Transcript Original powerpoint (~3.9 MB)

Chapter 14
Chemical Kinetics
Dr. Peter Warburton
[email protected]
http://www.chem.mun.ca/zcourses/1051.php
Rates
Rates can be measured as the
concentration change of a chemical (X)
over a period (change) of time
Rate = D[X] / Dt
Rate 
Xt  Xt
1
0
t1  t 0
Often use molL-1 (or M) as units of concentration
so rate often has units molL-1s-1 (or Ms-1 )
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Rate
Rates can be expressed in two different
ways. We can look at the
formation of a product
(increase in the concentration over time)
or the
disappearance of a reactant
(decrease in the concentration over time).
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Rate of formation of a product
2 Fe3+(aq) + Sn2 +(aq) → 2 Fe2+(aq) + Sn4+(aq)
t0 = 0.0 s
[Fe2+]t0 = 0.0000 M
t1 = 38.5 s
[Fe2+]t1 = 0.0010 M
Δt = 38.5 s Δ[Fe2+] = (0.0010 – 0.0000) M
rateof formationof Fe
2


Δ Fe 2
0.0010M


 2.6 x 105 M  s 1
Δt
38.5s
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Rate of disappearance of a reactant
2 Fe3+(aq) + Sn2 +(aq) → 2 Fe2+(aq) + Sn4+(aq)
t0 = 0.0 s
[Sn2+]t0 = 0.0015 M
t1 = 38.5 s
[Sn2+]t1 = 0.0010 M
Δt = 38.5 s Δ[Sn2+] = (0.0010 – 0.0015) M
rateof disappearance of Sn
2


Δ Sn 2
(-0.0005M)


 1.3 x 105 M  s 1
Δt
38.5s
The word “disappearance” implies the negative
sign. Because of this, all rates are considered to be
POSITIVE!
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Reaction rates
aA+bBcC+dD
1 ΔA 
1 ΔB
1 ΔC
1 ΔD
reaction rate  +
+
a Δt
b Δt
c Δt
d Δt
Reaction rates are always positive, so we must
put negative signs in front of reactant
concentration changes!
To account for the stoichiometric relationships
and their effect on rate, we must always divide
by the stoichiometric coefficient for the chemical.
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Reaction rate
2 Fe3+(aq) + Sn2 +(aq) → 2 Fe2+(aq) + Sn4+(aq)

3

1
Δ
Fe
 
reactionrate   
Δt
 2 

    1   ΔSn 
2

  1 
Δt




1.3 x 10 5 M s 1
slide 5
2
4
 1  Δ Fe   1  Δ Sn
   
  
Δt   1  Δt
 2 




2.6 x 10 5 Ms 1
slide 4
reactionrate  1.3 x 105 M  s 1
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Measuring rates
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Measuring rates
1) Average rates
2) Slopes
3) Time = 0
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Problem
At some point in the reaction
2A+B3C
[A] = 0.3629 M. At a time 8.25 minutes
later [A] = 0.3187 M. What is the average
rate of the disappearance of A in Ms-1,
what is the average rate of the formation of
C in Ms-1, and what is the average rate of
reaction in Ms-1 over that time interval?
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Problem answers
5
averagerateof disappearance of A  8.93 x 10 M  s
1
averagerateof formationof C  1.34 x 10 4 M  s 1
averagereactionrate  4.46 x 105 M  s 1
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Instantaneous reaction rates
What’s happening at “this instant in time”?
We can use
instantaneous
reaction rates.
The initial rate is
the instantaneous
reaction rate for a
reaction at time
zero.
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Rate laws
The rate of a chemical reaction
depends on the concentration of
some or all of the reactants.
A reactant might not affect
the rate, regardless of its
concentration.
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Rate laws
rate law for a reaction
is the equation showing the
dependence of the reaction rate on
The
concentrations
of the reactants.
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a A + b B + …  products
m
n
rate = k [A] [B] …
k is a constant for the reaction
AT A GIVEN TEMPERATURE,
and is called the rate constant.
The stoichiometry of the balanced reaction
equation IS NOT ALWAYS the source of the
rate equation exponents!
m DOES NOT HAVE TO equal a
n DOES NOT HAVE TO equal b
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Reaction order
Reaction order with respect to a given
reactant
is the value of the exponent of the rate law
equation for the specific reactant only.
The overall reaction order is the
sum of the reaction orders for
all reactants.
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Reaction order example
2
rate = k [A] [B]
The reaction order with respect to A is 2
or the reaction is second order in A
The reaction order with respect to B is 1
or the reaction is first order in B
The overall reaction order is 3 (2 + 1 = 3)
or the reaction is third order overall
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“Sensitivity” to concentration change
Rate change of the
reaction if [A] is
doubled depends on 2m
rate
doubles
rate
quadruples
zeroth
order in A
first order
in A
second
order in A
rate no change
halves
NOTE:
negative or
non-integer
orders
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Method of initial rates
Reaction rate laws are often
determined experimentally!
We most commonly carry out a series of
experiments in which the
initial rate of the reaction is measured
as a function of
different initial concentrations of
reactants
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Method of initial rates
If you see a table like this with chemical concentrations
or pressures and rate data, chances are good the
question is a method of initial rates problem.
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Method of initial rates
IGNORE THE REACTION with this type
of problem. The chemicals in the TABLE
are the interesting ones.
You always require at least one more
experimental reaction than your number
of chemicals given in your table!
Sometimes we are given a table with an
extra experiment which we can use to
check if we’ve done everything correctly.
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2 NO (g) + O2 (g)  NO2 (g)
Since rate laws are always expressed in
terms of reactants (and sometimes
catalysts – we’ll see these later), lets
create a general form of the rate law for
this reaction based on what chemicals the
TABLE tells us are involved in the rate of
the reaction.
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2 NO (g) + O2 (g)  NO2 (g)
m
n
rate = k [NO] [O2]
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Method of initial rates
For our initial reactant order determination
we need to choose a pair of reactions where
only one reactant concentration changes.
Experiments #1 and #2 fulfill this condition.
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Method of initial rates
m
n
rate = k [NO] [O2]
Since k is a constant then
k for experiment 1
IS EQUAL TO
k for experiment 2!
m
n
k = rate / [NO] [O2]
rate1
rate2
rate1 NO O

so

m
n
m
n
NO1 O2 1 NO2 O2 2 rate2 NO O
m
1
m
2
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

n
2 1
n
2 2
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Reaction order w.r.t. NO
rate 1 NO 1 O 2 1

rate 2 NO m2 O 2 n2
m
m
n
0.048M  s -1 (0.015M) (0.015M)

-1
0.192M  s
(0.030M)m (0.015M)n
0.25  (0.50)m

log 0.25  log (0.50)m
log 0.25  m log 0.50

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n
log 0.25
log 0.50
 0.602
m
 0.301
m2
m
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Reaction order w.r.t. O2
rate 1 NO 1 O 2 1

rate 3 NO 32 O 2 3n
2
2
n
0.048M  s -1 (0.015M) (0.015M)

-1
0.096M  s
(0.015M)2 (0.030M)n
0.50  (0.50)n

log 0.50  log (0.50)n
log 0.50  n log 0.50

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n
log 0.50
log 0.50
 0.301
n
 0.301
n 1
n
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ALTERNATE reaction order w.r.t. O2
rate 1 NO 1 O 2 1

rate 4 NO 24 O 2 n4
2
0.048M  s

-1
0.384M  s
(0.030M)2 (0.030M)n
-1
(0.015M)2 (0.015M)n
0.125 (0.50)2 (0.50)n
0.125 [0.25](0.50)n
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n
0.125
 (0.50)n
0.25
0.50  (0.50)n
n 1
28
Our rate law
2
1
rate = k [NO] [O2]
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Rate constant using experiment 1
2
1
k = rate / [NO] [O2]
0.048M  s 1
k


2
2
NO O 2  0.015M  0.015M  3.38 x 106 M 3
rate
0.048M  s 1
k  1.42 x 104 M -2  s 1
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Rate constant using experiment 2
2
1
k = rate / [NO] [O2]
rate
0.192M  s 1
0.192M  s 1
k


2
2
NO O 2  0.030M  0.015M  1.35 x 105 M 3
k  1.42 x 10 M  s
4
-2
1
The rate constant is the same, as it should be!
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Check using extra experiment
rate = (1.42 x 10 M s ) [NO] [O2]
4

rate  1.4
rate  1.4
-2
-1
2
1

0.030M 0.030M
2.7 x 10 M 
rate  1.42 x 104 M -2  s 1 NO O 2 
2
1
2
x 10 M  s
2
x 104 M -2  s 1
4
-2
2
-5
3
rate  3.83 x 10-1 M  s 1
The rate is the same as the experimentally observed
rate (within rounding errors). We MUST have done
everything right!
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Units of rate constants
Rate always has units in terms of concentration per
time unit.
Usually it is molL-1·s-1 (or M·s-1)
To ensure we get the right units for rate means the
rate constant must have different units depending
on the overall reaction order.
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Problem
H2O2 (aq) + 3 I- (aq) + 2 H+ (aq)  I3- (aq) + 2 H2O (l)
The rate of formation of the red-coloured triiodide ion
D[I3-]/Dt (and therefore the reaction rate – why?) can
be determined by measuring the rate of appearance of
the colour.
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Problem
Expt
Initial [H2O2] (M) Initial [I-] (M)
Initial D[I3-]/Dt (Ms-1)
1
0.100
0.100
1.15 x 10-4
2
0.100
0.200
2.30 x 10-4
3
0.200
0.100
2.30 x 10-4
4
0.200
0.200
4.60 x 10-4
a) What is the rate law for the formation of I3-?
b) What is the value for the rate constant?
c) What is the initial rate of formation of triiodide
when the concentrations are [H2O2] = 0.300
M and [I-] = 0.400 M?
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Problem answers
a) rate = k [H2O2] [I-]
b) k = 1.15 x 10-2 M-1s-1
c) rate = 1.38 x 10-3 Ms-1
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Zero order reactions
In a zero order reaction the rate of the
reaction does NOT depend on the
concentration of ANY of the reactants.
2 NH3 (g)  N 2 (g)  3 H 2 (g)
1130 K
Pt catalyst
0
rate = k [NH3] = k (1) = k = constant
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Integrated rate law – 0th order rxn
rat e  k
dA 
 k
dt
dA    k dt
A t
At  A0  k t
OR
dA    k  dt

 
A
0

A t  
k t  A 0


t
0
y
m x
b
A t  A 0  k t - 0
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Features of zero order reactions
Concentration versus
time graph gives a
straight line.
Integrated rate law

A t  
k t  A 0


m x
y
b
Rate constant is the
negative slope (UNITS!)
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First order reactions
In a first order reaction the overall order of
the reaction is 1. A common type of first
order reaction is the decomposition of a
chemical.
1
H 2 O 2 (aq) 
 H 2 O (l)  O 2 (g)
2
rate = k [H2O2]
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1
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Integrated rate law – 1st order rxn
rate kA 
dA 
 kA 
dt
dA 
 k dt
A
A t
dA 
 k  dt

A 0 A 
0
lnA t  ln A 0  k t
t

A t
ln
 k t - 0
A0
OR
lnA t  
k t  ln A 0

 m x 

y
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b
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Integrated rate law – 1st order rxn
Natural logarithm of
concentration versus
time graph gives a
straight line.
lnA t  
k t  ln A 0

 m x 

y
b
Rate constant is the
negative slope
(UNITS!)
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Problem
The reaction
A2B+C
is first order. If the initial [A] = 2.80 M and
k = 3.02 x 10-3 s-1, what is the value of [A]
after 325 s?
Answer: [A] = 1.05 M
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Half-life of 1st order reactions

A t
ln
 k t
A0
1 A 0
ln 2
 k t 1
2
A0
From the integrated
rate law of a first order
reaction we can look at
what time [A] is onehalf of the initial
concentration (½ [A]0).
We call this time
the half-life (t½).
t1
2
ln 1  k t 1
2
2
ln 1

- 0.693
2


k
k
0.693
t1 
2
k
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Half-life of 1st order reactions

A t
ln
 k t
A0
1 A 0
ln 2
 k t 1
2
A0
The half-life (t½)
of a first order
reaction is
constant!
t1
2
ln 1  k t 1
2
2
ln 1

- 0.693
2


k
k
0.693
t1 
2
k
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Half-life of 1st order reactions
[A]t = 1/2 [A]0 at t½
[A]t = 1/4 [A]0 at 2 x t½
[A]t = 1/8 [A]0 at 3 x t½
and so on
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Problem
Consider the first order reaction
A  products
with k = 2.95 x 10-3 s-1. What percent of A
remains after 150 s?
Answer: % A remaining = 64.2%
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Problem
At what time (in minutes) after the start of
the reaction is a sample of H2O2 (aq) twothirds decomposed if k = 7.30 x 10-4 s-1?
Answer: time = 25.1 min
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1st order reactions of gases
Amounts of gases
are often measured
by pressure
instead of
concentration
molesA n A
A 

volume V
but P V  nRT
n
P
or 
V RT
PA
so A  
RT
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Problem
Start with DTBP at a
pressure of 800.0
mmHg
at 147 C.
What will the pressure
be at t = 125 min if
t½ = 8.0 x 10 min?
ln
A t
A 0
 k t
 PA

 RT 

t
ln
 k t
 PA

 RT 

0
P 
ln A t   k t
PA 0
Answer:
k = 8.66 x 10-3 min-1
pressure = 271 mmHg
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Radioactive decay as 1st order rxns
Many radioactive decay processes are first
order…
β - decay
14
14
 half-life of 5730 yrs
C  N  ν  e
6
131
53
7

I  Xe  ν  e half-life of 8.04 days
238
92
β - decay
131
54
U  Th He
α decay
234
90
4
2
2
half-life of 4.51 x 109 yrs
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Second order reactions
In a second order reaction the overall
order of the reaction is 2. If there is only a
single reactant chemical, like in
2 NO2 (g) 
 2 NO (g)  O2 (g)
Then the rate is second order with respect
to that chemical
rate = k [NO2]
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2
52
Integrated rate law – 2nd order rxn
rat e kA 
2
dA 
2
  kA 
dt
dA 
  k dt
2
A 
A t
dA 
1  1 
  k t
 
A t  A 0 
OR
 1 
1

 k t  




A
A
mx
t
0 

 1 
1




   k t - 0 
  A   A  
y
b

  A 
2
A
0
t
t
  k  dt
0
0
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Integrated rate law – 2nd order rxn
Inverse of
concentration versus
time graph gives a
straight line.
 1 
1

 k t  

A t m x  A 0 



y
b
Rate constant is the
slope (UNITS!)
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Half-life of 2nd order reactions
From the integrated rate
law of a second order
reaction we can look at
what time [A] is one-half of
the initial concentration
(½ [A]0).
We also call this time
the
half-life (t½).
1  1 
  k t
 
A t  A 0 
 1 
  k t 1
 
1 A 0  A 0 
2
2
1 1
    k t 1 A 0
1 1
2
2
2  1  k t 1 A 0
1
2
1
 t1
2
k A 0
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Half-life of 2nd order reactions
The half-life (t½)
of a second
order reaction is
NOT constant!
1  1 
  k t
 
A t  A0 
 1 
  k t 1
 
1 A 0  A 0 
2
2
1 1
    k t 1 A 0
1 1
2
2
2  1  k t 1 A 0
1
2
1
 t1
2
k A 0
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Half-life of 2nd order reactions
The greater the
initial conc. of A
([A]0),
the smaller the
half-life!
 1 
1
  k t
 
A t  A 0 
 1 
  k t 1
 
1 A 0  A 0 
2
2
1 1
    k t 1 A 0
1
2
1
2
2  1  k t 1 A 0
1
2
1
 t1
2
k A 0
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Half-life of 2nd order reactions
1
 t1
2
k A 0
Since
reactant
concentration
is halved
over a half-life, the
next half-life
is twice as long
compared to the
previous half-life.
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Pseudo first-order reactions
If we carry out some second (or higher)
order reactions under conditions where
some reactant concentrations do not
change significantly, then the reaction
may APPEAR to act like a lower order
reaction (pseudo-first order, …)
CH3COOC2H5  H2O 
CH3COOH  C2H5OH
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Kinetics summary - rates
•You can calculate rate with known rate
law
•If you don’t know a rate law, the rate can
be determined from the tangent of a [A]
versus time graph or by –D[A]/Dt over a
small DT
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Kinetics summary – rxn orders
•Method of initial rates
•Plot graphs to find a straight line
•First order reactions have constant half-life
•Use data in integrated rate laws – k MUST NOT
change
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Kinetics summary – rate constants
•Rate constants are related to slope of
appropriate straight line graphs
•Use data in integrated rate laws to get k.
Once you know k you can use the
integrated rate law to solve for
concentrations or times.
•Use half-life of a first order reaction to
determine k
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Problem
Without graphing, find the order and the rate
constant for the following reaction using the
given data
B  products
Time (s)
[B] (M)
Time (s)
[B] (M)
0
0.88
100
0.44
25
0.74
150
0.31
50
0.62
200
0.22
75
0.52
250
0.16
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Problem answer
Reaction is first order.
-3
-1
k = 6.93 x 10 s
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64
Bumper cars
A gasp of surprise
could be a “reaction”
when riding in a
bumper car
“Reactions” generally occur when the
bumps are very hard and when they
come from behind
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Collision theory
A + BC  AB + C
At some point in time, the B-C bond starts to
break, while the A-B bond starts to form.
At this point, all three nuclei are weakly
linked together.
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Collision theory
Molecules tend to repel each other
when they get close.
It takes energy to force the molecules close
together! This is like forcing together the north
poles of two magnets.
This energy is the kinetic energy of the
molecules. It converts to potential energy as
the molecules get closer.
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Collision theory
A---B---C
has a higher potential energy than
either A + B-C or A-B + C
A---B---C
is the transition state
or the activated complex
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The difference in
energy between
products and
reactants is DE
The difference in energy between the
transition state and the reactants is
Ea – the activation energy
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Temperature is the average kinetic energy
of molecules.
Collisions between molecules at higher
temperatures are more likely to have
energy GREATER THAN the activation
energy.
Higher temperatures mean higher rates
of reaction!
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Collisions
1 billion collisions per molecule per
second
Every reaction should be almost
instantaneous.
This is not the case.
Not every collision breaks the
activation energy barrier!
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Collisions
The fraction of collisions that have enough energy
to overcome the activation energy barrier is
f = e-Ea/RT
e is approximately 2.7183,
Ea is the activation energy,
T is the temperature in Kelvin,
R is the gas law constant
(8.3145 JK-1mol-1)
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Bumper cars and energy
bumper car - a more
energetic collision is
more likely to make us
gasp (our “reaction”)
molecular collisions higher energy collisions
are more likely to lead to
reaction
(by overcoming the
activation energy)
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Bumper cars and orientation
You are also more likely to gasp if you
are hit from behind by another bumper
car.
The orientation of how the collision occurs
is also important to get a “reaction.”
The same is true for molecules where the
fraction of collisions that have the right
orientation is p. We call this fraction p the
steric factor.
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General reaction A + BC  AB + C
Atom A MUST collide with the B side of BC to form
the transition state A---B---C.
If atom A hits the C side, we get a different
transition state A---C---B.
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General reaction A + BC  AB + C
About half of our collisions won’t give the right
reaction EVEN IF THEY BREAK THE ACTIVATION
ENERGY BARRIER!
The steric factor p will be about 0.5.
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General reaction A + BC  AB + C
Collision rate = Z [A] [BC]
Z is a constant related to the collision
frequency.
Recall only a fraction (f) of the collisions have
a collision energy greater than or equal to
the activation energy.
Of those collisions, only a fraction (p) have
the correct orientation to proceed through
the transition state to the products.
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General reaction A + BC  AB + C
Reaction rate = p x f x Collision rate
Reaction rate = pfZ [A] [BC]
If for our general reaction
rate = k [A] [BC]
k = pfZ = pZ e-Ea/RT = A e-Ea/RT
(where A = pZ)
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Arrhenius Equation
pZ = A
As T increases
k increases
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Problem
AB + CD  AC + BD
What is the value of the activation energy for
this reaction? Is the reaction endothermic or
exothermic?
Suggest a
plausible
structure for
the transition
state.
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Using the Arrhenius Equation
If we know the rate constants for
a reaction at two different
temperatures, we can then
calculate the activation energy.
k = A e-Ea/RT
ln k = ln (A e-Ea/RT)
ln k = ln (A) + ln (e-Ea/RT)
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ln k = ln (A) – (Ea/RT)
This is the equation for a straight line!
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slope = -Ea/R so Ea = - slope x R
A graph of the
natural logarithm
of the rate
constant versus
inverse
temperature
will give a
straight line
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If we know
the rate constants (k1 and k2)
at two temperatures (in Kelvin!) T1 and T2
we will see the rate constant has changed
because the temperature has changed.
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One of the things that
HAS NOT CHANGED
with the temperature is the
ACTIVATION ENERGY
-Ea/R (at T1) = -Ea/R (at T2)
It is constant just like slope is constant
for a straight line!
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ln k2 = ln (A) – (Ea/RT2)
ln k1 = ln (A) – (Ea/RT1)
Subtract the bottom from the top on both sides
ln k2 – ln k1 = [-(Ea/RT2)] – [-(Ea/RT1)]
Pull out –Ea/R and put it in front
ln k2 – ln k1 = (–Ea/R) (1/T2 – 1/T1)
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Your textbook says
ln (k2/k1) = (Ea/R) (1/T1 – 1/T2)
This is absolutely correct as well!
Use whichever form of the relation that
you feel more comfortable with
mathematically. There are more forms
given in the text.
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Problem
Rate constants for the decomposition of
gaseous dinitrogen pentaoxide are
3.7 x 10-5 s-1 at 25 °C
and 1.7 x 10-3 s-1 at 55 °C
2 N2O5 (g)  4 NO2 (g) + O2 (g)
What is the activation energy of this
reaction in kJmol-1?
What is the rate constant at 35 C?
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90
Problem answer
-1
kJmol
Ea = 104
-4
-1
k35C = 1.4 x 10 s
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Reaction Mechanisms
A reaction mechanism is the sequence
of molecular events (elementary steps
or elementary processes) that defines
the pathway from the reactants to the
products in the overall reaction.
The elementary processes describe the
behaviour of individual molecules while
the overall reaction tells us stoichiometry.
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NO2 (g) + CO (g)  NO (g) + CO2 (g)
(Overall Reaction)
The reaction actually takes place in
two elementary processes!
 2 NO2  NO and NO3
NO3 + CO NO2 and CO2
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NO2 (g) + CO (g)  NO (g) + CO2 (g)
(Overall Reaction)
Elementary processes must add
together to give the overall
equation!
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NO2 (g) + CO (g)  NO (g) + CO2 (g)
(Overall Reaction)
Some of the “crossed-out” chemicals are
neither reactants nor products in the overall
reaction.
For example, in the above reaction NO3 is
formed in one elementary step and
consumed in a later elementary step.
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Reaction intermediate
A reaction intermediate is a species that
is formed in an elementary process, that
is consumed in a later elementary
process.
We never see reaction intermediates
in the overall reaction!
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Molecularity
The molecularity of an elementary
process is the number of molecules
on the reactant side
of the elementary process reaction.
A one molecule elementary reaction is unimolecular.
A two molecule elementary reaction is bimolecular.
A three molecule elementary reaction is termolecular.
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Molecularity
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Chances for molecularity
The chances of a unimolecular reaction
occurring only depends on the one molecule, and
are good.
A bimolecular reaction requires that two
molecules collide with each other. This isn’t
difficult and happens quite often.
A termolecular reaction requires that three
molecules collide with each other at the same
time. The chances of this happening are not
very good.
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Bumper cars
Consider bumper cars. Very often, you
will hit one other bumper car. It is a
very rare occurrence
to have a “bumper
car” pile-up where
three or more cars
hit at exactly the
same time.
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Problem
A suggested mechanism for the reaction of
nitrogen dioxide and molecular fluorine is shown
a) Give the chemical
equation for the
overall reaction, and
identify any reaction
intermediates.
b) What is the
molecularity of each
of the elementary
processes?
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Problem answer
2 NO2 (g) + F2 (g)  2 NO2F (g)
F is a reaction intermediate (formed in
step 1 then consumed in step 2)
Step 1 is bimolecular.
Step 2 is bimolecular.
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Rate Laws and Reaction Mechanisms
UNLIKE an overall reaction the
rate law for an elementary process
follows directly from the
molecularity of the elementary
reaction!
For a general elementary process
a A + b B  products
rate = k [A]a [B]b
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Ozone
Unimolecular decomposition of ozone.
O3 (g)  O2 (g) + O (g)
The rate law will
respect to ozone
be first order with
rate = k [O3]
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Bimolecular reaction
A + B  products
Reaction depends on collisions between
molecules A and B
Increase A, you increase # collisions
Increase B, you increase # collisions
rate = k [A] [B]
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Bimolecular reactions for one reactant
A + A  products
Increase A, you increase # collisions for
EACH A
rate = k [A] [A] = k [A]2
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Mechanisms and overall rate law
The mechanism of the overall
reaction is predicted through the
elementary processes therefore
the elementary processes will
determine the rate law of the
overall reaction!
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One step mechanisms
If the overall reaction occurs in
one elementary process, then
the two reactions are the
same.
The rate law for the overall
reaction is the same as the
elementary reaction rate law.
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Single step mechanism
rate = k [CH3Br]
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110
Rate-determining step
For reaction mechanisms of
more than one elementary
step, one of the elementary
step reactions MAY HAVE a
much slower rate than any of
the other steps.
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Rate-determining step
The rate-determining step
of an overall reaction is an
elementary step reaction
which has a much slower rate than
any other step.
The overall rate law will match the
rate law of the rate-determining step
in this case.
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Proposing mechanisms
A proposed mechanism must satisfy two
criteria to be considered plausible:
1) the elementary processes must add
up to give the appropriate overall
reaction
2) the rate law that arises from the
proposed mechanism must be
consistent with the observed rate law
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H2 (g) + 2 ICl (g)  I2 (g) + 2 HCl (g)
Observed
rate law
rate = k [H2][ICl]
SLOW : H 2  ICl 
 HI  HCl
FAST : HI  ICl 
 I 2  HCl
H 2  2 ICl 
 I 2  2 HCl
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H2 (g) + 2 ICl (g)  I2 (g) + 2 HCl (g)
The elementary
SLOW : H 2  ICl 
 HI  HCl
reactions DO add up
FAST : HI  ICl 
 I 2  HCl
to the overall
reaction.
H 2  2 ICl 
 I 2  2 HCl
The rate law of the rate-determining step is
rate = k [H2][ICl]
It turns out this is the same as the
experimentally observed rate law, so this
mechanism is plausible.
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2 NO (g) + O2 (g)  2 NO2 (g)
Observed rate = k [NO]2 [O2]
The rate law is termolecular.
The chances of a single slow termolecular
step reaction occurring are very poor!
We need a plausible mechanism where
the slow step cannot be the first step.
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Fast reversible step – slow step
A reversible reaction is one where
reactants can become products
and the
products can become reactants
AT THE SAME TIME.
Eventually both reactions
have the same rate
(the system is at equilibrium)
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Fast reversible step – slow step
FAST :
2 NO
k1


k2
N 2O 2
k3
SLOW : N 2 O 2  O 2 
2 NO2
k
2 NO  O 2 

2 NO2
Rate law of slow step
rate3 = k3 [N2O2] [O2]
BUT N2O2 is a reaction intermediate
(concentration is effectively constant!)
We can’t have it in the rate law!
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Fast reversible step – slow step
k
Rate law of fast

FAST :
2 NO
N 2O 2
forward step

k
k
rate1 = k1 [NO]2 SLOW : N 2O 2  O 2 
2 NO2
k
Rate law of fast
2 NO  O 2 

2 NO2
reverse step
rate2 = k2 [N2O2]
1
2
3
At equilibrium rate1
= rate2
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Fast reversible step – slow step
FAST :
2 NO
k1


k2
N 2O2
k3
SLOW : N 2 O 2  O 2 
2 NO 2
k3
2 NO  O 2 
2 NO 2
k1 [NO]2 = k2 [N2O2]
so [N2O2] = k1/k2 [NO]2
Let’s say that
k1/k2 = K
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Fast reversible step – slow step
FAST :
2 NO
k1


k2
N 2O 2
k3
SLOW : N 2 O 2  O 2 
2 NO2
2 NO  O 2 
 2 NO2
k
Rate law of slow step
rate3 = k3 [N2O2] [O2]
We can substitute in [N2O2] from fast rxn!
rate3 = k3 K [NO]2 [O2]
If we let k = k3 K then we see the rate law
of the slow reaction does match the
observed rate law!
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Just because a mechanism is
plausible doesn’t mean it is right!
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Problem
In a proposed two-step mechanism
for CO + NO2  CO2 + NO, the
second, fast step is
NO3 + CO  NO2 + CO2.
What must be the slow step? What
would you expect the rate law of the
rxn to be?
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Problem answer
Slow step
NO2 + NO2  NO3 + NO.
Rate law: rate = k [NO2]2
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Problem
Show that the proposed mechanism is
plausible for the reaction
2 NO2 (g) + F2 (g)  2 NO2F (g)
if the rate law is rate = k [NO2][F2]


FAST : NO2 (g)  F2 (g)
NO2 F2 (g)


SLOW :
FAST :
NO2 F2 (g) 
 NO2 F (g)  F (g)
F (g)  NO2 
 NO2 F (g)
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Steady-state approximation
More than one elementary process
may determine the rate of an overall
reaction.
No rate-determining step!
Can use the steady state
approximation in this case.
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2 NO (g) + O2 (g)  2 NO2 (g)
Observed rate = k [NO]2 [O2]
We’ve already seen this as the fast
reversible step – slow step
mechanism example.
Let’s propose a mechanism
WITHOUT making any initial
assumptions about the step rates!
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2 NO (g) + O2 (g)  2 NO2 (g)
P rocess1 :
P rocess2 :
k1
NO  NO 
N 2O2
N 2 O 2  NO  NO
k2
P rocess3 : N 2 O 2  O 2  2 NO 2
k3
This is essentially the same as the fast
reversible step – slow step mechanism,
but we treat the reversible reaction as two
separate forward reactions.
Each step has its own rate law.
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2 NO (g) + O2 (g)  2 NO2 (g)
P rocess1 :
P rocess2 :
NO  NO  N 2 O 2
k1
N 2 O 2  NO  NO
k2
P rocess3 : N 2 O 2  O 2  2 NO 2
k3
We know a reaction intermediate
ALWAYS has a near-zero concentration
that does not change much with time
(otherwise we would see it build up!)
This means D[N2O2]/Dt  0 (steady state)
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2 NO (g) + O2 (g)  2 NO2 (g)
P rocess1 :
P rocess2 :
NO  NO  N 2 O 2
k1
N 2 O 2  NO  NO
k2
P rocess3 : N 2 O 2  O 2  2 NO 2
k3
Choose our starting point as the step which
has a rate law closest to the observed one,
while including the reaction intermediate.
rate = k3 [N2O2] [O2]
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2 NO (g) + O2 (g)  2 NO2 (g)
D[N2O2]/Dt  0
rateof formationN 2 O 2  rateof disappearance N 2 O 2  0
rateof formationN 2 O 2  - rateof disappearance N 2 O 2 
We form the intermediate in process 1 and
consume it in processes 2 and 3
rateof formationN 2 O 2  rateprocess1  k1 NO
2
rateof disappearance N 2 O 2  - rateprocess2  rateprocess3

BE CAREFUL! These are
two different negative signs!
negative because we used positive
reactionrates for disappearance!
 k 2 N 2 O 2   k 3 N 2O 2 O 2 
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2 NO (g) + O2 (g)  2 NO2 (g)
rateof formationN 2 O 2  - rateof disappearance N 2 O 2 
k1 NO  -  k 2 N 2 O 2   k 3 N 2 O 2 O 2 
2
k1 NO  k 2 N 2 O 2   k 3 N 2 O 2 O 2 
2
Solve for [N2O2]
k 1 NO  N 2 O 2  k 2  k 3 O 2 
2
k 1 NO
N 2 O 2  
k 2  k 3 O 2 
2
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2 NO (g) + O2 (g)  2 NO2 (g)
Substitute into starting point
rate k 3 N 2 O 2 O 2 
 k 1 NO

O 2 
rate k 3 





k

k
O
3
2 
 2


k 1k 3
2
NO O 2 
rate 
 k 2  k 3 O 2  
2
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2 NO (g) + O2 (g)  2 NO2 (g)


k1k 3
2
NO O 2 
rate 
 k 2  k 3 O 2  
Very close to the observed rate law!
How do we make the first part constant?
Now we assume something
about step rates!
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2 NO (g) + O2 (g)  2 NO2 (g)
Assume rate2 >> rate3
so k2 [N2O2] >> k3 [N2O2] [O2]
k2 >> k3 [O2] SO k2 + k3 [O2]  k2
 k1k 3 
2
NO O2 
rate 
k2 


k
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Catalysis
Reaction rates are not just affected
by reactant concentrations and
temperatures.
A catalyst is a substance that
increases the rate of a reaction
WITHOUT undergoing permanent
change in the reaction.
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How does a catalyst work?
A catalyst makes available a
different reaction pathway that
is more efficient than the
uncatalyzed mechanism because
this pathway has a lower
activation energy.
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Homogeneous catalysts
A homogeneous catalyst exists in the
same phase as the reactants.
I- is a homogeneous catalyst for the
decomposition of hydrogen peroxide
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Decomposition of H2O2
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Decomposition of H2O2
Catalysts don’t appear in the
overall balanced equation!
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Heterogeneous catalysts
A heterogeneous catalyst
exists in a different phase
(usually solid) than the
reactants.
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Heterogeneous catalysts
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Heterogeneous catalysts
Most catalysts used in industry
are heterogeneous
It is much easier to separate a solid from a
gas or liquid (for example) than two liquids
or gases).
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Enzymes are catalysts
In living beings catalysts are
usually called enzymes
Carbonic anhydrase catalyzes the reaction
of carbon dioxide with water
CO2 (g) + H2O (l)  H+ (aq) + HCO3- (aq)
The enzyme increases the rate of this
reaction by a factor of 106. Equivalent to
about a 200 K increase …
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Enzymes
Lockand-key
model
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