Chapter 13: Kinetics
Renee Y. Becker
Valencia Community College
Introduction
1.
Chemical kinetics is the study of reaction rates
2.
For a chemical reaction to be useful it must occur at a
reasonable rate
3.
It is important to be able to control the rate of reaction
4. Factors that influence rate
a) Concentration of reactants (molarity)
b) Nature of reaction, process by which the reaction takes place
c) Temperature
d) Reaction mechanism (rate determining step)
e) Catalyst
Reaction rate
Reaction rate- Positive quantity that expresses how the conc. of a
reactant or product changes with time
Gen eq.
N2O5(g)  2NO2(g) + ½ O2(g)
1. [ ] = concentration, molarity, mole/L
2. [N2O5] decreases with time
3. [NO2] increases with time
4. [O2] increases with time
5. Because of coefficients the concentration of reactant and
products does not change at the same rate
6. When 1 mole of N2O5 is decomposed, 2 moles of NO2 and
½ mole of O2 is produced
N2O5(g)  2NO2(g) + ½ O2(g)
-[N2O5] = [NO2] = [O2]
2
½
coef. From gen eq
-[N2O5] because it’s concentration decreases, other
positive because they increase
Rate of reaction can be defined by dividing by the change
in time, t
rate = - [N2O5] =  [NO2] =  [O2]
t
2t
½ t
Reaction Rates
Generic Formula:
aA + bB  cC + dD
rate = -  [A] = -  [B] =  [D] =  [C]
at
b t
d t
ct
Reaction Rate and Concentration
The higher the conc. of starting reactant the more rapidly a reaction
takes place
1. Reactions occur as the result of collisions between reactant
molecules
2. The higher the concentration of molecules, the greater the #
of collisions in unit time and a faster reaction
3. As the reactants are consumed the concentration decreases,
collisions decrease, reaction rate decreases
4. Reaction rate decreases with time and eventually = 0, all
reactants consumed
5. Instantaneous rate, rate at a particular time
6. Initial rate at t = 0
Rate Expression and Rate Constant
Rate expression / rate law:
rate = k[A]
where k = rate constant, varies w/ nature and temp.
[A] = concentration of A
Order of rxn involving a single reactant
General equation
A  prod
rate expression
rate = k[A]m
where m=order of reaction
m=0 zero order
m=1 first order
m=2 second order
m, can’t be deduced from the coef. of the balanced eq.
Must be determined experimentally!
Order of rxn involving a single reactant
Rate of decomposition of species A measured at 2
different conc., 1 & 2
rate2 = k[A2]m rate1 = k[A1]m
By dividing we can solve for m, to find the order of the
reaction
Rate2 = [A2]m
Rate1 [A1]m
(Rate2/Rate1) = ([A2]/[A1])m
Example 1
CH3CHO(g)  CH4(g) + CO(g)
[CH3CHO] .10 M
.20 M
.30 M
.40 M
Rate (mol/L .085
s)
.34
.76
1.4
Using the given data determine the reaction order
Example 1
Rate2 = [A2]m
Rate1 [A1]m
second order
4 = 2m
m=2
rate = k[CH3CHO]2
once the order of the rxn is known, rate constant can be
calculated, let’s calculate the rate constant, k
Example 1
rate = k[CH3CHO]2
k=
rate = .085 mol/L s
conc = .10 mol/L
rate
= .085 mol/L s = 8.5 L/mol s
[CH3CHO]2
(.10 mol/L)2
now we can calc. the rate at any concentration, let’s try
.55 M
Example 1
rate = k[CH3CHO]2
rate = 8.5 L/mol s [.55]2 = 2.6 mol/ L s
Rate when [CH3CHO] = .55 M
is 2.6 mol/L s
Order of rxn with more than 1 reactant
aA + bB  prod
rate exp:
rate = k [A]m [B]n
m = order of rxn with respect to A
n = order of rxn with respect to B
Overall order of the rxn is the sum, m + n
Key
• When more than 1 reactant is involved the
order can be determined by holding the
concentration of 1 reactant constant while
varying the other reactant. From the
measured rates you can calculate the order
of the rxn with respect to the varying
reactant
Example 2
(CH3)3CBr + OH-  (CH3)3COH + Br-
[(CH3)3CBr]
Exp.
1
.5
Exp. Exp. Exp.
2
3
4
1.0 1.5 1.0
Exp.
5
1.0
[OH-]
.05
.05
.05
.10
.20
.01
.015 .01
.01
Rate (mol/L s) .005
Find the order of the reaction with respect to both
reactants, write the rate expression, and find the overall
order of the reaction
Reactant concentration and time
Rate expression
rate = k[A]
Shows how the rate of decomposition of A
changes with concentration
More important to know the relation between
concentration and time
Using calculus: Integrated rate equations
relating react conc. to time
For the following rate law, what is the overall order of the
reaction?
Rate = k [A]2 [B]
1.
1
2.
2
3.
3
4.
4
Order Rate
Conc-Time
Expression Relation
0
Rate = k
[A]0 – [A] = kt
1
Rate = k[A]
ln [A]0 = kt
[A]
2
Rate = k[A]2
1 – 1 = kt
[A] [A]0
Half-life
Linear
Plot
[A]0
2k
[A] vs. t
0.693
k
1
k[A]0
ln [A] vs. t
1 vs. t
[A]
First Order
First Order:
rate = k[A]
A  Products
ln [A]o/[A] = kt
[A]o = original conc. of A
[A] = Conc. of A at time, t
k = first order rate constant
ln = natural logarithm
t½ = .693/k
Second Order
Second order:
rate = k[A]2
1 – 1 = kt
[A] [A]0
A  Products
t½ = 1
k[A]0
Zero Order
Zero order:
rate = k
A  Products
[A]0 – [A] = kt
t½ = [A]0
2k
Example 3
The following data was obtained for the gas-phase
decomp. of HI
Time (h) 0
2
4
6
[HI]
0.50
0.33
0.25
1.00
Is this reaction zero, first, or second order in HI?
Hint: Graph each Conc. Vs. time corresponding to
correct [A], ln [A], or 1/[A]
[HI] vs. time not linear so it is not zero order
[HI] vs. t
[HI]
1.5
1
0.5
0
0
2
4
Time (h)
6
8
ln [HI] vs. time not linear so it is not first order
ln [HI] vs. t
ln [HI]
0
-0.5
0
2
4
-1
-1.5
Time (h)
6
8
1/[HI] vs. time is linear so it is second order
1/[HI] vs. t
1/[HI]
6
4
2
0
0
2
4
Time (h)
6
8
Activation Energy
Activation Energy: Ea (kJ)
For every rxn there is a certain minimum energy that
molecules must possess for collisions to be effective.
1. Positive quantity (Ea>0)
2. Depends only upon the nature of reaction
3. Fast rxn = small Ea
4. Is independent of temp and concentration
For the following reaction:
A + B  C
If I double the concentration of A and hold B
constant, the rate doubles. What is the order of
the reaction with respect to A?
1. 0
2. 1
3. 2
Activation Energy
Reaction Rate and Temp
Reaction Rate and Temp
1. As temp increases rate increases, Kinetic
Energy increases, and successful
collisions increase
2. General rule for every 10°C inc. in temp,
rate doubles
The Arrhenius Equation
The Arrhenius Equation
f = e-Ea/RT
f = fraction of molecules having an En. equal to or greater than Ea
R = gas constant
A = constant T = temp in K
ln k = ln A –Ea/RT
plot of ln k Vs. 1/T linear
slope = -Ea/R
Two-point equation relating k & T
ln k2 = Ea [1/T1 – 1/T2]
k1
R
Example 4
For a certain rxn the rate constant doubles
when the temp increases from 15 to 25°C.
a) Calc. The activation energy, Ea
b) Calc. the rate constant at 100°C, taking k
at 25°C to be 1.2 x 10-2 L/mol s
If I increase the temperature of a reaction
from 110 K to 120 K, what happens to the
rate of the reaction?
1. Stay the same
2. Doubles
3. Triples
Reaction Mechanism
Reaction Mechanism
Description of a path, or a sequence of steps, by which a
reaction occurs at the molecular level.
Simplest case- only a single step, collision between
two reactant molecules
Reaction Mechanism
“Mechanism” for the reaction of CO with NO2 at high temp (above
600 K)
CO(g) + NO2(g)  NO(g) + CO2(g)
“Mechanism” for the reaction of CO with NO2 at low temp
NO2(g) + NO2(g)  NO3(g) + NO(g) slow
CO(g) + NO3(g)  CO2(g) + NO2(g) fast
CO(g) + NO2(g)  NO(g) + CO2(g) overall
The overall reaction, obtained by summing the individual steps is
identical but the rate expressions are different.
High temp:
Low temp:
rate = k[CO][NO2]
rate= k[NO2]2
Reaction Mechanism
Elementary Steps:
Individual steps that constitute a reaction mechanism
Unimolecular
A B+C
Bimolecular
A+A B+C
rate = k[A][A] = [A]2
Termolecular
A+B+C D+E
rate = k[A][B][C]
rate = k[A]
The rate of an elementary step is equal to a rate constant, k,
multiplied by the concentration of each reactant molecule. You
can treat all reactants as if they were first order. If a reactant is
second order it will appear twice in the general equation.
Reaction Mechanism
Slow Steps- A step that is much slower than any other in a
reaction mechanism.
Rate-determining step - The rate of the overall reaction can
be taken to be that of the slow step
Step 1:
Step 2:
Step 3:
A B
B C
C D
fast
slow
fast
A D
The rate A  D (overall reaction) is approx. equal to the
rate of B  C the slow step
Deducing a Rate Expression from a Proposed Mechanism
1. Find the slowest step and equate the rate of the
overall reaction to the rate of that step.
2. Find the rate expression for the slowest step.
NO2(g) + NO2(g)  NO3(g) + NO(g)
CO(g) + NO2(g)  CO2(g) + NO2(g)
slow
fast
CO(g) + NO2(g)  CO2(g) + NO(g)
Rate of overall reaction = rate of 1st step
= k[NO2] [NO2] = k[NO2]2
Elimination of Intermediates
Intermediate
1. A species produced in one step of the mechanism and
consumed in a later step.
2. Concentration too small to determine experimentally
3. Must be eliminated from rate expression
4. The final rate expression must include only those
species that appear in the balanced equation for the
overall reaction
Example 5
Find the rate expression for the following reaction
mechanism
Step1:
NO(g) + Cl2(g)  NOCl2(g)
fast
Step2:
NOCl2(g) + NO(g)  2NOCl(g)
slow
2 NO(g) + Cl2(g)  2NOCl(g)
Example 6
The decomposition of ozone, O3, to diatomic oxygen, O2, is
believed to occur by a two-step mechanism:
Step 1:
O3(g) 
Step 2:
O3(g) + O(g)  2 O2(g)
2 O3(g)
O2(g) + O(g)
 3 O2(g)
• Find the rate expression for this reaction
fast
slow
Which is an intermediate for the following multi step mechanism?
2A + 2B  C + 2D
Step 1
Step 2
Step 3
1.
2.
3.
4.
E
F
E&F
A
2B  E
E + A  D + F
F + A  C + D
Catalysts
Catalysis
A catalyst increases the rate of a reaction without being consumed
by it. Changes the reaction mechanism to one with a lower
activation energy.
1.
Heterogeneous catalysis
a) Catalyst is in a different phase from the reaction mixture.
Most common, solid catalyst with gas or liquid mixture.
b) Solid catalyst is easily poisoned, foreign material
deposited on the surface during reaction reduce or destroy
its effectiveness.
2.
Homogeneous Catalysis
a) Same phase as the reactants
Which is the catalyst for the following multi step mechanism?
2A + 2B  C + 2D
Step 1
Step 2
Step 3
1.
2.
3.
4.
E
G
E&F
A
2 B + G E
E + A  D + F
F + A  C + D+G
Summary Problem and
Homework Problems
Chapter 12: Kinetics
Hydrogen peroxide decomposes to water and oxygen according to
the following reaction
H2O2(aq)  H2O + ½ O2(g)
It’s rate of decomposition is measured by titrating samples of the
solution with potassium permanganate (KMnO4) at certain
intervals.
a) Initial rate determinations at 40C for the decomposition
give the following data:
[H2O2]
0.10
0.20
0.30
Rate (mol/L min)
1.93 x 10-4
3.86 x 10-4
5.79 x 10-4
1. Order of rxn?
2. Rate expression?
3. Calc. k @ 40°C
4. Calc. half-life @ 40°C
b) Hydrogen peroxide is sold commercially
as a 30.0% solution. If the solution is kept
at 40C, how long will it take for the
solution to become 10.0% H2O2?
c) It has been determined that at 50C, the
rate constant for the reaction is
4.32 x 10-3/min. Calculate the activation
energy for the decomposition of H2O2
d) Manufacturers recommend that solutions
of hydrogen peroxide be kept in a
refrigerator at 4C. How long will it take
for a 30.0% solution to decompose to 10.0%
if the solution is kept at 4C?
e) The rate constant for the uncatalyzed reaction
at 25C is 5.21 x 10-4/min. The rate constant for
the catalyzed reaction at 25C is 2.95 x 108/min.
1) What is the half-life of the uncatalyzed
reaction at 25C?
2) What is the half-life of the catalyzed
reaction?
1) Express the rate of reaction
2HI(g)  H2(g) + I2(g)
a) in terms of [H2]
b) in terms of [HI]
3) Dinitrogen pentaoxide decomposes according
to the following equation:
2N2O5(g)  4NO2(g) + O2(g)
a) write an expression for reaction rate in
terms of [ N2O5], [NO2], and [O2]
4) What is the order with respect to each
reactant and the overall order of the reactions
described by the following rate expressions?
a)
rate = k1[A]3
b)
rate = k2[A][B]
c)
rate = k3[A][B]2
d)
rate = k4[B]
e)
rate = k
5) Complete the following table for the
reaction, which is first order in both reactants
A(g) + B(g)  products
[A]
[B]
K (L/mol s) Rate (mol/L s)
.2
.3
1.5
.029
.78
.45
.520
.025
.033
7) The decomposition of ammonia on tungsten at
1100C is zero-order, with a rate constant of 2.5 x 10-4
mol/L min
a) write the rate expression
b) calculate the rate when the concentration of
ammonia is 0.080M
c) At what concentration of ammonia is the
rate equal to the rate constant?
8) In solution at constant H+ concentration, I- reacts with H2O2 to
produce I2
H+(aq) + I-(aq) + ½ H2O2(aq)  ½ I2(aq) + H2O
The reaction rate can be followed by monitoring iodine
production. The following data apply:
[I-]
.02
.04
.06
.04
[H2O2]
.02
.02
.02
.04
Rate (mol/L s)
3.3 x 10-5
6.6 x 10-5
9.9 x 10-5
1.3 x 10-4
a) Order of Ib) Order of H2O2
c) Calc. k
d) Rate? When [I-] =
.01 M [H2O2] = .03
M
9) In the first-order decomposition of acetone at
500C it is found that the concentration is
0.0300 M after 200 min and 0.0200M after 400
min.
H3C-CO-CH3(g)  products
Calculate
a) The rate constant
b) The half-life
c) The initial concentration
10) The decomposition of hydrogen iodide is
second-order. Its half-life is 85 seconds
when the initial conc. is 0.15M
HI(g)  ½ H2(g) + ½ I2(g)
a) What is k for the reaction?
b) How long will it take to go from 0.300M
to 0.100M?
11) Write the rate expression for each of the
following elementary steps:
a)
K+ + HCl  KCl + H+
b)
NO3 + CO  NO2 + CO2
c)
2NO2  2NO + O2
12) For the reaction
2H2(g) + 2NO(g)  N2(g) + 2H2O(g)
the experimental rate expression is rate = k[NO]2[H2]
The following mechanism is proposed:
2NO  N2O2
N2O2 + H2  H2O + N2O
N2O + H2  N2 + H2O
fast
slow
fast
Show that the mechanism is consistent with the rate expression.