-
Ionic bondingIONIC BONDWhen atoms combine they do so by trying
to achieve an inert gas configuration. Ionic compounds are formed
when electronsare transferred from one atom to another to form ions
with complete outer shells of electrons. In an ionic compound the
positiveand negative ions are attracted to each other by strong
electrostatic forces, and build up into a strong lattice. Ionic
compoundshave high melting points as considerable energy is
required to overcome these forces of attraction.
The classic example of an ionic compound is sodium chloride
Na+CI-, formed when sodium metal burns in chlorine. Chlorine isa
covalent molecule, so each atom already has an inert gas
configuration. However, the energy given out when the ionic
latticeis formed is sufficient to break the bond in the chlorine
molecule to give atoms of chlorine. Each sodium atom then transfers
oneelectron to a chlorine atom to form the ions.
2.B.1 I[NeI3s') 2.B.71[NeI3s'3p')11 protons 1 7 protons11
electrons 17 electrons
2.81[Nell 2.8.81[ArD11 protons 1 7 protons10 electrons 18
electrons
The charge carried by an ion depends on the number of electrons
the atom needed to lose or gain to achieve a full outer shell.
Cations Anions
Group 1 Group 2 Group 3 Group 5 Group 6 Group 7+1 +2 +3 -3 -2
-1
l.i" Na" K+ Mg2+Ca1+ A13+ N3- p3- 02-S1- F- CI- Br
Thus in magnesium chloride two chlorine atoms each gain one
electron from a magnesium atom to form Mg1+CI-1. Inmagnesium oxide
two electrons are transferred from magnesium to oxygen to give
Mg1+02-. Transition metals can form morethan one ion. For example,
iron can form Fe1+ and Fe3+ and copper can form Cu" and Cu2+.
FORMULAS OF IONIC COMPOUNDSIt is easy to obtain the correct
formula as the overall charge of the compound must be zero.
lithium fluoride Li+F- magnesium chloride Mg1+CI-2 aluminium
bromide AI3+Br-3sodium oxide Na+20
2- calcium sulfide Ca1+S2- iron(1I1) oxide Fe3+202-3potassium
nitride K+3N3- calcium phosphide Ca2+3p3-1 iron(ll) oxide
Fe2+02-
Note: the formulas above have been written to show the charges
carried by the ions. Unless asked specifically to do this it
iscommon practice to omit the charges and simply write LiF, MgCI2,
etc.
IONS CONTAINING MORE THAN ONE ELEMENTIn ions formed from more
than one element the charge is often spread (delocalized) over the
whole ion. An example of a positiveion is the ammonium ion NH/, in
which all four N-H bonds are identical. Negative ions are sometimes
known as acid radicalsas they are formed when an acid loses one or
more H+ ions.
hydroxide OH-nitrate NO)sulfate SO/-hydrogensulfate HS04-
(from nitric acid, HN03)
{ from sulfuric acid, H1S04 }
carbonate CO/-hydrogencarbonate HC03-ethanoate CH3COO-
{ from carbonic acid, H1C03 }
(from ethanoic acid, CH3COOH)
The formulas of the ionic compounds are obtained in exactly the
same way. Note: brackets are used to show that the subscriptcovers
all the elements in the ion.
sodium nitrate Na+N03-ammonium sulphate (NH/)2S0/-
calcium carbonate Ca2+CO/-magnesium ethanoate Mg2+(CH3COO-)2
aluminium hydroxide AP+(OH-)3
IONIC OR COVALENT?Ionic compounds are formed between metals on
the left of the Periodic Table and non-metals on the right of the
Periodic Table;that is, between elements in groups 1, 2, and 3 with
a low electronegativity (electropositive elements) and elements
with a highelectronegativity in groups 5,6, and 7. Generally the
difference between the electronegativity values needs to be greater
thanabout 1.8 for ionic bonding to occur.
AI F AI ° AI CI AI BrElectronegativity 1.5 4.0 1.5 3.5 1.5 3.0
1.5 2.8~ '---v----' '-------v--' ~Difference in electronegativity
2.5 2.0 1.5 l.3Formula AIF3 AI103 A12C16 AI2Br6Type of bonding
ionic ionic intermediate between covalent
ionic and covalent
M. pt / °C 1265 2050 Sublimes at 180 97
Bonding 19
-
Covalent bondingSINGLE COVALENT BONDSCovalent bonding involves
the sharing of one or more pairs of electrons so that eachatom in
the molecule achieves an inert gas configuration. The simplest
covalentmolecule is hydrogen. Each hydrogen atom has one electron
in its outer shell. Thetwo electrons are shared and attracted
electrostatically by both positive nucleiresulting in a directional
bond between the two atoms to form a molecule. When onepair of
electrons is shared the resulting bond is known as a single
covalent bond.Another example of a diatomic molecule with a single
covalent bond is chlorine, C12.
LEWIS STRUCTURESIn the Lewis structure (also known as electron
dot structure) all the valence electrons are shown. There are
various differentmethods of depicting the electrons. The simplest
method involves using a line to represent one pair of electrons. It
is alsoacceptable to represent single electrons by dots, crosses or
a combination of the two. The four methods below are all
correctways of showing the Lewis structure of fluorine.
IF-FIxx xx~nnxx xx
: F: F:.. ..xx ••
~HF:xx ••
Sometimes just the shared pairs of electrons are shown, e.g.
F-F. This gives information about the bonding in the molecule, but
itis not the Lewis structure as it does not show all the valence
electrons.
SINGLE COVALENT BONDSIFI
- I _IF-C-FI- I -
1£1tetrafluoro-
methane
HI
H-C-HI
Hmethane
H-FIH-N-HI
Hammonia
H-OII
Hwater hydrogen
fluoride
The carbon atom (electronic configuration 2.4) has fourelectrons
in its outer shell and requires a share in four moreelectrons. It
forms four single bonds with elements that onlyrequire a share in
one more electron, such as hydrogen orchlorine. Nitrogen (2.5)
forms three single bonds withhydrogen in ammonia leaving one
non-bonded pair ofelectrons (also known as a lone pair). In water
there aretwo non-bonded pairs and in hydrogen fluoride
threenon-bonded pairs.
BOND LENGTH AND BOND STRENGTHThe strength of attraction that the
two nuclei have for theshared electrons affects both the length and
strength of thebond. Although there is considerable variation in
the bondlengths and strengths of single bonds in
differentcompounds, double bonds are generally much stronger
andshorter than single bonds. The strongest covalent bonds areshown
by triple bonds.
Length StrengthInm IkJ mol-1
Single bonds CI-CI 0.199 242C-C 0.154 348
Double bonds C=C 0.134 6120=0 0.121 496
Triple bonds C=C 0.120 837N=N 0.110 944
e.g. ethanoic acid: 0.1240nm "II-0 C 0.143nm,/~
,
-
Shapes of simple molecules and ionsVSEPR THEORY
No. of charge Shape Name of BondThe shapes of simple molecules
and ions can be centres shape angle(s)determined by using the
valence shell electron pair 2 0-0-0 linear 180°repulsion (VSEPR)
theory. This states that pairs ofelectrons arrange themselves
around the central 0atom so that they are as far apart from each
other as 1
trigonalpossible. There will be greater repulsion between 3 0
120°non-bonded pairs of electrons than between
0/ ""0 planar
bonded pairs. Since all the electrons in a multiple 0bond must
lie in the same direction, double and 4 1 tetrahedral 109.5°0triple
bonds count as one pair of electrons. Strictly Q-j'/ ""0speaking
the theory refers to negative charge 0
centres, but for most molecules this equates to pairs 00'-,1
trigonalof electrons. 5 90°, 120°, 180°'0-0 bipyramidalThis results
in five basic shapes depending on the 0/1
number of pairs. 0
00'-.1 octahedral 90°, 180°6 0-'0-0
I~o0
WORKING OUT THE ACTUAL SHAPETo work out the actual shape of a
molecule calculate the number of pairs of electronsaround the
central atom, then work out how many are bonding pairs and how
manyare non-bonding pairs. (For ions the number of electrons which
equate to the chargeon the ion must also be included when
calculating the total number of electrons.)
2 NEGATIVE CHARGE CENTRES
CI-Be-CIO=C=O
'Uble bondcounts as one pair
H-C==C-H H-C==N
~ triple bond ~counts as one pair
3 NEGATIVE CHARGE CENTRES3 bonding pairs - trigonal planar
F1
[0 ]2-
O/~""Ocarbonate ion
2 bonding pairs, 1 non-bonded pair - bent or V-shaped
nitrite ion
5 AND 6 NEGATIVE CHARGE CENTRES5 and 6 negative charge
centres
F FF", 1 F'"I /F;P-F ;s~F 1 F 1 F
F F
CICI", \
'S:)C( I
CIdistorted
tetrahedraltrigonal
bipyramidaloctahedral square planar
non-bonding pairs asfar apart as possible
above and below plane
even greaterrepulsion by twonon-bondingpairs so bond angleeven
smaller
4 NEGATIVE CHARGE CENTRES
4 bonding pairs - tetrahedral
CI1
,C"/,
-
Intermolecular forces and allotropes of carbonMOLECULAR
POLARITYWhether a molecule is polar, or not, depends both on the
relative electronegativities of the atoms in the molecule and on
itsshape. If the individual bonds are polar then it does not
necessarily follow that the molecule will be polar as the resultant
dipolemay cancel out all the individual dipoles.
0- 20+ 0-o=c=onon-polar
(resultant dipole zero)
resultant
70-dipole0+ __W, 0+H
polar
0- 1resultantCI dipole
0+1H'--'C""
, HHpolar
0-CI140+
°CI/C"", 0-Clo-CI
non-polar (resultant dipole zero)
Van der Waals' forcesEven in non-polar molecules the electrons
can at anyone moment beunevenly spread. This produces temporary
instantaneous dipoles. Aninstantaneous dipole can induce another
dipole in a neighbouring particleresulting in a weak attraction
between the two particles. Van der Waals' forcesincrease with
increasing mass.
increasing van der Waals' forces
CI270.9-34.0
Br216058.0
F2M, 38.0b. pt I °C -188
increasing van der Waals' forces
CH4M, 16.0b. pt/ °C -162
C3HS44.0-42.2
INTERMOLECULAR FORCESThe covalent bonds between the atoms within
a moleculeare very strong. The forces of attraction between
themolecules are much weaker. These intermolecular forcesdepend on
the polarity of the molecules.
Hydrogen bondingHydrogen bonding occurs when hydrogen is bonded
directly to asmall highly electronegative element, such as
fluorine, oxygen, ornitrogen. As the electron pair is drawn away
from the hydrogen atomby the electronegative element, all that
remains is the proton in thenucleus as there are no inner
electrons. The proton attracts a non-bonding pair of electrons from
the F, N, or 0 resulting in a muchstronger dipole:dipole
attraction. Water has a much higher boilingpoint than the other
group 6 hydrides as the hydrogen bondingbetween water molecules is
much stronger than the dipole:dipolebonding in the remaining
hydrides. A similar trend is seen in thehydrides of group 5 and
group 7. Hydrogen bonds between themolecules in ice result in a
very open structure. When ice melts themolecules can move closer to
each other so that water has itsmaximum density at 4 "C.
boilingtemperature / K
400
300
200
100
the ice lattice
~= hydrogen~ bond
3period
22 Bonding
254183
Dipole:dipole forcesPolar molecules are attracted to each
otherby electrostatic forces. Although stillrelatively weak the
attraction is strongerthan van der Waals' forces.
non-polar
H H H HI I I I
H-C-C-C-C-HI I I IH H H H
polar0-
H ~ HI C IH-9/0+"9-HH H
butane Mr = 58b. pt -0.5°C
propanone Mr = 58b. pt 56.2°C
identical masses(different intermolecular forces)
ALLOTROPES OF CARBONAllotropes occur when an element can exists
indifferent crystalline forms. In diamond each carbonatom is
covalently bonded to four other carbonatoms to form a giant
covalent structure. All thebonds are equally strong andthere is no
plane of weakness in diamondthe molecule so diamond isexceptionally
hard and becauseall the electrons are localized itdoes not conduct
electricity.Both silicon and silicondioxide, Si02, form
similargiant tetrahedral structures.
In graphite each carbon atom has very strong bonds tothree other
carbon atoms to give layers of hexagonalrings. There are only very
weak bonds between thelayers. The layers can slide over each other
so graphiteis an excellentlubricant andbecause theelectrons
aredelocalizedbetween the layersit is a goodconductor
ofelectricity.
A third allotrope of carbon is buckminsterfullerene.Sixty carbon
atoms are arranged in hexagons andpentagons to give a geodesic
spherical structuresimilar to a football. Following the initial
discoveryof buckminsterfullerene many other similar carbonmolecules
have been isolated. This has led to anew branch of science called
nanotechnology.
-
Metallic bonding and physical properties related tobonding
type
METALLIC BONDINGThe valence electrons in metals become
detachedfrom the individual atoms so that metals consist ofa close
packed lattice of positive ions in a sea ofdelocalized electrons. A
metallic bond is theattraction that two neighbouring positive ions
havefor the delocalized electrons between them.Metals are
malleable, that is, they can be bent andreshaped under pressure.
They are also ductile,which means they can be drawn out into a
wire.
delocal izedelectrons
Metals are malleable and ductile because the close-packed layers
of positive ions can slide over eachother without breaking more
bonds than are made.
••88888888888888888+
Impurities added to the metal disturb the lattice andso make the
metal less malleable and ductile. This iswhy alloys are harder than
the pure metals they aremade from.
nucleus andinner shells
TYPE OF BONDING AND PHYSICAL PROPERTIESMelting and boilinq
pointsWhen a liquid turns into a gas the attractive forces between
the particles arecompletely broken so boiling point is a good
indication of the strength ofintermolecular forces. When solids
melt the crystal structure is broken down, butthere are still some
attractive forces between the particles. Melting points areaffected
by impurities. These weaken the structure and result in lower
meltingpoints.
SOlubility'Like tends to dissolve like'. Polarsubstances tend to
dissolve in polarsolvents, such as water, whereas non-polar
substances tend to dissolve innon-polar solvents, such as heptaneor
tetrachloromethane. Organicmolecules often contain a polar headand
a non-polar carbon chain tail. Asthe non-polar carbon chain
lengthincreases in an homologous series themolecules become less
soluble inwater. Ethanol itself is a good solventfor other
substances as it containsboth polar and non-polar ends.
Covalent macromolecular structures have extremely high melting
and boilingpoints. Metals and ionic compounds also tend to have
relatively high boilingpoints due to ionic attractions. Hydrogen
bonds are in the order of fa th thestrength of a covalent bond
whereas van der Waals' forces are in the order of lessthan ,bo of a
covalent bond. The weaker the attractive forces the more volatile
thesubstance.
CH30H
C2HsOH
C3H70H
C4HgOH
NaCIdecreasingsolubility in water
tDiamond (melting point over 4000 °C)All bonds in the
macromolecularstructure covalent
Sodium chloride (melting point 801°C)Ions held strongly in ionic
lattice
H toI II
H-C-C-HI b+
H
H H HI I I
H-C-C-C-HI I IH H H
H HI I b-Oo+OH-C-C-O-HI IH H
CompoundMrM. pt I °CPolarityBonding type
ethanal4420.8polardipole:dipole
ethanol4678.5polarhydrogen bonding
propane44
-42.2non-polarvan der Waals' Ethanol is completely miscible with
water as it
can hydrogen-bond to water molecules.
ConductivityFor conductivity to occur the substance must possess
electrons or ions that are free to move. Metals (and graphite)
containdelocalized electrons and are excellent conductors. Molten
ionic salts also conduct electricity, but are chemicallydecomposed
in the process. Where all the electrons are held in fixed
positions, such as diamond or in simple molecules,no electrical
conductivity occurs.
When an ionic compound melts, the ions are free to move to
oppositely chargedelectrodes. Note: in molten ionic compounds it is
the ions that carry the charge,not free electrons.When a potential
gradient is applied to the metal, the delocalized
electrons can move towards the positive end of the
gradientcarrying charge. 8 88 + 8
heat 8 88 8NaCI 8 88 8
+
Bonding 23
-
Molecular orbitals and hybridization (1)
COMBINATION OF ATOMIC ORBITALS TO FORM MOLECULAR
ORBITALSAlthough the Lewis representation is a useful model to
represent covalent bonds it does make the false assumption that all
thevalence electrons are the same. A more advanced model of bonding
considers the combination of atomic orbitals to formmolecular
orbitals.
a bondsA a (sigma) bond is formed when two atomic orbitals on
differentatoms overlap along a line drawn through the two nuclei.
Thisoccurs when two s orbitals overlap, an s orbital overlaps with
a porbital, or when two p orbitals overlap 'head on'.
Jt bondsA Jt (pi) bond is formed when two p orbitals
overlap'sideways on'. The overlap now occurs above andbelow the
line drawn through the two nuclei. A Jt bondis made up of two
regions of electron density.
rt bond a bond
~ ~o,'=""\§below line of centress p
([X)t
CDt
HYBRIDIZA TlON (1)
Sp3 hybridizationMethane provides a good example of Sp3
hybridization. Methane contains four equal C-H bonds pointing
towards thecorners of a tetrahedron with bond angles of 109.5°. A
free carbon atom has the configuration 1s22s22p2. It cannot retain
thisconfiguration in methane. Not only are there only two unpaired
electrons, but the p orbitals are at 90° to each other and willnot
give bond angles of 109.5° when they overlap with the s orbitals on
the hydrogen atoms.
When the carbon bonds in methane one of its 2s electrons is
promoted to a 2p orbital and then the 2s and three 2p
orbitalshybridize to form four new hybrid orbitals. These four new
orbitals arrange themselves to be as mutually repulsive aspossible,
i.e. tetrahedrally. Four equal a bonds can then be formed with the
hydrogen atoms.
lower in energy than2p orbitals, so moreenergetically
favourable
\11 1 11 11 11 1
~ 2p CD 2p25
electron25 25 and three 2p.
promoted orbitals hybridiz~
0 free carbon ~15 atom 15
11 11 11 1
sp3
o15
24 Bonding
-
Molecular orbitals and hybridization (2)
HYBRIDIZA TlON (2)
Sp2 hybridizationSp2 hybridization occurs in ethene. After a 2s
electron on the carbon atom is promoted the 2s orbital hybridizes
with two ofthe 2p orbitals to form three new planar hybrid orbitals
with a bond angle of 1200 between them. These can form 0 bondswith
the hydrogen atoms and also a 0 bond between the two carbon atoms.
Each carbon atom now has one electronremaining in a 2p orbital.
These can overlap to form a it bond. Ethene is thus a planar
molecule with a region of electrondensity above and below the
plane.
11 11 12p: ,
one 2p orbitalremains
it bond (above and below plane)
1L~oo0~CCJDc~ond
&O~O~
CD2s
2s and two2p orbitalshybridize
ethene
sp hybridizationsp hybridization occurs when the 2s orbital
hybridizes with just one of the 2p orbitals to form two new linear
sp hybridorbitals with an angle of 1800 between them. The remaining
two p orbitals on each carbon atom then overlap to form two
itbonds. An example is ethyne.
Two it bonds at90° to each other
1l
CD2s
11 11 1:2p: two 2p orbitals
remain
2s and one2p orbitalshybridize
RELA TlONSHIP BETWEEN TYPEOF HYBRIDIZA TION, LEWIS Hybridization
Regular bond angle Examples
HSTRUCTURE, AND MOLECULAR Sp3 109S I 0. 0. 0.SHAPES ,C N ,OJ
,N____N-'H/J "H H/~ "H H/~ H/ ~ / ~Molecular shapes can be
arrived at either
H H H H H H •by using the VSEPR theory or by knowing
hydrazinethe type of hybridization. Hybridization
H" /H H" l\i0 °can take place between any sand p orbital Sp2
IIin the same energy level and is not just
120°/C=C" N= /C"(::. "
restricted to carbon compounds. If the H H H H H
shape and bond angles are known fromusing Lewis structures then
the type ofhybridization can be deduced. Similarly if sp 180°
H-C=C-H (N=N)the type of hybridization is known theshape and bond
angles can be deduced.
Bonding 25
-
Delocalization of electrons
RESONANCE STRUCTURESWhen writing the Lewis structures for some
moleculesit is possible to write more than one correct
structure.For example, ozone can be written:
These two structures are known as resonance hybrids.They are
extreme forms of the true structure, which liessomewhere between
the two. Evidence that this is truecomes from bond lengths, as the
bond lengths betweenthe oxygen atoms in ozone are both the same and
areintermediate between an 0=0 double bond and an0-0 single bond.
Resonance structures are usuallyshown with a double headed arrow
between them.Other common compounds which can be written
usingresonance structures are shown here.
DELOCALlZA TION OFELECTRONSResonance structures can also
beexplained by the delocalization ofelectrons. For example, in
theethanoate ion the carbon atom and thetwo oxygen atoms each have
a porbital containing one electron afterthe 0 bonds have been
formed. Insteadof forming just one double bondbetween the carbon
atom and one ofthe oxygen atoms the electrons candelocalize over
all three atoms. This isenergetically more favourable thanforming
just one double bond.
Delocalization can occur wheneveralternate double and single
bondsoccur between carbon atoms. Thedelocalization energy in
benzene isabout 150 k] mol:", which explainswhy the benzene ring is
so resistant toaddition reactions.
26 Bonding
o -0II I
CH3 - C - o-+----; CH3 - C = 0
H HI I
H.....C
CC/H H
-
IB QUESTIONS - BONDING
1. Which compound contains both covalent and ionicbonds?
A. sodium carbonate, Na2C03B. magnesium bromide, MgBr2C.
dichloromethane, CH2CI2D. ethanoic acid, CH3COOH
2. Which pair of elements is most likely to form a
covalentlybonded compound?
A. Li and CI
B. P and °C. Ca and S
D. Zn and Br
3. Given the following electronegativities,H: 2.2 N: 3.0 0: 3.5
F: 4.0
which bond would be the most polar?
A. O-H in H20
B. N-F in NF2
C. N-O in N02D. N-H in NH3
4. What is the correct Lewis structure for methanal?
A. H:C:::O:H C. HC::O:H
D. :C:O:HH
B. HH:C::O:
5. When CH4, NHy H20, are arranged in order of increasingbond
angle, what is the correct order?
A. CH4, NHy H20 C. NHy CH4, H20
B. NHy H20, CH4 D. H20, NHy CH4
6. When the H-N-H bond angles in the species NH2, NH3,NHr are
arranged in order of increasing bond angle(smallest bond angle
first), which order is correct?
A. NH2 < NH3 < NH~ C. NH3 < NH2 < NH~
B. NH~ < NH3 < NH:; D. NH:; < NHr < NH3
7. In which of the following pairs does the second substancehave
the lower boiling point?
A. F2, CI2B. H20, H2S
C. C2H6, C3HaD. CH30CHy CH3CH20H
8. In which of the following substances would hydrogenbonding be
expected to occur?
I. CH4
II. CH3COOH
III. CH30CH3
A. II only
B. I and III only
C. I and III only
D.I, II and III
9. Which one of the following statements is correct?
A. The energy absorbed when liquid ammonia boils isused to
overcome the covalent bonds within theammonia molecule.
B. The energy absorbed when sol id phosphorus (P4) meltsis used
to overcome the ionic bonds between thephosphorus molecules.
C. The energy absorbed when sodium chloride dissolvesin water is
used to form ions.
D. The energy absorbed when copper metal melts is usedto
overcome the non-directional metallic bondsbetween the copper
atoms.
10. A solid has a melting point of 1440 "C. It conducts heatand
electricity. It does not dissolve in water or in organicsolvents.
The bond between the particles is most likely tobe
A. covalent.
B. dipole:dipole.
C. ionic.
D. metallic.
11.What are the types of hybridization of the carbon atoms inthe
compound
H2CIC-CHrCOOH ?123
3.Sp2
sp
12. Which molecule or ion does not have a tetrahedral shape?
A. XeF4
B. SiCI4
c. BF:!D. NH4
13.When the substances below are arranged in order ofincreasing
carbon-carbon bond length (shortest bond first),what is the correct
order?
I. H2CCH2
A. I < II < IIIB. I < III < II
III.©II. H3CCH3
C. II < I < III
D.III < II < I
14.Which of the following species is considered to involve
sp-'hybridization?
I. BCI3
A. lonly
B. II only
II. CH4 III. NH3
C. I and III only
D. II and III only
15. How many J1 bonds are present in CO2?
A. OneB. Two
C. Three
D. Four
16. When the following substances are arranged in order
ofincreasing melting point (lowest melting point first) thecorrect
order is
A. CH3CH2CH3, CH3COCH3, CH3CH2CH20H
B. CH3CH2CHy CH3CH2CH20H, CH3COCH3C. CH3COCHy CH3CH2CH20H,
CH3CH2CH3D. CH3CH2CH20H, CH3CH2CHy CH3COCH3
IB questions - Bonding 27
-
Wt &
Enthalpy changes
enthalpy, H
EXOTHERMIC AND ENDOTHERMIC REACTIONSEnergy is defined as the
ability to do work, that is, move a force through a distance. It is
measured in joules.
Energy = force x distance(J) (N x m)
In a chemical reaction energy is required to break the bonds in
the reactants, and energy is given out when new bonds areformed in
the products. The most important type of energy in chemistry is
heat. If the bonds in the products are stronger thanthe bonds in
the reactants then the reaction is said to be exothermic, as heat
is given out to the surroundings. Examples ofexothermic processes
include combustion and neutralization. In endothermic reactions
heat is absorbed from the surroundingsbecause the bonds in the
reactants are stronger than the bonds in the products.
The internal energy stored in the reactants is known as its
enthalpy,H. The absolute value of the enthalpy of the reactants
cannot beknown, nor can the enthalpy of the products, but what can
bemeasured is the difference between them, !1H. By convention
!1Hhas a negative value for exothermic reactions and a positive
valuefor endothermic reactions. It is normally measured under
standardconditions of 1 atm pressure at a temperature of 298 K.
Thestandard enthalpy change of a reaction is denoted by !1H".
TEMPERATURE AND HEATIt is important to be able to distinguish
between heat andtemperature as the terms are often used
loosely.
• Heat is a measure of the total energy in a given amount
ofsubstance and therefore depends on the amount ofsubstance
present.
• Temperature is a measure of the 'hotness' of a substance.It
represents the average kinetic energy of the substance,but is
independent of the amount of substance present.
\\ (r-- 50°C (' Two beakers of water.
r- 50°C ~Both have the sametemperature, but the100 cm-' of water
contains
~50 cm-'twice as much heat as the
~ 100 ern-50 crn '.
simple calorimeter
polystyrene
reactionmixture
reactants
!1H = Hproducts - Hreactants(value negative)
products (more stable than reactants)
Representation of exothermic reaction usingan enthalpy
diagram.
enthalpy, H
products (less stable than reactants)
!1H = Hproducts - Hreactants(value positive)
reactantsRepresentation of endothermic reaction usingan enthalpy
diagram.
CALORIMETRYThe enthalpy change for a reaction can be
measuredexperimentally by using a calorimeter. In a
simplecalorimeter all the heat evolved in an exothermic reaction
isused to raise the temperature of a known mass of water.
Forendothermic reactions the heat transferred from the water tothe
reaction can be calculated by measuring the lowering oftemperature
of a known mass of water.
To compensate for heat lost by the water in exothermicreactions
to the surroundings as the reaction proceeds a plotof temperature
against time can be drawn. By extrapolatingthe graph, the
temperature rise that would have taken placehad the reaction been
instantaneous can be calculated.
Compensatingfor heat lostTo = initial temperature of
reactantsTl = highest temperature
actually reachedT2 = temperature that would have
been reached if no heat:- - - - ___ lost to surroundings, --
extrapolation atsame rate ofcooling
!1Tfor reaction = T2 - To
reactants mixed
time
28 Energetics