CI 10.1
The collision theory of reactions
Reactions occur when particles of reactants collide with a certain
minimum kinetic energy. (Activation enthalpy)
Low concentration
Higher concentration
Any factor which increases the number of collisions will increase
the rate of reaction.
As particles approach and collide, K.E. is converted to P.E. and
the P.E. of reactants rises.
X
E
n
t
h
a
l
p
y
Activation enthalpy
Reactants
H
Products
Progress of reaction
In the above exothermic reaction, existing bonds start to stretch
and break and new bonds form. Only Reactants with sufficient
K.E. will overcome the activation enthalpy and form products.
At higher temperatures, more of the colliding particles have
enough energy to react.
Enthalpy profiles
X
E
n
t
h
a
l
p
y
Activation enthalpy
Reactants
H
Products
Progress of reaction
The curved line is the energy pathway for a pair of colliding
molecules (called the energy or enthalpy profile for the reaction).
X corresponds to the arrangement of atoms where old bonds are
stretched and new bonds are starting to form.
Eg. For chlorine atoms and ozone, the oxygen – oxygen bonds in
ozone break and a new bond between oxygen and chlorine is
forming.
Cl + O3  Cl------O------O
O  ClO + O2
At X
(this is a very unstable arrangement and only lasts for a very short
time).
The enthalpy profile is shown as a single curve (this is only true
for single-step reactions). Other reactions have several steps, so
the enthalpy profile would show several steps, too.
 do problems for 10.1 ‘Rates of Reactions’ p. 222-223 questions
1 to 4.
Questions
1. For each of the following reactions, say how, if at all, you
would expect its rate of reaction to be affected by the
following factors:
a) Temperature
b) Total pressure of gas.
c) Concentration of solution
d) Surface area of solid
a
b
C
A
Rate will
Rate will
Rate will
increase with increase with increase with
temperature
temperature
temperature
B
Rate of
Rate increases Rate of
forward
forward
reaction not
reaction not
affected
affected
C
Increasing the Solutions not Increasing the
concentration involved
concentration
of acid will
of peroxide
increase the
will increase
rate
the rate
D
The more
The more
Solids not
finely divided finely divided involved
the
the catalyst the
magnesium,
faster the rate
the faster the
rate
2.) Both the acid and the enzyme can act as catalysts for the
hydrolysis of a protein.
3. a) The greater the concentration of reactants, the greater the
rate of collisions and hence the faster the reaction proceeds.
3 b) A change of temperature has little effect. Most collisions
result in a reaction.
4 a) B and C
b) A and D
c) D
d) B
e) B f) D
CI 10.2: The effect of temperature on rate
For many reactions, increasing the temperature by 10oC will
DOUBLE the rate of reaction.
Extending the collision theory of reactions
Consider the reaction in the Haber process for making ammonia:
N2 (g) + 3H2 (g)
2NH3 (g)
The collision theory says that N2 and H2 will only react when they
collide. The more frequently they collide, the faster the rate of
reaction. Increasing the pressure brings H2 and N2 closer together,
so they collide more often. Increasing the temperature makes
molecules move faster, increasing the frequency of collisions.
How much more frequently do molecules collide, when
temperatures rise?
The average speed of molecules is proportional to the square root
of the absolute temperature.
Example: Temperature rise: 300 K to 310 K
Increase in average speed of molecules: (310/300)1/2
= 1.016
(this is 101.6%...so an increase of 1.6% faster)
Particles must also react with a certain minimum kinetic energy.
In the Haber process, (reaction between H2 and N2)
at 300 K only 1 in 1011 collisions between H2 and N2 results in a
reaction!
Even at 800 K only 1 in 104 collisions results in a reaction.
The collision theory says:
Reactions occur when molecules collide with a certain minimum
kinetic energy. The more frequent these collisions, the faster the
rate of reaction.
The energy needed to overcome the energy barrier is called the
activation enthalpy, for the reaction. It is the energy needed to
start breaking bonds in the colliding molecules, so that collisions
can lead to reactions.
The distribution of energies
Molecules are moving at different speeds, at any given time. Some
have higher energies, some medium energies, others have lower
energies.
Number of molecules with kinetic energy E
Maxwell-Boltzmann distribution: distribution of kinetic energies
in a gas, at a given temperature.
300 K
Kinetic energy (E)
Number of molecules with kinetic energy E
As the temperature increases, more molecules move at higher
speeds and have higher kinetic energies.
300 K
310 K
Kinetic energy (E)
Number of molecules with kinetic energy E
What is the significance of this for reaction rates?
Eg. the activation enthalpy for a reaction is: Ea is +50 kJ mol-1.
How many collisions have MORE energy than +50 kJ mol-1?
300 K
Kinetic energy (E)
Only collisions with energies in this region can lead to a reaction.
Number of molecules with kinetic energy E
At higher temperatures, a significantly higher proportion of
molecules will have energies above +50 kJ mol-1.
300 K
310 K
Kinetic energy (E)
About twice as many molecules have enough energy to react –so
the reaction goes twice as fast.
SL: What is removing the ozone?
Several radicals remove ozone, in the stratosphere:
Chlorine atom, Cl
Bromine atom, Br
From the oceans, burning coal / vegetation
Chloromethane, CH3Cl
(eg forest fires); they are the source of Br
Bromomethane, CH3Br
and Cl in the atmosphere.
Several reactions occur in the stratosphere:
Chlorine-containing molecules absorb high energy solar radiation
and break down to give chlorine atoms.
Reaction 6: Cl + O3  ClO + O2
(chlorine atoms react with ozone, forming radicals.)
Reaction 7: ClO + O  Cl + O2
(The ClO radical then reacts with oxygen atoms.)
We now have TWO reactions, competing to remove ozone:
Reaction 4
O + O3  O2 + O2
and
Reaction 6
Cl + O3  ClO + O2
The concentration of Cl atoms in the stratosphere is much less
than the concentration of O atoms. How significant is reaction 6?
 do assignment 7, p.69
Chlorine atoms are particularly effective at removing ozone. A
single atom can remove about 1 million ozone molecules.
Add equations 6 and 7 together to produce the equation for the
overall reaction caused by chlorine atoms.
Comment on the result. What role are Cl atoms playing in the
overall reaction?
 (O + O3  O2 + O2 )
 The radicals are regenerated, and so acting as catalysts.
It is important for chemists to know which reaction (6 or 7) is
happening fastest, to understand whether oxygen atoms or
chlorine atoms are responsible for the removal of ozone.
In fact, Cl atoms react 1500 faster with ozone, compared to O
atoms. Even though Cl atoms have a much lower concentration in
the stratosphere than O atoms, the fact that they can be
regenerated in a catalytic cycle can have devastating effects.
Bromine is 100 times more effective at destroying ozone than
chlorine, despite being much less concentrated than chlorine.
CI 10.5 How do catalysts work?
Collision theory and enthalpy profiles can help us to understand
how catalysts work.
Bond breaking is endothermic (requires an energy input to
overcome the activation enthalpy barrier). Bonds first stretch and
then break, reactants are made and new bonds instantly form. If
few molecules have enough energy to overcome this activation
enthalpy, then the reaction will be slow.
Catalysts provide an alternative reaction pathway for breaking and
remaking bonds that has a lower activation enthalpy.
Activation enthalpy
uncatalysed reaction
E
n
t
h
a
l
p
y
Activation enthalpy
catalysed reaction
H
Reactants
Products
Progress of reaction
Catalysts and equilibrium
Catalysts only affect the RATE and not the position of
equilibrium in a reversible reaction. The composition of the
reaction mixture remains unchanged.
Homogeneous catalysts
A homogeneous catalyst forms an intermediate compound before
breaking down to the final product and reforming the catalyst
again. This is why the enthalpy profile shows TWO humps.
Enthalpy
Intermediate compound
Final product
H
Reactants
Products
Progress of reaction
CFCs act as homogeneous catalysts in the stratosphere, breaking
down ozone. Cl atoms catalyse the reaction, forming the
intermediate ClO:
Cl + O3  O2 + ClO  intermediate
ClO + O  Cl + O2
O3 + O  O2 + O2  overall change
A single Cl atom can catalyse the reaction of many ozone
molecules through a catalytic cycle.
Industry uses mostly heterogeneous catalysts. However
homogeneous catalysts can be more specific and controllable.
Eg. Methanol Rhodium (aq)
ethanoic acid
Conversion is 99% with soluble rhodium compounds.
 Do problems for 10.5 p.243 questions 1 and 2.
SL: Other ways ozone is removed
Radicals such as hydroxyl and nitrogen monoxide can destroy
ozone, as well as chlorine and bromine.
In general:
X + O3  XO + O2
XO + O  X + O2
Overall reaction:
O + O3  O2 + O 2
Hydroxyl radicals (HO) form in the stratosphere when water
molecules react with oxygen. The reaction with ozone is:
HO + O3  HO2 + O2
HO2 + O  HO + O2
The reformed HO radicals can react with more ozone, in a
catalytic cycle.
Nitrogen monoxide (NO) forms nitrogen dioxide and dioxygen
when it reacts with ozone. NO and NO2 are relatively stable
radicals which can be collected in ordinary ways.
 do assignment 8, p.71
a) Write an equation to show the formation of HO radicals from
O atoms and water.
H2O(g) + O(g)  HO + HO
b) Write equations to show how nitrogen monoxide can destroy
ozone in a catalytic cycle.
NO + O3  NO2 + O2
NO2 + O  NO + O2
SL A4: The CFC story
In the early 1970s there was concern about jet aircraft releasing
NO in their exhausts. Levels were not significant at the time. In
1974 CFCs became a concern…
 read about it p.71-74 and do assignments 9 and 10.
CI 13.1 Halogenoalkanes
Are man-made compounds with one or more halogen atoms (F,
Cl, Br, I) attached to a carbon atom. The attached halogen changes
the chemical properties of alkane chains…they are very
unreactive, and so have been very useful to humans.
Naming halogenoalkanes (haloalkanes)
(similar rules to naming alcohols, just add the halogen as a prefix):
 halogens are in alphabetical order.
 lowest numbers possible are used.
CH3CH2CH2Cl
is
1-chloropropane
CH3CHClCH2Cl
is
1,2-dichloropropane
CH3CHBrCH2CH2Cl
is 3-bromo, 1-chlorobutane.
CH3CHICHBrCH2Cl
2-bromo,1-chloro,3-iodobutane
2-bromo, 3-chloro, 1-iodopentane
CH2ICHBrCHClCH2CH3
Physical properties of halogenoalkanes
 immiscible with water
 The bigger the halogen atom /the larger the number of
halogen atoms the higher the boiling point.
 Larger halogen atoms (Br or Cl) cause greater environmental
damage than smaller halogen atoms (F); this is important
when designing replacements for CFCs.
Chemical reactions of halogenoalkanes
Carbon –halogen (C-Hal) bonds can break either homolytically or
heterolytically.
Homolytic Fission forms radicals eg when a halogenoalkane
absorbs radiation of the right frequency.
H
H
H
C
Cl
+
hv

H
H
C
+
Cl
H
Chloromethane
methyl radical
Shorthand is: CH3-Cl + hv  CH3 + Cl
chlorine radical
(occurs in stratosphere).
Heterolytic fission is more common in lab conditions using polar solvents
such as ethanol or ethanol and water. The polar C-Hal bond can break,
leaving a negative halide ion and positive carbocation.
CH3
CH3
C
CH3
Cl
+
hv

H
CH3
C+ + ClH
2-chloro-2-methylpropane
carbocation
chloride ion
(negatively charged substances may react with the positive carbocation
causing a substitution reaction).
Importance of reaction conditions…for determining how bonds break
Eg. Bromoethane C-Br bonds break:
 Heterolytically, forming ions when dissolved in a polar solvent
(say a mixture of ethanol and water) BUT
 Homolytically, in the gas phase at high temp. or when dissolved in a
non-polar solvent, such as hexane.

Different halogens, different reactivity.
All reactions with halogenoalkanes involve breaking the C-Hal bond. The
C-F bond is the strongest (bond enthalpy 467 kJmol-1) and therefore the
hardest to break, whereas the C-I bond is relatively weaker (228 kJmol-1)
and therefore easier to break. C-Hal bonds get weaker, and so more
reactive, down group 7.
Chloro compounds are fairly unreactive and remain in the troposphere
long enough to reach the stratosphere, where they react with and destroy
the ozone layer.
Substitution reactions of halogenoalkanes
Halogenoalkanes can hydrolysed by hydroxide ions to form alcohols.
Eg. Bromobutane forms butanol:
CH3–CH2–CH2–CH2–Br + OH-  CH3–CH2–CH2–CH2–OH + BrThe C-Br bond is polar



C–Br
The oxygen atom on OH- is –vely charged.




H–O
The partial positive charge on the carbon atom attracts the negatively
charges oxygen of the hydroxide ion. A lone pair of electrons on the O
atom forms a bond with the C atom as the C__Br bond breaks.
H H H H
H__C__C__C__C__Br
H H H H
H H H H
H__C__C__C__C__O__H
H H H H
_
O
H
_
+
Br
Heterolytic fission results in IONS and not radicals.
Curly arrows show the movement of electrons (full headed
arrows for a pair of electrons…unlike radical reactions).
Halogenoalkanes can give substitution reactions with hydroxide ions
and other NUCLEOPHILES. Nucleophiles can donate a pair of electrons
to a positively charged carbon atom to create new covalent bonds.
Some common nucleophiles:
Name
Formula
Hydroxide ion
OH-
Structure showing
lone pairs
_
H__O
CH3COO-
Ethanoate ion
_
__
__
CH3 C O
O
C2H5O-
Ethoxide ion
_
__
CH3CH2 O
Water molecule
H2O
O
H
Ammonia molecule
H
NH3
N
H
H
H
-
Cyanide ion
CN
_
N
C
The carbon atom attacked by the nucleophile may be part of a
carbocation and carry a full positive charge, or it may be part of a
neutral molecule (as in the above example with bromobutane) and carry a
partial positive charge as a result of bond polarisation.
If X- represents a nucleophile, the nucleophilic substitution process is:
X
 
C–Hal 





C__X + Hal-







Water as a nucleophile
Nucleophiles may be neutral or have negative charge, so long as it has a
lone pair of electrons which can form a bond to a carbon atom.
Eg. Water has 2 lone pairs of electrons on the O atom. First it attacks the
halogenoalkane (bromobutane in this case):
H H H H
H C__C__C__C__Br
H H H H
H H H H
H C__C__C__C__O__H
H H H H H
__
__
_
+
Br
+
H+
O
H
H
The resulting ion loses H+ to form an alcohol:
H H H H
+
__ __ __ __
H C C C C O
H H H H H
__
H H H H
H C__C__C__C__O__H
H H H H
__
The overall equation for the reaction of water with a genera;
halogenoalkane R__Hal is:
R__Hal + H2O  + R__OH + H+ + Hal-
Ammonia as a nucleophile
A lone pair of electrons on the N (similar to water) attacks the
halogenoalkane to produce an AMINE with an NH2 group:
R__Hal + NH3  R__NH3+ Hal-
R__NH2 + H+ + Hal-
Using nucleophilic substitution to make halogenoalkanes
Halogenoalkane + OH-
alcohol
Halogenoalkanes can be made via the reverse reaction of making alcohols;
the nucleophile is Hal-.
Eg. 1-bromobutane is made using a nucleophilic substitution reaction
between butan-1-ol and Br- ions, in the presence of a strong acid.
Ist step: H+ ions bond to O atom on the alcohol:
H H H H
H H H H
+
__ __ __ __
__ __ __ __ __ __
H C C C C O
H C C C C O H
H H H H
H H H H H
+
H
This gives the C atom to which the O is attached a greater partial positive
charge. It is now more readily attacked by Br- ions, forming bromobutane.
__
H H H H
+
__ __ __ __
H C C C C O
H H H H H
_
Br
__
H H H H
H C__C__C__C__Br
H H H H
__
+
H2O
The overall equation for the reaction is:
CH3CH2CH2CH2OH + H+ + Br-  CH3CH2CH2CH2Br + H2O
 Activity A4.2
 Problems for 13.1 pages 303- 304
questions 1- 9.