Chemical Kinetics - Arkansas Tech University

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Transcript Chemical Kinetics - Arkansas Tech University

Chapter 12
Chemical Kinetics
• The area of chemistry that concerns
reaction rates.
• Goals: To understand the steps (reaction
mechanism) by which a reaction takes
place. Allows us to find ways to facilitate
the reaction
Reaction Rate
Kinetics deals with the speed (rate) at which
changes occur. The quantity that changes is the
amount or concentration of a reactant or a product.
Change in concentration (conc.) of a reactant or
product per unit time.
conc of A at time t2  conc of A at time t1
Rate =
t2  t1
  A
2NO2(g)  2NO(g) + O2(g)
The concentration of the reactant (NO2)
decreases with time and the concentration of the
products (NO and O2) increase with time.
A change can be positive (increase) or negative
(decrease) leading to a positive or negative
reaction rate. However, we will always define
the rate as a positive quantity.
Figure 12.1 Definition of Rate (Conc. vs.Time)
change in [NO2]  [ NO2]
Rate =
Time elapsed
The concentration of NO2 decreases with time,
[NO2] is a negative quantity. The reaction rate is
a positive quantity
 [NO2]
Rate = t
The concentration of reactants always decrease
with time. The rate expression involving reactant
will include a negative sign.
2NO2(g)  2NO(g) + O2(g)
The rate can also be defined in terms of the products. In
doing so we must take into account the coefficients in the
balance equation for the reaction. Stoichiometry
determines the relative rates.
Rate of consumption = rate of production = 2(rate of production
of NO2
of NO
of O2)
 [NO2]  [NO]
  [O2]
 2
 t 
Rate Laws
2NO2(g)  2NO(g) + O2(g)
The reaction rate will depend on the concentrations of
the reactants
Rate = k[NO2]n
The above expression shows how the rate depends on
the concentration of reactants is called a rate law.
k = rate constant (proportionality constant)
n = rate order (integer or fraction including zero)
 [NO2]
Rate =  k[NO2]
Types of Rate Laws
Differential Rate Law: expresses how
rate depends on concentration.
Integrated Rate Law: expresses how
concentration depends on time.
• We consider reactions where the reverse
reaction is unimportant, rate laws involve only
concentrations of reactants.
• Differential and integrated rate laws for a given
reaction are related in a well-defined way, the
experimental determination of either of the rate
laws is sufficient.
• Experimental convenience dictates which types
of rate law is determined experimentally.
Method of Initial Rates
Initial Rate: the “instantaneous rate”
just after the reaction begins.
The initial rate is determined in several
experiments using different initial
Figure 12.3 A Plot of the Concentration of N2O5 as a Function of
Time for the Reaction
Overall Reaction Order
Sum of the order of each component in the
rate law.
rate = k[H2SeO3][H+]2[I]3
The overall reaction order is 1 + 2 + 3 = 6.
First-Order Rate Law
For aA  Products in a 1st-order reaction,
 A
Rate =
k A
Integrated first-order rate law is
ln[A] = kt + ln[A]o
ln[A] = -kt + ln[A]o
• The equation shows how the concentration of
A depends on time. If the initial concentration
of A and the rate constant k are known, the
concentration of A at any time can be
• The above equation is the equation of straight
line of the form y = mx + b, where a plot of y
versus x is a straight line with slope m and
intercept b.
• The reaction is first order in A if a plot
of ln[A] versus t is a straight line.
• The integrated rate law for a first order
reaction can also be expressed in terms
of a ratio of [A] and [A]o as follow:
 kt
Figure 12.4 A Plot of In(N2O5) Versus Time
Half-Life of a First-Order Reaction
The time required for a reactant to reach half its
original concentration is called the half-life of a
reactant and is designated by the symbol t1/2.
 kt when t = t1 / 2 then [A] =
 kt1 / 2  ln(2) = kt1 / 2
[A]o / 2
t1 / 2 =
 t1 / 2 =
t1/2 = half-life of the reaction, k = rate constant
For a first-order reaction, the half-life does not depend
on concentration.
Figure 12.5 A Plot of (N2O5) Versus Time for the Decomposition Reaction of N2O5
Second-Order Rate Law
For aA  products in a second-order reaction,
   A
Rate =
 k  A
Integrated rate law is
 kt +
 A
 A o
A plot of 1/[A] versus t will produce a straight line
with a slope equal to k
The above equation shows how [A] depends on time
and can be used to calculate [A] at any time t,
provided k and [A]o are known
Half-Life of a Second-Order Reaction
When one half-life of the second order reaction has
elapsed (t = t1/2), by definition, [A] = [A]o/2 then the
integrated rate law becomes
 kt1 / 2 +
 kt1 / 2 
 kt1 / 2
[A]o / 2
[A]o [A]o
solving for t1 / 2 gives the expression
t1 / 2 =
t1/2 = half-life of the reaction, k = rate constant,
Ao = initial concentration of A
The half-life is dependent upon the initial concentration.
Figure 12.6 (a) A Plot of In(C4H6) Versus t
(b) A Plot of 1/(C4H6) Versus t
Zero-Order Rate Laws
The rate law for a zero-order reaction is
Rate = k[A]o = k(1) = k
For a zero-order reaction, the rate is constant. It
does not change with concentration as it does
for first-order or second-order reactions.
The integrated rate law for a zero-order
reaction is
[A] = -kt + [A]o
[A] = -kt + [A]o
In this case a plot of [A] versus t gives a
straight line of slope –k.
[A] = [A]o/2, when t = t1/2
[A]o/2 =-kt1/2 + [A]o
Solving for t1/2 gives,
t1/2 = [A]o/2k
Figure 12.7 A Plot of (A) Versus t for a Zero-Order Reaction
Rate Laws for Reactions with More Than One Reactant
A + B + C  Product
Rate = k[A]n[B]m[C]p
For such reaction, concentration of one reactant remain
small compared with the concentrations of the others.
So the rate law reduce to
Rate = k`[A]n
Where, k` = k[B]m[C]op and [B]o>>[A]o and
The value of n can be obtained by determining whether
a plot of [A] versus t is linear (n = 0), a plot of ln[A]
versus t is linear (n = 1), or a plot of 1/[A] versus t is
linear (n = 2). The value of k` is determined from the
A Summary
Simplification: Conditions are set such
that only forward reaction is important.
Two types of rate law: differential rate
law and integrated rate law
Which type? Depends on the type of
data collected - differential and
integrated forms can be interconverted.
A Summary
Most common: method of initial rates.
Concentration v. time: used to
determine integrated rate law, often
For several reactants: choose
conditions under which only one
reactant varies significantly (pseudo
first-order conditions).
Reaction Mechanism
 The
series of steps by which a chemical
reaction occurs.
chemical equation does not tell us how
reactants become products - it is a summary
of the overall process.
 The
purpose for studying kinetics is to learn
as much as possible about the steps involved
in a reaction.
Reaction Mechanism
The reaction
NO2(g) + CO(g)  NO(g) + CO2(g)
has many steps in the reaction mechanism.
The rate law for this reaction is known from
experiment to be
Rate = k[NO2]2
The balanced equation tells us the reactants,
the products, the stoichiometry.
NO2(g) + NO2(g)
NO3(g) + NO(g)
NO3(g) + CO(g) 
NO2(g) + CO2(g)
Often Used Terms
• Intermediate: formed in one step and used up in a
subsequent step and so is never seen as a product.
(neither a reactant nor a product)
• Molecularity: the number of species that must
collide to produce the reaction indicated by that step.
• Elementary Step: A reaction whose rate law can be
written from its molecularity.
uni, bi and termolecular
• The sum of the elementary steps must give the
overall balanced equation
• The mechanism must agree with the experimentally
determined rate law.
Rate-Determining Step
Multistep reaction often have one step that is
much slower than all the others. Reactants can
become products only as fast as they can get
through this slowest step. The overall reaction
can be no faster than the slowest or rate
determining step.
In a multistep reaction, it is the slowest step. It
therefore determines the rate of reaction.
NO2(g) + NO2(g)  NO3(g) + NO(g) slow(rate determining)
NO3(g) + CO(g)  NO2(g) + CO2(g) fast
Overall rate = k1[NO2]2
Collision Model
Key Idea: Molecules must collide to react.
However, only a small fraction of collisions
produces a reaction. Why?
Arrhenius: An activation energy (threshold
energy) must be overcome to produce a
chemical reaction.
2BrNO(g)  2NO(g) + Br2(g)
Figure 12.11 Change in Potential Energy
Arrhenius Equation
 Collisions
must have enough energy to
produce the reaction (must equal or exceed
the activation energy).
 Orientation
of reactants must allow
formation of new bonds necessary to
produce products.
Figure 12.13 Several Possible Orientations for a Collision Between Two BrNO Molecules
Arrhenius Equation
k  Ae
 Ea / RT
k = rate constant, A = frequency factor
Ea = activation energy, T = temperature
R = gas constant
ln(k) = -Ea/R(1/T) + ln(A)
A plot of ln(k) versus 1/T gives a straight
line, slope = -Ea/R and intercept = ln(A)
Figure 12.14 Plot of In(k) Versus 1/T for the Reaction 2N2O5(g) 2(g) + O2(g)
Arrhenius Equation
Activation energy (Ea) can also be calculated from the
values of k at only two temperatures
At temperature T1, where the rate constant is k1,
ln(k1) =  ln(A)
At temperature T2, where the rate constant is k2,
ln (k 2 )  
 ln (A)
RT 2
 Ea
ln (k 2 )- ln (k 1 )=  ln (A) 
 RT 2
 k 2  Ea  1
ln  
  
 k 1
R  T1 T 2 
 Ea
  RT 1  ln (A)   RT 2  RT 1
Ea can be calculated from k1 and k2 at temperature T1 and T2
• Catalyst: A substance that speeds up a
reaction without being consumed it self.
• Enzyme: A large molecule (usually a
protein) that catalyzes biological reactions.
• Homogeneous catalyst: Present in the same
phase as the reacting molecules.
• Heterogeneous catalyst:
Present in a
different phase than the reacting molecules.
Figure 12.15 Energy Plots for a Catalyzed and an Uncatalyzed Pathway for a Given Reaction