Download Chapter 15: Kinetics

17/11/2016
Chemical kinetics studies:
Lecture 7
Kinetics

rates of chemical reactions

the factors that affect rate reactions

the mechanisms (the series of steps) by which
reactions occur
1
Kinetics, rate of the reaction, reaction
mechanism



2
Rate of reaction
The speed with which the reactants disappear
and the products form is called the rate of
the reaction.
A study of the rate of reaction can give
detailed information about how reactants
change into products.
The series of individual steps that add up to
the overall observed reaction is called the
reaction mechanism.
We are all familiar with processes in which some
quantity change with time
 Car travels at 40 km/hour (miles/hour)
 Faucet delivers water at 30 l/min (gallons/minute)
 Factory produces 32,000 tires/day
Each of these ratios is called a rate
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A spectroscopic method for determining reaction
rates.
Light of the wavelength that is absorbed by the
investigated substance is passed through a
reaction chamber. The changes in the reactant or
product concentration as reaction progresses
cause the decrease or increase of the light
intensity.
Blue dye is reacting with bleach, which converts it
into a colourless products.
The colour decreases and eventually disappears.
The rate of the reaction could be determined by
repeatedly measuring both the colour intensity and
the elapsed time.
The concentration of the dye could be calculated
from the intensity of the blue colour.
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Quiz


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Experience tells us that different chemical
reactions occur at very different rates, e.g.
Identify which of the following are rates:
A) 15 cm
B) 30 m / s2
C) 25oC
D) 5oC/min
E) 1.2 mol /min
F) 45 min
1. combustion reactions – burning methane
(component of natural gas) or combustion
of isooctane (C8H18)in gasoline – proceed
very rapidly, even explosively.
2. Erosion of the rocks – proceed very slowly.
3. Similarly the iron rusting is slow process.
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The reactions of strong acids with strong bases are
thermodynamically favoured (spontaneous) and occur at
very rapid rates.
3) C(diamond) + O2(g)→CO2(g)
1) 2HCl + Mg(OH)2 → MgCl2 + 2H2O
ΔG0rxn = -591.8 kJ/mol
Similarly reaction of some compound with oxygen (e.g.
This reaction does not occur at an observable rate.
4) C(graphite) + O2(g) →CO2(g)
burning) is thermodynamically favoured e.g.
ΔG0rxn = - 397 kJ/mol
ΔG0rxn = - 394 kJ/mol.
This reaction occurs rapidly.
2) CH4 + 2O2 → CO2 + 2H2O
The difference in the reaction speed of 3 and 4 reactions is
ΔG0rxn = -800.8 kJ/mol
explained by kinetics, not thermodynamics.
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Rate, formula
Rate of reaction, v,



The rate of reaction describes how fast
reactants are used up and products are
formed.
Knowledge of the rate of a reaction can be an
invaluable tool in helping us to understand
how chemical compounds behave when they
interact.

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A rate, is always expressed as a ratio.
One way to describe a reaction rate is to
select one component of the reaction and
describe the change in its concentration per
unit of time:
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rate with respect to X 



Quiz
(conc. of X at time t2  conc. of X at time t1 )
(t 2  t1 )
An 8.00 g piece of magnesium was placed into
6.0 M HCl . After 25 s. 3.50 g of unreacted
magnesium remained. The average rate at
which magnesium was consumed is:

A.
0.14 g/s

B.
0.18 g/s

C.
0.32 g/s

D.
4.50 g/s
(conc. of X )
t
Molarity (mol/L) is normally presented as
the concentration unit whereas the second
(s) is the most often used unit of time.
Typically, the reaction rate has the unit
mol/L
or mol L-1 s-1
s
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
Consider the following reaction:

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Consider the following reaction at a constant
temperature in an open system:
MgCO3(s) + 2HCl(aq) → CO2(g) + H2O(l) + MgCl2(aq)
Which of the following properties could be used to
determine reaction rate?
A.
Mass of the system
B.
Pressure of the gas
C.
Concentration of H2O
D.
Concentration of MgCO3
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2CaCrO4(s) +2H+(aq)  2Ca2+(aq) + H2O(l) + Cr2O72- (aq)

(orange)





The progress of the reaction could be followed by
observing the rate of
A.
mass loss
B.
decrease in pH
C.
precipitate formation
D.
formation of orange colour in the
solution
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a A+ b B → c C+ d D
If no other reaction takes place, the changes in
rate= -Δ[A]/ Δt;
concentration are related to one another.
rate = -Δ[B]/ Δt, or
For every a mol/L that described decrease of [A], [B]
rate = Δ[C]/ Δt;
must decrease by b mol/L, [C] must increase by c
rate= Δ[D]/ Δt
mol/L and so on…
The reaction rate must be positive because it describes the
The number of moles of reaction that occur per litter
forward reaction, which consumes A and B.
in a given time describe the rate of reaction.
The concentration of reactants A and B decrease in time
interval Δt.
Δ[A]/ Δt and Δ[B]/ Δt are negative quantities.
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a A+ b B → c C+ d D
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Quiz
2A + B →3C +D
If the rate of disappearance of A is equal to -0.084
mol/L s at the start of the reaction what are the rates
of change for B, C and D at this time?
Rate of change of B =
Rate of change of C =
Rate of change of D =
a) B= 0.042 M/s; C= 0.056 M/s; D= - 0.042 mol/L s
b) B = -0.042M/s; C = 0.112 M/s; D = 0.042 mol/L s
c) B= -0.042 M/s; C= - 0.112 M/s; D= 0.042 mol/L s
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Consider the following reaction:
N2H4(l) + 2H2O2(l) 
C3 H8 ( g )  5O2 ( g )  3CO2 ( g )  4H 2O( g )
N2(g) + 4H20(l)

Compared to the rate with respect to
propane:
◦ Rate with respect to oxygen is five times faster
◦ Rate with respect to carbon dioxide is three
times faster
◦ Rate with respect to water is four times faster
In 1.0 seconds, 0.015 mol of H2O2 is consumed.
The rate of production of N2 is
A.1.5 x 10-3 mol/s

B. 7.5 x 10-3 mol/s
C. 6.0 x 10-3 mol/s
Since the rates are all related any may be
monitored to determine the reaction rate
D. 1.5 x 10-2 mol/s
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Quiz
Determine relative reaction rates of the four
substrates involved in the following chemical
reaction. Give the appropriate numbers instead w, x,
y and z letters:


2C2H2(g) + 5O2(g) → 4CO2 + 2H2O(l)


A reaction rate is generally not constant
throughout the reaction.
Since the most of reactions depend on the
concentration of reactants, the rate
changes as they are used up.
The rate at any particular moment is called
the instantaneous rate.
It can be calculated from a concentration
versus time plot.
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Plot of [H2] vs time for the reaction of 1.000 M H2 with 2.000 M ICl.
The instantaneous rate of reaction at any time, t, equals the negative
slope of the tangent to this curve at time t. The initial ratio of the
reaction is equal to the negative of the initial slope (t=0). The
determination of the instantaneous rate at t=2 second is illustrated.
A plot of the concentration of HI versus time for the reaction:
2HI(g)  H2(g) + I2(g). The slope is negative because we are
measuring the disappearance of HI. When used to express the
rate it is used as a positive number.
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A rate law is a mathematical equation that describes the
concentration
Quiz
Based on the graph below determine the
instantaneous rate of change of [X] at 8 seconds
[X] =…..
progress of the reaction.
In general, rate laws must be determined experimentally.
Unless a reaction is an elementary reaction, it is not
M/s
possible to predict the rate law from the overall chemical
equation.
0.10
There are two forms of a rate law for chemical kinetics:
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0.04
the differential rate law and the integrated rate law.
2
8
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s
time
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Rate constant, k
For a generic reaction:
rate = k [reactant 1]n [reactant 2]m
reactant 1 + reactant 2 → product
where k is the constant of proportionality
with no intermediate steps in this reaction
named rate constant. Its value is generally
mechanism, the rate law is given by
constant provided that reaction is performed at
Rate = k [reactant 1]n [reactant 2]m
constant temperature T.
Values of k are always positive, although it
may be an exception of this rule.
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Consider the following reaction:

Order of reaction
H 2 SeO3  6 I   4 H   Se  2 I 3  3H 2O


From experiment, the rate law (determined
from initial rates) is
rate  k[ H 2 SeO3 ]1[ I  ]3[ H  ]2



At 0oC, k equals 5.0 x 105 L5 mol-5 s-1
Thus, at 0oC
The exponents in the rate law are generally
unrelated to the chemical equation’s
coefficients.
The exponent in a rate law is called the
order of reaction with respect to the
corresponding reactant.
◦ Never simply assume that the exponents and
coefficients are the same.
◦ The exponents must be determined from the
results of experiments.
rate  (5.0 105 L5 mol -5 s 1 )
 [ H 2 SeO3 ][I  ]3[ H  ]2
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Reaction order quiz

For the rate law:
Analyse the following rate equations, and
determine the orders of reactions with
respect to reactant and overall reaction order:
 1. rate = k [Cu2+] [NH3]
 2. rate = k [OH-]
 3. rate = k [NO]2[O2]
 4. rate = k [A]3[B]2
rate  k[ H 2 SeO3 ]1[ I  ]3[ H  ]2

We can say
◦
◦
◦
◦

The
The
The
The
reaction is
reaction is
reaction is
reaction is
first order with respect to H2SeO3
third order with respect to Isecond order with respect to H+
sixth order overall
Exponents (orders of reactions) in a rate
law can be fractional, negative, and even
zero.
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Looking for patterns in experimental data provide
way to determine the exponents in a rate law.
One of the easiest ways to reveal patterns in data
is to form ratios of results using different sets of
conditions.
 This technique is generally applicable.
 Consider the hypothetical reaction:


Suppose the experimental concentration-rate data
for five experiments is:
Inital Conc.
[ A]
[ B]
Expt
1
2
3
4
5
A  B  products
rate  k[ A]m [ B]n
Initial Rate
(mol L-1 ) (mol L-1 ) (mol L-1 s -1 )
0.10
0.10
0.20
0.20
0.10
0.40
0.30
0.10
0.60
0.30
0.20
2.40
0.30
0.30
5.40
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

For experiments 1, 2, and 3 [B] is held constant, so any
change in rate must be due to changes in [A]
The rate law says that at constant [B] the rate is
proportional to [A]m


m
rate2  [ A]2 


rate1  [ A]1 
rate2  0.40 mol L-1 s -1 
2

rate1  0.20 mol L-1 s -1 
m
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
For experiments 3, 4, and 5 [A] is held
constant, so any change must be due to
changes in [B]
The rate law says that at constant [A] the
rate is proportional to [B]n
Using the results from experiment 3 and 4:
m
 [ A]2 
 0.20 mol L-1 

  
  2m
-1 
[
A
]
 0.10 mol L 
 1
Thus m=1
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n
rate 4  [ B]4 


rate3  [ B]3 
rate 4  2.40 mol L-1 s -1 
4

rate3  0.60 mol L-1 s -1 
n


Thus n=2
k
n
 [ B]4   0.20 mol L-1 

  
  2n
-1 
 [ B]3   0.10 mol L 

The rate constant (k) can be determined
using data from any experiment
Using experiment 1:
rate
0.20 mol L-1 s -1

2
[ A][B]
(0.10 mol L-1 )(0.10 mol L-1 ) 2
 2.0  102 L2 mol -2 s -1
The reaction is second order in respect to
B and
rate=k[A]1[B]2

Using data from a different experiment
might give a slightly different value of k
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

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Quiz
Quiz
For 2NO + O2 --> 2NO2 , initial rate data are:
The reaction has the rate law: rate = k[C][D]2.
What will happen to the reaction rate when
the following change in conditions is
performed?
 doubling [C]
 tripling [D]
Experiment
[NO]
M/L
[O2 ]
M/L
Rate
[mM/ s]
1
0.010
0.010
2.5
2
0.010
0.020
5.0
3
0.030
0.020
45.0
Determine the reaction rates in terms of [NO], [O2] and k
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The relationship between concentration and
time can be derived from the rate law and
calculus.
 Integration of the rate laws gives the
integrated rate laws, which present
concentration as a function of time.
 Integrated laws can be very complicated,
so only a few simple forms will be
considered.
In this reaction the speed of the reaction does
not depend on the reagent concentrations.
𝑣= −
∆𝑐
= 𝑘𝑐 0 = 𝑘
∆𝑡
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v  k c0  k
v
v
•photochemical reactions
•heterophaseous reaction in which the slowest process
t
C
c
is connected with the phase change
•burning ethanol by living organism
C0
• decomposition of ammonia
tg   a  - k
• synthesis of HCl from hydrogen and chlorine

performed on the sun light
t
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First order reaction examples
First order reaction
◦ Formula for the rate law is: rate = k [A]


◦ The integrated rate law can be expressed as:
ln

[ A]0
 kt or [ A]t  [ A]0 e  kt
[ A]t


 [A]0 is [A] at t (time) = 0
 [A]t is [A] at t = t
 e = base of natural logarithms = 2.71828…

Absorption, distribution, elimination rates
Microbial death kinetics
Photo dissociation of ozone with UV light
Decomposition of hydrogen peroxide at room
temperature
Hydrolysis of sucrose (sugar) to glucose and
fructose
e.g. SO2Cl2→ SO2 + Cl2 at 320 oC k = 2 x 10-5 kJ/ s
C2H6→ 2 CH3• at 700 oC k = 5.36 x 10-4 kJ/ s
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Example of calculation
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Solution
For certain first-order reaction the initial
[A]6 =[A]0 e-kt
concentration of reactant A is equal 2M/L and
the constant of this reaction k = 0.15M/ L s.
What is the concentration of A after time 6s ?
[A]6 = [2] (2.718-0.15 x6) = 0.813M/L
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Quiz


If a reaction is first order with a rate constant
of 5.48 x 10-2 sec-1, how long is required for
3/4 of the initial concentration of reactant to
be used up?
Graphical methods can be used to
determine the first-order rate constant,
note
ln
[ A]0
 kt
[ A]t
ln[ A]0  ln[ A]t  kt
ln[ A]t  ln[ A]0   kt
ln[ A]t   kt  ln[ A]0

y


 mx  b
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
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A plot of ln[A]t versus t gives a straight line
with a slope of -k
Determination of reaction constant
from plot
The decomposition of N2O5. (a) A graph of concentration versus
time for the decomposition at 45oC. (b) A straight line is obtained
from a logarithm versus time plot. The slope is negative the rate
constant.
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ln of initial
Decreasing function, abtuse angle
reactant
concentration
1.4
Plot ln [A] vs. time
1.4
1.2
1.2
1
1
0.8
0.8
0.6
0.6
a
0.4
ln CA
ln CA
0.4
0.2
0
-0.2
tg α = a/b
a = 1.2- (-0.28) =
1.48 =
b= 125
tg α = - 0.0118 =
-k
k = 0.0118
0
25
50
75
100
125
150
175
0.2
0
200
0
25
50
75
100
125
150
175
200
-0.2
-0.4
-0.4
-0.6
-0.6
-0.8
-0.8
-1
-1
t [min]
b
t [min]
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Second order reaction

Second order reaction examples
The simplest second-order rate law has
the form
2
rate  k[ B]


The integrated form of this equation is

1
1

 kt
[ B]t [ B]0

[ B]0  the initial concentration of B

[ B]t  the concentration of B at time t
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Decomposition of HI without catalyst
Decomposition of NO2 to NO and O2
ClO- + Br-→ BrO- + Clat 25 oC k = 4.2 x 10-7 kJ/l mol-1s-1
H+ + OH-→ H2O
at 25 oC k = 1.35 x 1011 kJ/l mol-1s-1
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

Half life t1/2
Graphical methods can also be applied to
second-order reactions
A plot of 1/[B]t versus t gives a straight
line with a slope of k

The amount of time required for half of a
reactant concentration to disappear is called
the half-life, t1/2
Second-order kinetics. A plot of 1/[HI] versus time
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◦The half-life of a first-order
 𝑡1
2

=
It is not affected by the initial concentration
𝐴0
2𝑘
First - order rate law : ln
[ A]0
 kt
[ A]t
1
at t  t1/ 2 , [ A]t  [ A]0 , substituti ng
2
[ A]0
ln 2
ln 1
 kt1/ 2 or t1/ 2 
k
[
A
]
0
2
For zero-order reactions, the half-life
depends on the initial concentration of
reactant and the rate constant.
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◦ The half-life of a second-order reactions does
depend on the initial concentration
The equation for the half-life of a second order reaction:
𝑡1/2 =
1
𝑘 ∙ 𝐵0
This inverse relationship suggests that as
the initial concentration of reactant
is increased, there is a higher probability
of the two reactant molecules interacting
to form product.
Consequently, the reactant will be consumed
in a shorter amount of time,
i.e. the reaction will have a shorter half-life.
1
1

 kt
[ B]t [ B]0
Second - order rate law :
1
at t  t1/ 2 , [ B]t  [ B]0 , substituti ng
2
1
1

 kt1/ 2
1
[ B]0 [ B]0
2
1
 kt1/ 2
[ B]0
or t1/ 2 
ln 2
k[ B]0
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(2) Graphical method
Reagent
concentration
to
t1
t2
t3
t4
...
co
c1
c2
c3
c4
...
n=0
n=1
c
co
tg   a  - k

ln co
tg   a  - k
c  c  k t
1
c


t
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ln c
n=2
t
ln c  ln c  k t
1
c
tg   a  k
t
1 1

k t
c c
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Quiz

Substance A decomposes by a first-order
reaction. Starting initially with [A] = 2.00 M,
after 150 min [A] = 0.50 M. For this reaction
what is t1/2 ?
First-order radioactive decay of iodine-131. The
initial concentration is represented by [I]0.
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Quiz
120
100
The table below presents plote of concentration
of biologically active metabolite T-IDA vs time.
Using of these data determine graphically the
half-time life (t1/2) of this metabolite.
Concentra
tion
0
10
20
30
40
50
60
70
100
50
25
12,5
6.25
3.13
1.56
0.781
concentration [mol /L]
Time
[min]
80
[mol /L]
60
40
20
0
0
10
20
30
40
50
Time [min]
60
70
80
T1/2 = 10 min
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
There are five principle factors that
influence reaction rates:
1) Chemical nature of the reactants
2) Ability of the reactants to come in contact
with each other
3) Concentration of the reactants
4) Temperature
The progress of the reaction A  B. The number of A molecules (in
red) decreases with time while the number of B molecules (in blue)
increases. The steeper the concentration versus time curve, the
faster the reaction rate is. The film strip represents the relative
number of A and B molecules at each time.
5) Availability of rate-accelerating agents called
catalysts
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
78
Chemical nature of the reactants
◦ Bonds break and form during reactions
 The most fundamental difference in reaction rates lies in the
molecules, or ions) collide before the reaction can
 Some reactions are fast by nature whereas others are slow
occur.
ΔG0rxn = -397 kJ/mol
◦ This depends on the phase in which the reactants
This reaction one does not occur.
C(graphite) + O2(g) →CO2(g)
Ability of the reactants to meet
◦ Most reactions require that particles (atoms,
reactants themselves.
C(diamond) + O2(g)→CO2(g)

are.
ΔG0rxn = -394 kJ/mol.
This reaction occurs rapidly.
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◦ In a homogeneous reaction the reactants are in the
same phase:
Effect of crushing a solid.
When a single solid is
subdivided into much smaller
pieces, the total surface area on
all of the pieces becomes very
large.
 For example both reactants in the gas (vapour) phase.
◦ In a heterogeneous reaction the reactants are in
different phases:
 For example one reactant is present in the liquid
whereas the second is in the solid phase.
◦ In heterogeneous reactions the reactants meet only at
the intersection between the phases.
◦ Thus the area of contact between the phases
determines the rate of the reaction.
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
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Concentration of the reactants
H2O2(aq) → H2O(l) + ½ O2 (g)
Volume
O2 [cm3]
◦ Both homogeneous and heterogeneous reaction
rates are affected by reactant concentration
Concentration of H2O2
0,5 mol/l of
0,4 mol/l
 For example, red hot steel wool bursts into
flames in the presence of pure oxygen
0,3 mol/l
0,2 mol/l
0,1 mol/ l
Time [s]
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Temperature
Example
◦ The rates for almost all chemical reactions
Consider this reaction:
enhance as the temperature is increased
 Cold-blooded creatures, such as insects and
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
reptiles, become sluggish at lower temperatures
as their metabolism slows down.
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Condition
Condition
Concentration
Temperature
Affect on Rate
Explanation
Increasing the
More HCl particles means there
concentration of HCl will will be more collisions between
increase the reaCtion rate.
HCl and Zn.
Increasing temperature
increases the reaction
rate.
HCl particles will gain more
kinetic energy increasing the
number of collisions with Zn
atoms.
More Zn and HCl particles will
have sufficient energy to react
resulting in more successful
collisions.
Affect on Rate
Explanation
Particle Size
Reducing the size of the Zn
Reducing the size of
particles increases the surface area
Zn particles will
available for reaction with HCl
increase the rate of
molecules resulting in more
reaction.
collisions.
Stirring Rate
Stirring will keep small Zn particles
Increasing the
in suspension, increasing the
stirring rate of this
surface area available for
mixture will increase
collisions, resulting in an increased
the reaction rate.
reaction rate.
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Quiz

Catalyst
Student performed experiment on dissolution
of Mg in HCl in four test tubes according to
the following conditions.
Test tube
Mg size

HCl concentration
1
cube
1.0 M
2
cube
0.5 M
3
powder
1.0 M
4
powder
0.5 M
A catalyst is a substance that changes the
rate of a chemical reaction without itself
being used up.
◦ Positive catalysts speed up reactions
◦ Negative catalysts or inhibitors slow reactions


Determine the order of test tubes according to the
reaction time decreasing
(Positive) catalysts speed reactions by
allowing the rate-limiting step to proceed
with a lower activation energy.
Thus a larger fraction of the collisions are
effective.
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
Catalysts can be divided into two groups
◦ Homogeneous catalysts exist in the same phase
as the reactants.
◦ Heterogeneous catalysts exist in a separate
phase.

NO2 is a homogeneous catalyst for the
production of sulfuric acid in the lead
chamber process.

The mechanism is:
(a) The catalyst provides an alternate, low-energy path from the
reactants to the products.
(b) A larger fraction of molecules have sufficient energy to react
when the catalyzed path is available.
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S  O 2  SO 2

SO 2  12 O 2  SO 3

SO3  H 2 O  H 2SO 4

Heterogeneous catalysts are typically
solids
Consider the synthesis of ammonia from
hydrogen and nitrogen by the Haber
process
3H2  N 2  2NH3
The second step is slow, but is catalyzed
by NO2:
NO 2  SO 2  NO  SO3
NO 12 O2 NO2
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Mechanism theories
The Haber process. Catalytic formation of ammonia molecules
from hydrogen and nitrogen on the surface of a catalyst.
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


The reaction’s mechanism is the series of
simple reactions called elementary processes.

The rate law of an elementary process can be

written from its chemical equation.

One of the simplest models explaining
reaction rates is collision theory.
According to collision theory, the rate of
reaction is proportional to the effective
number of collisions per second among the
reacting molecules.
An effective collision is one that actually gives
product molecules.
The number of all types of collisions increase
with concentration, including effective
collisions.
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There are a number of reasons why only a
small fraction of all the collisions leads to
the formation of product:
◦ Only a small fraction of the collisions are
energetic enough to lead to products.
◦ Molecular orientation is important because a
collision on the “wrong side” of a reacting
species cannot produce any product.
 This becomes more important as the complexity of
The key step in the decomposition of NO2Cl to NO2 and Cl2 is
the collision of a Cl atom with a NO2Cl molecules. (a) A poorly
orientated collision. (b) An effectively orientated collision.
the reactants increases.
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◦ The minimum of kinetic energy of the
colliding particles must have is called the
activation energy, Ea .
◦ In a successful collision, the activation
energy changes to potential energy as the
bonds rearrange to form products.
◦ Activation energies can be large, so only a
small fraction of the well-orientated,
colliding molecules have it.
◦ When temperature increases the average
kinetic energy of the reacting particle also
increases.
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2
Kinetic energy distribution for a reaction at two different
temperatures. At the higher temperature, a larger fraction of the
collisions have sufficient energy for reaction to occur. The shaded
area under the curves represent the reacting fraction of the
collisions.

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Activation energies and heats of reactions
can be determined from potential-energy
diagrams
Potential-energy
diagram for an
endothermic
reaction. The heat
of reaction and
activation energy
are labeled.
The potential-energy diagram for an exothermic
reaction. The extent of reaction is represented as
the reaction coordinate.
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
Reactions generally have different activation
energies in the forward and reverse direction
A unsuccessful (a) and successful (b) collision for an exothermic
reaction.
Activation energy barrier for the forward and reverse reactions.
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
Transition state theory explains what
happens when reactant particles come
together.
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
The brief moment during a successful
collision that the reactant bonds are
partially broken and the product bonds are
partially formed is called the transition
state.
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

The overall rate law determined for the
mechanism must agree with the observed
rate law.
The exponents in the rate law for an
elementary process are equal to the
coefficients of the reactants in chemical
equation
Elementary process :
2NO 2  NO 3  NO
Formation of the activated complex in the reaction between
NO2Cl and Cl.
NO2Cl+ClNO2+Cl2




rate  k[NO 2 ]2
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Multistep reactions are common.
The sum of the element processes must
give the overall reaction.
Slow set in a multistep reaction limits how
fast the final products can be formed and
is called the rate-determining or ratelimiting step.
Simultaneous collisions between three or
more particles are extremely rare.
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


A reaction that depended on a three-body
collision would be extremely slow.
Thus, reaction mechanism seldom includes
elementary process that involves more
than two-body or bimolecular collisions.
Consider the reaction
2NO  2H 2  N 2  2H 2 O
rate  k[NO] 2 [H 2 ] (experimental)

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The mechanism is thought to be
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2NO 
 N 2O 2
(fast)
N 2O 2  H 2  N 2O  H 2O (slow)
N 2O  H 2  N 2  H 2O



(fast)
rate(forward)  k f [ NO]2
The second step is the rate-limiting step,
which gives
rate(reverse)  k r [ N 2 O 2 ] thus
rate  k[ N 2O2 ][H2 ]

The first step reaches a fast equilibrium
At equilibrium, the rates of the forward
and reverse reaction are equal
k f [ NO]2  k r [ N 2 O 2 ] or
N2O2 is a reactive intermediate, and can be
eliminated from the expression.
[ N 2O 2 ] 
kf
kr
[ NO]2
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
Substituting, the rate law becomes

rate  k[ N 2 O 2 ][H 2 ]
The activation energy is related to the rate
constant by the Arrhenius equation
k  Ae  Ea / RT
 kf 
rate  k  [ NO]2 [H 2 ] or
 kr 
rate  k '[ NO]2 [H 2 ]

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k = rate constant
Ea = activation energy
e = base of the natural logarithm
R = gas constant = 8.314 J mol-1 K-1
T = Kelvin temperature
A = frequency factor or pre-exponential factor
Which is consistent with the experimental
rate law.
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
The Arrhenius equation can be put in
standard slope-intercept form by taking
the natural logarithm
ln k  ln A  Ea / RT

k  E  1 1 
ln 2   a   
 k1  R  T2 T1 
or
ln k  ln A  ( Ea / R)  (1 / T )


y  b


 m
The activation energy can be related to the
rate constant at two temperatures

x
A plot of ln k versus (1/T) gives a straight
line with slope = -Ea/RT
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Quiz

The activation energy of the first order
reaction is 50,2 kJ/mol at 25oC. At what
temperature will the rate constant double?
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True or false quiz
1. A catalyst alters the rate of a chemical reaction by:
a) always providing a surface on which molecules react
b) changing the products formed in the reaction
c) inducing an alternate pathway for the reaction
with generally lower activation energy
d) changing the frequency of collisions between molecules
2. The Rate of a Chemical Reaction
a) usually is increased when the concentration of one of the
reactants is increased
b) is dependent on temperature
c) may be inhibited by certain catalytic agents
d) will be very rapid if the activation energy is large
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