TOPIC 5

TOPIC 5

ENERGETICS/THERMOCHEMISTRY

5.3

BOND ENTHALPIES

By: Merinda Sautel Alameda Int’l Jr/Sr High School Lakewood, CO [email protected]

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.

INTERNATIONAL- MINDEDNESS

Stratospheric ozone depletion is a

particular concern in the polar regions of the planet, although the pollution that causes it comes from a variety of regions

and sources. International action and cooperation have helped to ameliorate

the ozone depletion problem.

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

extent of reaction

Endothermic reaction Enthalpy diagram

products

enthalpy

H

extent of reaction

Sample problem

• Which of the following processes are endothermic?

• 2Cl Cl

• Na Na

+ e

• Na

+ Cl

NaCl

• Na

Na

• The answer is the 2

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.

Average Bond Enthalpies

• Bond enthalpies are calculated by separating the atoms.

• Cl

(g) 2Cl(g) ∆H⁰ = +242 kJ/mol

• It is more complicated in molecules that contain more than 2 atoms.

• H

O(g) H(g) + OH(g) ∆H⁰ = +502 kJ/mol

• OH(g) H(g) + O(g) ∆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.

CH

+ 2O

CO

+ 2H

O

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.

• Ref: schooltutoring.com

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

O

O

+ O

light wavelength <242 nm

Ozone dissociation

1.O3(g)

O

+ 2O

2. O

+ O

2O

exothermic reaction

H=neg

• The O

bonds are stronger and harder to break than the O

bonds of ozone.

• The stronger O

bonds require UV light energy of shorter wavelengths

because they need the higher energy radiation to break the bonds.

• The bond energy of O

with a bond order of 2 is 498 kJ/mol.

• The bond energy of O

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 UV-C 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 1: O

O

+ O

λ<242 nm STEP 2: O

+ 2O

O

exo rxn

STEP 3: O

O

+ 2O

λ <330 nm STEP 4: O

+ O

2O

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, NO

•

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:

CCl

F

CClF

+ Cl

Cl

+ O

O

+ ClO

ClO

+ O

O

+ Cl

• Here the chlorine radical acts as the catalyst and the net reaction is again:

O

+ O

2O

• 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

International Baccalaureate Organization. Chemistry Guide, First assessment 2016. Updated 2015.

Brown, Catrin, and Mike Ford. Higher Level Chemistry. 2nd ed. N.p.: Pearson Baccalaureate, 2014. Print.

ISBN 978 1 447 95975 5 eBook 978 1 447 95976 2

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 Brown and Ford and is used for direct instructional purposes only.