in the flask
decreases. Rate=change in conc. of reactant
time interval
-∆[HI]/∆t
We put a negative sign because the concentration of HI is
decreasing.
Instantaneous reaction rate
The instantaneous reaction rate is the (positive) slope of a tangent drawn
to the graph of concentration as a function of time, it varies as the reaction
proceeds.
Rate Laws
 The rate of a chemical reaction is the amount of
substance reacted or produced per unit time.
 The rate law is an expression indicating how the rate
depends on the concentrations of the reactants and
catalysts. The power of the concentration in the rate
law expression is called the order with respect to the
reactant or catalyst.
Order of a reaction
 The order of a chemical reaction with respect to a
reactant is an indication of how the reaction takes
place, not its stoichiometry.
Identifying the order of the reactions.
 First Order Rate Law :
The simplest reaction mechanism is that of uni molecular
decomposition. In such a process, a single reactant
undergoes a transformation at a constant probability per
unit time. Such a mechanism leads to a first-order
reaction rate law. Let's assume the reaction has a simple
stoichiometry: AB
 A First-Order Rate Law is called such because the rate of
product formation is proportional to the first power of
the number of available reactants (or reactant
concentration): rate=k[A]
 where [A] represents the concentration of species A in
the sample.
 Second Order Rate Law
If two molecules undergo a bimolecular reaction
such as a reaction that involves a collisional
encounter to produce products, and has a
stoichiometry like this: A+AB+C
 we expect a reaction rate law that is Second Order in
the concentration of [A])
 rate=k[A]2
 Zero Order Rate Law
If a reaction is catalyzed by a surface and has enough
(excess) reactant, the rate of the reaction depends on
the area of the catalyst, not on how much reactant is
present. This is an unusual circumstance outside of
the realm of catalyzed reactions and is described by a
Zero Order rate law: rate = k
 Many reactions can be classified according to their
order in a particular species. The order is the power
to which a species is raised in the rate law . The
overall order is the sum of all the individual orders.
Integrated rate law
 The differential rate law describes how the rate of reaction varies
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with the concentrations of various species, usually reactants, in the
system. The rate of reaction is proportional to the rates of change in
concentrations of the reactants and products; that is, the rate is
proportional to a derivative of a concentration.
To illustrate this point, consider the reaction
A B
The rate of reaction, r, is given by
r = - d [A]/dt
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Suppose this reaction obeys a first-order rate law:
r = k [A]
This rate law can also be written as
r = - d [A]/d t = k [A]
 This previous equation is a differential equation that
relates the rate of change in a concentration to the
concentration itself. Integration of this equation
produces the corresponding integrated rate law,
which relates the concentration to time.
 d [A]/[A] = - k d t
 At t = 0, the concentration of A is [A]0. The
integrated rate law is thus
[A] = [A]0 e- k t
 [A] molar concentration of A.
 Initial concentration is [A₀];k is the rate constant , t is
the elapsed time; exponential e , is the base of natural
logarithm. The variation of concentration with time
predicted by this equation is shown in the fig.
 The behavior shown in the illustration is called an
exponential decay. The change in concentration is initially
rapid but the concentration changes more slowly as time
goes on and the reactant is used up.
 Calculate the concentration of N2O₅ remaining
600.s(10.0 min) after the decomposition at 65⁰C when its
concentration was 0.040 mol/L. The reaction and its rate
law are 2N2O₅(g) 4NO2 (g)+O2(g)
rate of disappearance of N2O₅=k[N2O₅] with k=5.2x10-3 /s
 In a first order reaction the concentration of reactant
decays exponentially with time. To verify that a
reaction is first order plot the natural logarithm of its
concentration as a function of time and expect a
straight line, the slope of the straight line is –k.
Home work
 Page 609
 13.5 all
 13.13
 13.28 all
Half lives for First –Order Reactions
 The half life of a reaction is defined as the time it
takes for one half of a reactant to disappear. The half
life is given the symbol t1/2 to denote that it is the
time at which the concentration of reactant is one
half its initial value. For the first order reaction, you
can plug the definition of the half life into the
concentration-time reaction to obtain a neat
relationship:
ln ([A]0/[A]t) = kt
[A]t1/2 = 1/2[A]0
ln ([A]0 /1/2[A]0) = kt1/2
ln 2 = 0.693 = kt1/2
 The half life of a first order reaction is the
characteristic of the reaction and independent of the
initial concentration. A reaction with a large rate
constant has a short half life.
Second order integrated rate law.
 For second order reaction with rate law
 Rate=k[A]2
 The integrated rate law is [A]t =[A]₀/t+[A]t kt or
1/[A]t =1/[A]₀+kt
 For a second order reaction the plot of 1/[A] against t is
linear with a slope k.
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 For first order reaction
 For second order reaction
First order reaction
Second order reaction
 Rate equation Rate=k[A]
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 Integrated form
 [A]t =[A]₀e-kt or ln[A]t
=ln[A]₀-kt
 Half life of a first-order
reaction is independent
of the starting
concentration and is
given t1/2 =ln2/k =0.693/k
Rate equation Rate=k[A]2
Integrated form
[A]t =[A]₀/1+[A]₀kt or
1/[A]t =1/[A]₀ +kt
Half life equation for a
second-order reaction
dependent on one secondorder reactant is t1/2
=1/k[A]₀
 For a second order reaction
half lives progressively
doubles.
Class practice
 A pollutant escapes into a local picnic site. Students
had shown that the pollutant decays by a first order
reaction with rate constant 3.8x10-3/h. calculate the
time needed for the concentration to fall to one half
and one fourth its initial value.
Controlling reaction rates
 Effect of temperature: Most reactions go faster as
the temperature is raised. An increase in 10⁰C from
room temperature doubles the rate of reaction.
 An Arrhenius plot displays the logarithm of a rate
ln(k) plotted against inverse temperature 1 / T.
Arrhenius plots are often used to analyze the effect of
temperature on the rates of chemical reactions. For a
single rate-limited thermally activated process, an
Arrhenius plot gives a straight line, from which the
activation energy and the pre-exponential factor can
both be determined.
 The Arrhenius equation given in the form:k=Ae-Ea/RT
 can be written equivalently as: ln(k)=ln(A)-Ea/R(1/T)
 When plotted in the manner described above, the value
of the "y-intercept" will correspond to ln(A), and the
gradient of the line will be equal to − Ea / R.
 The pre-exponential factor, A, is a constant of
proportionality that takes into account a number of
factors such as the frequency of collision between and the
orientation of the reacting particles.
 The expression − Ea / RT represents the fraction of the
molecules present in a gas which have energies equal to
or in excess of activation energy at a particular
temperature.
Slope=-Ea/R
An Arrhenius plot of ln k vs 1/T is used to determine
the Arrhenius parameters of a reaction, a large
activation energy signifies a high sensitivity of the rate
to change in temperature.
Activation energy
 A factor that influences whether reaction will occur is
the energy the molecules carry when they collide.
Not all of the molecules have the same kinetic
energy, as shown in the figure below. This is
important because the kinetic energy molecules carry
when they collide is the principal source of the
energy that must be invested in a reaction to get it
started.
 ClNO2(g) + NO(g) NO2(g) + ClNO(g) ∆Go = -23.6
kJ/mol
 Before the reactants can be converted into products, the
free energy of the system must overcome the activation
energy for the reaction, as shown in the figure below. The
vertical axis in this diagram represents the free energy of
a pair of molecules as a chlorine atom is transferred from
one to the other. The horizontal axis represents the
sequence of infinitesimally small changes that must
occur to convert the reactants into the products of this
reaction.
 According to the collision theory of gas phase
reactions, molecules react when they collide. In
collision theory the activation energy is the energy
needed for the reaction to occur. It is represented as
a energy barrier between the reactants and the
products, with a height equal to the activation
energy.
 In activated complex theory, a reaction occurs only if
two molecules acquire enough energy, perhaps from
the surrounding solvent, to form an activated
complex and cross an energy barrier.
Catalysis and rate of reaction
 Catalysts increase the rates of reactions by providing
a new mechanism that has a smaller activation
energy, as shown in the figure below. A larger
proportion of the collisions that occur between
reactants now have enough energy to overcome the
activation energy for the reaction. As a result, the
rate of reaction increases.
 To illustrate how a catalyst can decrease the activation
energy for a reaction by providing another pathway for
the reaction, let's take the mechanism for the
decomposition of hydrogen peroxide catalyzed by the Iion. In the presence of this ion, the decomposition of
H2O2 doesn't have to occur in a single step. It can occur
in two steps, both of which are easier and therefore
faster. In the first step, the I- ion is oxidized by H2O2 to
form the hypoiodite ion, OI-.
 H2O2(aq) + I-(aq) H2O(aq) + OI-(aq)
 In the second step, the OI- ion is reduced to I- by H2O2.
 OI-(aq) + H2O2(aq) H2O(aq) + O2(g) + I-(aq)
 Because there is no net change in the concentration
of the I- ion as a result of these reactions, the I- ion
satisfies the criteria for a catalyst. Because H2O2 and
I- are both involved in the first step in this reaction,
and the first step in this reaction is the rate-limiting
step, the overall rate of reaction is first-order in both
reagents.
Rate laws and elementary reactions
 An elementary reaction is a chemical reaction in
which one or more chemical species react directly to
form products in a single reaction step and with a
single transition state.
 In a unimolecular elementary reaction a molecule, A,
dissociates to form the product(s). Aproducts
 The rate of such a reaction, at constant temperature,
is proportional to the concentration of the species A
 d[A]/dt=-k[A]
 In a bimolecular elementary reaction, two atoms,
molecules, ions or radicals, A and B, react together to
form the product (s). A+B products
 The rate of such a reaction, at constant temperature,
is proportional to the product of the concentrations
of the species A and B.
 d[A]/ dt=d [B]/ dt =-k [A] [B]
Home work
 Page 612
 13.36,13.39, 13.49,13.57,13.61,13.69