TOPIC 5
ENERGETICS/THERMOCHEMISTRY
5.3
BOND ENTHALPIES
ESSENTIAL IDEA
Energy is absorbed when bonds are
broken and is released when bonds
are formed.
NATURE OF SCIENCE (2.2)
Models and theories – measured energy changes
can be explained based on the model of bonds
broken and bonds formed. Since these explanations
are based on a model, agreement with empirical
data depends on the sophistication of the model
and data obtained can be used to modify theories
where appropriate.
UNDERSTANDING/KEY IDEA
5.3.A
Bond-forming releases energy
and bond-breaking requires
energy.
UNDERSTANDING/KEY IDEA
5.3.B
Average bond enthalpy is the
energy needed to break one
mole of a bond in a gaseous
molecule averaged over
similar compounds.
GUIDANCE
Average bond enthalpies are
only valid for gases so
calculations involving bond
enthalpies may be inaccurate
because they do not take into
account intermolecular forces.
• A common error is to fail to
indicate that all species have to
be in their gaseous state.
APPLICATION/SKILLS
Be able to sketch and evaluate
potential energy profiles in
determining whether reactants
or products are more stable
and if the reaction is
exothermic or endothermic.
• Chemical reactions involve the
breaking of bonds in reactants
and the making of bonds in
products.
• To understand energy changes
in chemical reactions, we need
to look at the energy needed to
break bonds in reactants and
the energy released when bonds
are formed in products.
• Endothermic processes involve
the separation of particles
which are held together by a
force of attraction.
• Exothermic processes involve
the bringing together of
particles which have an
attractive force between them.
Exothermic reaction
Enthalpy diagram
reactants
enthalpy
H
ΔH = negative
products
extent of reaction
Endothermic reaction
Enthalpy diagram
products
enthalpy
H
ΔH = positive
reactants
extent of reaction
Sample problem
• Which of the following
processes are endothermic?
•
•
•
•
2Cl
Cl2
Na
Na+ + eNa+ + ClNaCl
Na(g)
Na(s)
• The answer is the 2nd equation
which is the only one separating
into particles.
• A covalent bond is due to the
electrostatic attraction
between the shared pair of
electrons and the positive
nucleus.
• Energy is needed to separate
the atoms in a bond.
GUIDANCE
Average bond enthalpies are
only valid for gases and
calculations involving bond
enthalpies may be inaccurate
because they do not take into
account intermolecular forces.
Average Bond
Enthalpies
• Bond enthalpies are calculated by
separating the atoms.
• Cl2(g)
2Cl(g)
∆H⁰ = +242 kJ/mol
• H2O(g)
• OH(g)
H(g) + OH(g)
H(g) + O(g)
• It is more complicated in molecules
that contain more than 2 atoms.
∆H⁰ = +502 kJ/mol
∆H⁰ = +427 kJ/mol
• In order to compare bond
enthalpies, average bond enthalpies
are tabulated.
• All tabulated bond enthalpies
refer to reactions in their
gaseous state so that the
enthalpy changes caused by the
formation and breaking of
intermolecular forces can be
ignored.
• Multiple bonds have higher bond
enthalpies and are shorter than
single bonds.
LIMITATIONS OF USING
AVERAGE BOND ENTHALPIES
• As said before, everything has to be in the
gaseous state.
• If water were a liquid as a product, then more
heat would be evolved because it would also
have to include the heat of vaporization of
water.
• Average bond enthalpies are used which are
obtained considering a number of similar
compounds.
• In reality, the energy of a particular bond varies slightly
in different compounds and it is affected by neighboring
atoms.
• So ∆H values obtained from using average bond
enthalpies will not necessarily be very accurate.
• Explain, in terms of bond
enthalpies, why some reactions
are exothermic and some are
endothermic.
• The reaction is exothermic
overall if the bonds which are
formed are stronger than the
bonds which are broken.
• The reaction is endothermic
when the bonds broken are
stronger than the bonds that are
formed.
APPLICATION/SKILLS
Be able to calculate the
enthalpy changes from known
bond enthalpy values and
compare them to
experimentally measured
values.
GUIDANCE
Bond enthalpy values are
given in the data booklet in
section 11.
Sample problem
• Use bond enthalpies to calculate
the heat of combustion of methane.
First write and balance the equation.
CH4 + 2O2
CO2 + 2H2O
Next figure out the structural bonds.
4 C-H + 2O=O
2C=O + 4O-H
• Very important:
When using bond enthalpies, you use
reactants minus products, not
products minus reactants as in the
standard heat of formation formula.
• ∆H = ∑(bonds broken) – ∑(bonds formed)
reactants
products
• Now look up the values in a
table or they will be given to
you. Be very careful not to
confuse double or triple bonds
with single bonds.
• ∆H = [4C-H + 2O=O] – [2C=O + 4O-H]
= [4(412) + 2(496)]-[2(743)+4(463)]
= (2640)-(3338)
= -698 kJ/mol
• The literature value is -890 kJ/mol
which is the value for standard
conditions.
• The standard state for water is
liquid and the bond enthalpy
calculation assumed the gaseous
state so this is not the most
accurate method.
APPLICATION/SKILLS
Be able to discuss the bond
strength of ozone relative to
oxygen in its importance to
the atmosphere.
OZONE REVIEW
• Ozone, O3, has a bent shape with a bond angle of 117◦
• It has 2 resonance structures.
• The double bond consists of one pi and one sigma
bond.
• The electrons in the pi bond are held less tightly so
they become delocalized giving rise to the resonance
structure.
• The bond order is 1.5 which means the length is
intermediate and the strength is between a double and
single bond.
• www.chemwiki.ucdavis.edu
• The ozone molecule is polar which is
explained by formal charge and the uneven
distribution of electrons.
• The lower part of the stratosphere, known
as the ozone layer, contains 90% of the
atmospheric ozone.
• Ozone levels are maintained through a
cycle of reactions involving the formation
and breakdown of oxygen and ozone.
• Oxygen and ozone form a protective screen
which ensures that the radiation that
reaches the earth is different from that
emitted by the sun.
OZONE CYCLE
There are 2 key steps in the ozone cycle. The
O. represents a free radical that has an unpaired
electron so it is highly reactive.
Oxygen dissociation
O2(g)
O.(g) + O.(g)
light wavelength <242 nm
Ozone dissociation
.
1. O3(g)
O
fast
(g) + 2O2(g) light wavelength <330 nm
2. O3(g) + O.(g)
slow
2O2(g)
exothermic reaction
H=neg
• The O2 bonds are stronger and
harder to break than the O3 bonds
of ozone.
• The stronger O2 bonds require UV
light energy of shorter wavelengths
because they need the higher
energy radiation to break the
bonds.
• The bond energy of O2 with a bond
order of 2 is 498 kJ/mol.
• The bond energy of O3 with a bond
order of 1.5 is 364 kJ/mol.
• The fact that ozone absorbs radiation of
wavelengths of 200 nm to 315 nm is very
important.
• This corresponds to the higher range of
ultraviolet light, known as UV-B and UVC which causes damage to living tissue.
• The ozone layer protects us from this
radiation.
• The absorption of UV radiation by ozone
is also a major source of heat in the
stratosphere and is the reason why the
temperature in the stratosphere rises
with height.
CHAPMAN CYCLE
STEP 1: Oxygen is broken into two free radicals with uv
light with wavelengths less than 242 nm. (Endo)
STEP 2: The free radical reacts with oxygen to form
ozone. (Exothermic and the heat given out heats up the
stratosphere.)
STEP 3: Ozone is broken into oxygen and a free radical
with uv light of lower energy less than 330nm. (Endo)
STEP 4: The oxygen free radical then reacts with another
ozone molecule to form two oxygen molecules.
(Exothermic so provides more heat to the stratosphere.)
The level of ozone stays at a constant level as the rate of formation of
ozone is balanced by the rate of its removal.
This cycle of reactions is significant because dangerous uv light has
been absorbed and the stratosphere has become warmer. Both
processes are essential for the survival of life on earth.
CHAPMAN CYCLE
STEP
STEP
STEP
STEP
1:
2:
3:
4:
O2(g)
O.(g) + O.(g) λ<242 nm
O.(g) + 2O2(g)
O3(g) exo rxn
O3(g) fast O.(g) + 2O2(g) λ <330 nm
O3(g) + O.(g) slow 2O2(g) exo rxn
APPLICATION/SKILLS
Be able to describe the
mechanism of the catalysis of
ozone depletion when
catalyzed by CFCs and NOx.
• Ozone’s ability to absorb UV
radiation also means that it is
unstable.
• It reacts easily with compounds
found in the stratosphere that have
been created by human activity.
• There are two types of compounds
that produce highly reactive free
radicals that catalyze the
decomposition of ozone to oxygen.
• Nitrogen oxides, NOx
• Chlorofluorocarbons, CFC’s
Nitrogen oxides
• Nitrogen monoxide is produced in vehicle engines.
• It is a free radical as it has an odd number of
electrons.
• Nitrogen dioxide forms from the oxidation of NO and is
also a free radical.
• The reactions of nitrogen oxides with ozone are:
NO.(g) + O3(g)
NO2.(g) + O2(g)
NO2.(g) + O.(g)
NO.(g) + O2(g)
NO. has acted as a catalyst because it regenerated
during the reaction and the net change is the breakdown
of O3.
O3(g) + O.(g)
2O2(g)
Chlorofluorocarbons
• Chlorofluorocarbons are widely used in
aerosols, refrigerants, solvents and plastics
due to their low reactivity and low toxicity in
the troposphere.
• However, when they reach the stratosphere,
the higher energy UV radiation breaks them
down releasing free chlorine atoms which are
reactive free radicals.
• The reaction of the CFC freon is:
CCl2F2(g)
CClF2.(g) + Cl.(g)
Cl.(g) + O3(g)
O2(g) + ClO.(g)
ClO.(g) + O.(g)
O2(g) + Cl.(g)
• Here the chlorine radical acts as the catalyst
and the net reaction is again:
O3(g) + O.(g)
2O2(g)
• These reactions upset the balance of the
ozone cycle and lead to the thinning of the
ozone layer.
• The UV radiation reaching the Earth is most
pronounced in the polar regions.
• This has been a global concern since the
1970’s.
Citations
Brown, Catrin, and Mike Ford. Higher Level
Chemistry. 2nd ed. N.p.: Pearson
Baccalaureate, 2014. Print.
Most of the information found in this power
point comes directly from this textbook.
The power point has been made to directly
complement the Higher Level Chemistry
textbook by Catrin and Brown and is used for
direct instructional purposes only.