chem 101

(1)

Chemical Bonding I:

Basic Concepts

Chapter 9

Chapter 9 General Introduction

General Introduction

•• Why d

Why do atom

o atoms of dif

s of differe

ferent elem

nt elements re

ents react?

act?

•

• Wha

Wha

t are th

e forc

es tha

t hold a

toms to

gethe

r in

molecules and ions in

ionic compounds?

•

• Wh

Wh

at sh

ape

s do th

ey a

ssum

e?

•

• Thes

Thes

e are som

e of the qu

estion

s addre

ssed in t

his

Chapter and in Chapter 10.

•

• We be

We be

gin by l

ookin

g at the

two ty

pes of b

onds Io

nic

and Cova

lent–

and the

forces that

stabilize them

Chapter Outline

9.1 Lewis Dot Symbols

9.2 The Ionic Bond

9.3 Lattice Energy of

Ionic Compounds

9.4 The Covalent Bond

9.5 Electronegativity

9.6 Writing Lewis

Structure

9.7 Formal Charge and

Lewis Structure

9.8 The Concept of Resonance

9.9 Exceptions to the Octet Rule

9.10 Bond Enthalpy

9.1 Lewis Dot Symbols

•

• The dev

The develo

elopmen

pment of the pe

t of the period

riodic tabl

ic tabl

e and conc

e and concept of e-

ept of e-

gave ch

gave chemi

emists

sts

a basis for molecule and compound formation.

•• This expl

This explanation,

anation, formul

formulated by

ated by Lewis, is

Lewis, is that at

that atoms com

oms combine in

bine in order

order

to achieve a more stable EC. Maximum stability results when an atom

is isoelect

is isoelectronic

ronic with a noble g

with a noble gas.

as.

•

• When atom

When atom

s inter

act to

form a

chemical

bond, only

their ou

ter regi

ons

are in contact. For this reason, when we study chemi

cal bonding,

we

are concerned prim

arily with the valence e-

of the atoms.

•• To kee

To keep tra

p track of va

ck of valenc

lence e-

e e- in a che

in a chemic

mical rea

al reactio

ction, an

n, and to ma

d to make sur

ke suree

that the total number of

e-that the total number of e- does not change, chemist

does not change, chemists use a system of

s use a system of

dots devised by Lewis and called Lewis dots symbols.

•• A

A Lewis dot sy

Lewis dot s

Lewis dot symbol

Lewis dot sy

ymbol consists of the symbol of an element and one dot

mbol

consists of the symbol of an element and one dot

for each valence

(2)

9.1 9.1

Valence electrons

Valence electrons are the outer shell electrons of an

are the outer shell electrons of an

atom.

atom. The valence electrons are the electrons that

The valence electrons are the electrons that

participate in chemical bonding.

1A

ns

11

1

1 2A

2A

ns

22

2

2 3A

3A

ns

22

_np

11

3

3 4A

4A

ns

22

_np

22

4

4 5A

5A

ns

22

_np

33

5

5 6A

6A

ns

22

_np

44

6

6 7A

7A

ns

22

_np

55

7

7 G

Grro

ou

up

p

_e

--

_{configuration}

#

of

o

f

v

va

alle

en

nc

ce

e

--9.1 9.1 The number of unpaired dots corresponds to the number of

The number of unpaired dots corresponds to the number of bonds an atom of thebonds an atom of the

element can form in a compound.

The number of valence

e-The number of valence e-each atom has is the same as the group number of theeach atom has is the same as the group number of the element except for He.

element except for He.

In this chapter we will learn to use EC and the periodic table to predict the type of bond

atoms will form, as

atoms will form, as well as the number of bonds an well as the number of bonds an atom of a particular element canatom of a particular element can

form and the stability of

form and the stability of the product.the product.

Lewis dot symbols for the

representative elements and the

noble gases.

Li

Li +

+ F

F

Li

++

_F

--9.2 The Ionic Bond

9.2 The Ionic Bond

1s

22

_2s

11

_1s

22

_2s

22

_2p

55

_1s

22

_1s

22

_2s

22

_2p

66

[He]

[Ne]

Li

++

₊

_{+ e}

_e

--e

e

--

+

F

--F

F

--Li

Li

++

₊

_Li

++

_F

--An ionic bond is the electrostatic force

An ionic bond is the electrostatic force that holds ions together in an that holds ions together in an ionic compound.ionic compound. Consider, for ex, the reaction between Lithium and fluorine to form lithium fluoride, a

Consider, for ex, the reaction between Lithium and fluorine to form lithium fluoride, a

poisonous white powder used in lowering the melting point of

poisonous white powder used in lowering the melting point of solders and insolders and in

manufacturing ceramics.

The EC of Lithium is

The EC of Lithium is 1s²2s1s²2s11_{, and that of}_{, and that of Fluorine is 1s²2s²2p}_{Fluorine is 1s²2s²2p}55_..

When Lithium and fluorine atoms come in contact

When Lithium and fluorine atoms come in contact with each other , the with each other , the outer 2souter 2s11_valence_valence

e-e-of Li is transferred to the fluorine atom. Usof Li is transferred to the fluorine atom. Using Lewis dot symboing Lewis dot symbols, we represent thels, we represent the

reaction like this:

Ionization of Lithium Ionization of Lithium Acceptance

Acceptance of an eof an e-- by Fby F Forma

Formation of LiFtion of LiF unitunit isoelectronic isoelectronic

9.2 9.2

(3)

(4)

(5)

Double bond

–

two atoms

share two pairs

of electron

s

O

O C

C O

O

or

O

C

O

8e

--

_8e

--

_8e

- -double bonds

double bonds double bondsdouble bonds

Triple bond

–

two atoms

share three p

airs of electr

ons

N

8e

--

_8e

-

-N

N

triple bond triple bond triple bond triple bond

or

9.4 9.4

Double bonds are found in molecules of Carbon Dioxide

(CO

₂₂

) and ethylene (C

₂₂

H

₄₄

)

A

Triple bond

arises when two atoms share three pairs of e-,

as in the nitrogen molecule (N

₂₂

)

116

116 C

C

≡

_N

138

138 C

C

=

N

143

143 C

C

_-

N

120

120 C

C

≡

C

133

133 C

C

=

C

154

154 C

C

_-

C

Bond

Length

(pm)

Bond

Type

Lengths of Covalent Bonds

Bond Lengths

Triple bond < Double Bond < Single Bond

9.49.4 Multiple bonds

Multiple bonds Multiple bonds

Multiple bonds are shorter than single covalent bonds.are shorter than single covalent bonds. Bond length

Bond length Bond length

Bond length is defined as the distance between theis defined as the distance between the nuclei of two covalently bonded atom in a nuclei of two covalently bonded atom in a molecule.molecule. For a given pair of a

For a given pair of atoms, such as carbon and nitrogen,toms, such as carbon and nitrogen, triple bonds are shorter than double bonds, which, in triple bonds are shorter than double bonds, which, in turn, are shorter than single

turn, are shorter than single bonds. The shorterbonds. The shorter multiple bonds are also more stable than single bonds, multiple bonds are also more stable than single bonds, as we will see later.

as we will see later.

9.4.1 Comparison of the Properties of

9.4.1 Comparison of the Properties of Covalent and Ionic CompoundsCovalent and Ionic Compounds There are two types of attractive forces in covalent compounds.

There are two types of attractive forces in covalent compounds.

The

The first type first type first type first type is the force that holds the atoms together in a molecule.is the force that holds the atoms together in a molecule. The

Thesecond type second type second type second type of attractive force operates between of attractive force operates between molecules and is called anmolecules and is called anintermolecularintermolecular

force.

force. Because intermolecular forces are usually quite wBecause intermolecular forces are usually quite weak compared with the feak compared with the forces holdingorces holding

atom together within a molecule, molecules of a covalent compound are not held together

tightly.

tightly.Consequently covalent compounds are usConsequently covalent compounds are usually gases, liquids, or low-melting solids.ually gases, liquids, or low-melting solids.

•• On thOn the othee other hand, r hand, the elthe electrostectrostatic atic forces forces holdiholding ionng ions togets together iher in an in an ionic comonic compound pound areare usually very strong.

usually very strong.

•

• SSo io io no niic cc co mo mppoouunnd sd s,,

are solids at RT and

have high melting

points. points. • • MMaanny y iioonniicc compounds are compounds are

soluble in water, and

the resulting the resulting aqueous solutions aqueous solutions conduct electricity, conduct electricity, because the because the compounds are compounds are strong electrolytes. strong electrolytes. _9.4_9.4

9.4.1 Comparison of the Properties of

9.4.1 Comparison of the Properties of Covalent and Ionic CompoundsCovalent and Ionic Compounds •

• Most Most covalent covalent compoundcompounds are s are insolinsoluble iuble in watn water, or er, or if thif they do ey do dissoldissolve, theve, their aquir aqueouseous

solutions generally do not conduct electricity, because the compounds are nonelectrolytes.

•

• MoltMolten ionien ionic compoc compounds counds conduct nduct electelectricity ricity because because they cthey contain ontain mobilmobile catie cations and ons and anions;anions;

liquid or molten covalent compounds do not conduct electricity because no ions are present.

•

• This This tabltable compe compares soares some of me of the the general general propertproperties ies of of typicatypical iol ionic nic compoundcompound, sod, sodiumium

chloride, with those of a covalent compound, Carbon Tetrachloride (CCl

(6)

(7)

Covalent Covalent share e share e- -Polar Covalent Polar Covalent partial transfer of e partial transfer of e- -Ionic Ionic transfer e transfer e- -Increasing difference in electronegativity

Increasing difference in electronegativity

Classification of bonds

Classification of bonds by difference in electronegativity

by difference in electronegativity

D Diiffffeerreennccee BBoonnd d TTyyppee 0 0 CCoovvaalleenntt ≥ ≥22 IonicIonic 0 0 < < aannd d <<22 PPoollaar r CCoovvaalleenntt 9.5 9.5 •• An ionic bond forms when the EN difference between the two bonding atoms iAn ionic bond forms when the EN difference between the two bonding atoms is 2.0 or more.s 2.0 or more. •• This rule applies to most but not all compounds. Sometimes chemists use the quaThis rule applies to most but not all compounds. Sometimes chemists use the quantityntity % ionic% ionic

character

characterto describe the nature of a bond. A purely ionic bond would have 100% ionic characterto describe the nature of a bond. A purely ionic bond would have 100% ionic character although no such bond is known, whereas a non polar or purely covalent bond has 0 % ionic although no such bond is known, whereas a non polar or purely covalent bond has 0 % ionic character.

character. •

•EN and EA are relEN and EA are related but different ated but different concepts. Both indicate theconcepts. Both indicate thetendency of an atom ttendency of an atom to attract e-.o attract e-.

However, EA refers to an isolated atom’s attraction for an additional e-, whereas EN signifies the

ability of an atom in a chemical bond to attract the shared e-.

•• EA is an experimentally measurable quantity, whereas EN is an estimated number EA is an experimentally measurable quantity, whereas EN is an estimated number that can notthat can not be measured.

be measured.

Classify the following bonds as ionic,

polar covalent,

or covalent:

The bond in CsCl; t

he bond in H

22

S; and

the NN bond in H

22

NNH

22

.

C

Cs

s

– 0

– 0.

.7

7 C

Cl – 3

l – 3.

.0

0

3

3. .0

0 – 0

– 0.

.7

7 =

=

2

2. .3

3 I

Io

on

ni

ic

c

H

– 2

.

1

1 S

S

– 2

.

5

2

2 .

.

5

5 – 2

– 2

.

1

1 =

=

0

0 .

.

4

4 P

P

o

l

a

r

C

o

v

a

l

e

n

t

N

–

3

3 .

.

0

0 N

N

–

3

3 .

.

0

3

3 .

.

0

0 –

–

3

3 .

.

0

0 =

=

0

0 C

C

o

v

a

l

e

n

t

9.5 9.5 We follow the 2.0 rule of EN We follow the 2.0 rule of EN difference and look up the values in difference and look up the values in the EN table

the EN table

1.

1. Draw skel

Draw skel

etal str

uctu

re of compo

und show

ing what ato

ms

are bonded to

each other.

Put least electron

egative

element in the center.

2.

2. Count

Count

tota

l nu

mber

of v

alenc

e e

--

_.

_{Add 1 for}

_{each negativ}

_e

charge.

Subtract 1

for each posit

ive charge.

3.

3. Compl

Compl

ete

an oc

tet

for

all a

toms

except

hydrogen

4. 4. If stru

If structur

cture contai

e contains too many el

ns too many electro

ectrons, for

ns, form double and

m double and

triple bonds on central atom as

triple bonds on central atom as needed.

needed.

9.6. Writing Lewis Structures

9.6 9.6 •

•Although the octet rule and Lewis structures do not present a complete picture of covalentAlthough the octet rule and Lewis structures do not present a complete picture of covalent

bonding, they do help to explain the bonding s

bonding, they do help to explain the bonding scheme in many compounds and account for cheme in many compounds and account for thethe

properties and reactions of molecules.

•• For this reason, you should practice For this reason, you should practice writing Lewis structures of compounds. The writing Lewis structures of compounds. The basic steps arebasic steps are as follows:

as follows:

Write the Lewis stru

cture of nitrog

en trifluoride

(NF

33

).

Step 1 –

Step 1 – N is less electronegative thN is less electronegative than F, put N in centeran F, put N in center

F

N

F

Step

Step 2 –2 – Count Count valenvalence elce electronectrons s N -N - 5 (2s5 (2s22_2p_2p33_{) an}_{) and F -}_{d F - 7 (2}_{7 (2s}_s22_2p_2p55₎₎

5 + (3 x 7) =

5 + (3 x 7) = 26 valence electrons26 valence electrons

Step 3 –

Step 3 – Draw single bonds between N and F atoms Draw single bonds between N and F atoms and completeand complete octets on N and F atoms.

octets on N and F atoms. Ste

Step 4 -p 4 - CheCheck, ack, are # ore # of ef e--_{in structure equal to number of valence e}_{in structure equal to number of valence e}--_?_?

3 single bonds (3x2) + 10 lone pairs (10x2) =

3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons26 valence electrons

9.6 9.6

(8)

(9)

H

C

O

H

C

O

H

C – C – 4 e4 e --O O –– 6 6 ee --2H – 2H – 2x2x1 e1 e --12 12 ee- -2 si

2 single ngle bonds bonds (2x2) (2x2) = = 44 1

1 double double bond bond = = 44 2

2 lone lone pairs pairs (2x2) (2x2) = = 44 Total = 12 Total = 12 Formal Formal Charge on C Charge on C

=

4 -

2

2 -

-

½

x 6

= -

1

Formal Formal Charge on O Charge on O

=

6

6 -

-

2

2 -

-

½

x 6

= +

1

Formal charge Formal charge on an atom in on an atom in a Lewis a Lewis structure structure

=

1

2

Total number Total number of bonding of bonding electrons electrons

((

))

Total number Total number of valence of valence electrons in electrons in

the free atom

--

Total numberTotal numberof nonbondingof nonbonding

electrons electrons

----1

1 +

+1

1

9.7 9.7 C – C – 4 4 ee --O – O – 6 e6 e --2H – 2H – 2x2x1 e1 e --12 12 ee- -2

2 single single bonds bonds (2x2) (2x2) = = 44 1

1 double double bond bond = = 44 2 l

2 lone one pairs (2xpairs (2x2) = 2) = 44 Total = 12 Total = 12

H

C

O

H

Formal charge Formal charge on C on C

=

4

4 -

-

0 -

½

x

8

8 =

=

0

Formal charge Formal charge on O on O

=

6

6 -

-

4 -

½

x

4

4 =

=

0

9.7 9.7 Formal charge Formal charge on an atom in on an atom in a Lewis a Lewis structure structure

=

1

2

Total number Total number of bonding of bonding electrons electrons

((

))

Total number Total number of valence of valence electrons in electrons in

the free atom

--

Total numberTotal numberof nonbondingof nonbonding

electrons

(10)

--O

O

+

O

--

O

+

O

9.8 9.8

9.8 The Concept of Resonance

Our drawing of the Lewis structure for

Our drawing of the Lewis structure for ozone (Oozone (O₃₃) satisfied the octet rule for the central) satisfied the octet rule for the central atom because we placed a double bond between it and one of

atom because we placed a double bond between it and one of the two end O the two end O atoms.atoms.

In fact, we can put

In fact, we can put the double bond at either end of the double bond at either end of the molecule, as shown by these twothe molecule, as shown by these two

equivalent Lewis Structures:

O

We would expect the O—O bond in

We would expect the O—O bond in OO₃₃to be longer than the O==O to be longer than the O==O bond becausebond because

double bond are known to be shorter than single bonds.

Yet experimental evidence shows that both

Yet experimental evidence shows that both oxygen-to-oxygeoxygen-to-oxygen bonds are n bonds are equal in lengthequal in length

(128 pm). We resolve this discrepancy by using both Lewis Structures to

(128 pm). We resolve this discrepancy by using both Lewis Structures to represent therepresent the

Ozone molecule:

O

+

O

--

O

+

O

Electrostatic potential map Electrostatic potential map Electrostatic potential map Electrostatic potential map of O

of O of O

of O₃₃₃₃. The electron density. The electron density. The electron density. The electron density is evenly distributed is evenly distributed is evenly distributed is evenly distributed between the two end O between the two end O between the two end O between the two end O

atoms atoms atoms atoms Each of these structures is called resonance structure

Each of these structures is called resonance structure Each of these structures is called resonance structure Each of these structures is called resonance structure

O

C

O

--

-

-O

O

C

O

-O

O

C

O

--

_9.8_9.8

9.8 The Concept of Resonance

What are the

resonance structures of the

carbonate (CO

332-2-

) ion?

Carbonate ion provides another ex of resonance Carbonate ion provides another ex of resonance

O

+

O

--

O

+

O

A

A resonance structure

resonance structure is one of

is one of two or more Lewis structures for

two or more Lewis structures for

a single molecule that cannot be

a single molecule that cannot be represented accurately by only

represented accurately by only

one Lewis s

(11)

(12)

The enthalpy change required to break a particular bond in

one mole of gaseous molecules is

one mole of gaseous molecules is the

the bond energy

bond energy ..

H

2 (2 (g g ))

H

((g g ))

+

+ H

H

((g g )) ∆∆

H

00

= 436.4 kJ

Cl

2 (2 (g g ))

Cl

((g g ))

+

+ Cl

Cl

((g g )) ∆∆

H

00

= 242.7 kJ

HCl

((g g ))

H

((g g ))

+

+ Cl

Cl

((g g )) ∆∆

H

00

= 431.9 kJ

O

2 (g 2 (g ))

O

((g g ))

+

+ O

O

((g g )) ∆∆

H

00

= 498.7 kJ

O

N

2 (g 2 (g ))

N

((g g ))

+

+ N

N

((g g )) ∆∆

H

00

=

= 9

94

41 1..4

4 k

kJ

J

N

Bond Energy

Bond Energies

Single bond < Double bond < Triple bond

9.10 9.10

9.10 Bond Enthalpy

Breaking the covalent bonds in 1 Breaking the covalent bonds in 1

mole of gaseous H mole of gaseous H22moleculesmolecules

requires 436.4 kJ of energy requires 436.4 kJ of energy Bond enthalpies can also be

Bond enthalpies can also be

directly measured for diatomic

molecules containing unlike

elements elements For For molecules molecules containing containing double double and triple and triple bonds bonds

Average

Average bond energy

bond energy in polyatomic molecules

in polyatomic molecules

H

22

O

((g g ))

H

((g g ))

+ OH

+

OH

((g g ))∆∆

H

00

= 502 kJ

OH

_((g_g ))

H

_((g_g ))

+

+ O

O

_((g_g )) ∆∆

_H

00

_{= 427 kJ}

Average OH bond energy =

502 + 427

2

2 = 464 kJ

= 464 kJ

Measuring the strength of covalent bon

Measuring the strength of covalent bonds in polyatomic moleculesds in polyatomic molecules is more complicated.is more complicated. For ex, measurements show that the energy needed to break the f

For ex, measurements show that the energy needed to break the f irst O-H bond in Hirst O-H bond in H₂₂OO is different from that needed to

is different from that needed to break the second O-H bond.break the second O-H bond. The first step is more

The first step is more endothermic.endothermic.

If the environment of bonds is

If the environment of bonds is different, so the energy needed to break down a different, so the energy needed to break down a bond isbond is different.

different. Ex: CH

Ex: CH33OH and HOH and H22OO

Thus for polyatomic molecules, we can measure the energy of the

O-Thus for polyatomic molecules, we can measure the energy of the O-H bond in 10H bond in 10 different polyatomic molecules and obtain the average O-H bond enthalpy by dividing different polyatomic molecules and obtain the average O-H bond enthalpy by dividing the sum of the bond enthalpies by 10.

the sum of the bond enthalpies by 10.

The following table lists the average bond enthalpies of a

The following table lists the average bond enthalpies of a number of diatomic andnumber of diatomic and

polyatomic molecules.

∆

_{H°= total energy}

_{input –}

_{total energy}

_released

=

ΣΣ

_{BE(reactants) –}

ΣΣ

_BE(products)

Use of Bond Enthalpies in

Thermo chemistry

Use of Bond Enthalpies in Thermo chemistry

•• In many cases, it is possible to predict the approximate enthalpy of reaction by usingIn many cases, it is possible to predict the approximate enthalpy of reaction by using the average bond enthalpies.

the average bond enthalpies. •

•Because energy is always requBecause energy is always required to break chemical bonds and chemical bondired to break chemical bonds and chemical bond

formation is always accompanied by a release of energy, we can estimate

formation is always accompanied by a release of energy, we can estimate the enthalpythe enthalpy

of a reaction by counting the total

of a reaction by counting the total number of bonds broken and formed in number of bonds broken and formed in the reactionthe reaction

and recording all the

and recording all the corresponding energy changes.corresponding energy changes.

•

•The enthalpy of reaThe enthalpy of reaction in the gas phase is gction in the gas phase is given byiven by

Where BE stands for average bond enthalpy and

Where BE stands for average bond enthalpy and ΣΣ is the summation signis the summation sign

9.10 Bond Enthalpy

(13)

∆

_{H°= total energy}

_{H°= total energy input –}

_{input – total energy}

_{total energy released}

_released

=

ΣΣ

_{BE(reactants) –}

ΣΣ

_BE(products)

Use of Bond Enthalpies in Thermo chemistry

Use of Bond Enthalpies in

Thermo chemistry

•

•As written, the below equaAs written, the below equation takes care of thetion takes care of the

sign convention for

sign convention for∆∆H°H°. Thus if the tot. Thus if the total energyal energy

input is greater than the t

input is greater than the total energy released,otal energy released, ∆

∆_{H°is positive and the reaction is Endothermic.}_{H°is positive and the reaction is Endothermic.}

•

•On the other hand, if more enOn the other hand, if more energy is releasedergy is released

than absorbed,

than absorbed,∆∆H°is negative and the reactionH°is negative and the reaction

is Exothermic.

Bond Energies (BE) and Enthalpy changes in reactions

∆

_H

00

_{= total ene}

_{rgy input –}

_{total energy r}

_eleased

=

ΣΣ

_{BE(reactants) –}

ΣΣ

_BE(products)

9.10 9.10 •• If reactants and products are all diatomic molecules, then the below equation will yieldIf reactants and products are all diatomic molecules, then the below equation will yield accurate results because the bond enthalpies of

accurate results because the bond enthalpies of diatomic molecules are accuratelydiatomic molecules are accurately known.

known.

•

•If some or all of the reactants and products are polyatomic molecules, the aboveIf some or all of the reactants and products are polyatomic molecules, the above

equation will yield only approximate results because the bond enthalpies used will be

averages. averages.

9.10 Bond Enthalpy

9.10 9.10

H

22

((g

g ) + Cl

) + Cl

22

((g

g )

)

2HCl

2HCl ((g

g ))

2H

22

((g

g ) + O

) + O

22

((g

g )

)

2H

22

O (

O (g

g ))

(14)

Use bond energies to calculate the enthalpy change for:

H

2 (2 (g g ))

+ F

2 (2 (g g ))

2HF

((g g ))

∆

_H

00

₌

ΣΣ

_BE(reactants

_{BE(reactants)}

_{) –}

_–

ΣΣ

_BE(products)

Type of Type of bonds broken bonds broken Number of Number of bonds broken bonds broken Bond energy Bond energy (kJ/mol) (kJ/mol) Energy Energy change (kJ) change (kJ)

H

1

4

43

36 6..4

4

43

36 6..4

4 F

F

1

15

56 6..9

9

1

15

56 6..9

9

Type of Type of bonds formed bonds formed Number of Number of bonds formed bonds formed Bond energy Bond energy (kJ/mol) (kJ/mol) Energy Energy change (kJ) change (kJ)

H

F

2

5

56

68 8..2

2

1

11

13

36 6..4

4

∆ ∆

H

00

_{= 436.4 + 15}

_{= 436.4 + 156.9 –}

_{6.9 – 2 x 568.2 =}

_{2 x 568.2 = -543.1 kJ}

_{-543.1 kJ}

9.10 9.10