CHAPTER 3
CHEMICAL BONDS
The world around us is composed almost entirely of
compounds and mixture of compounds. Most of the pure
elements also contain many atoms bound together, for
example, diamond is a native form of carbon, in which a
large number of carbon atoms are bound.
In compounds, atoms are held together by forces
known as chemical bonds. Electrons play a key role in
chemical bonding.
There are three ideal types of chemical bonds:
- ionic bond (between metals and nonmetals);
- covalent bond (between nonmetals);
- metallic bond (between metallic atoms).
The ionic bond is a type of chemical bond based on
the electrostatic attraction forces between ions having opposite
charges.
It can form between electropositive and electronegative
elements, e.g. between metal and non-metal ions.
The metal, with a few electrons on the last shell,
donates one or more electrons to get a stable electron
configuration and forms positively charged ions (cations).
These electrons are accepted by the non-metal to form a
negatively charged ion (anion) also with a stable electron
configuration. The electrostatic attraction between the anions
and cations causes them to come together and form a bond.
Example: the formation of ionic bond between Na and Cl.
For the sodium atom the electron configuration is:
1s22s22p63s1
The first and second shells of electrons are full, but
the third shell contains only one electron.
When this atom reacts, it gains the configuration of
the nearest rare gas in the periodic table: Ne
1s22s22p6
Na atom loses one electron from its outer shell:
Na → Na+ + e-
The chlorine atom has the configuration
1s22s22p63s23p5
It gains one electron and realizes the stable electron
configuration of Ar:
1s22s22p63s23p6
Cl + e-  ClWhen sodium and chlorine react, the outer electron of
the sodium atoms are transferred to the chlorine atoms to
produce sodium ions Na+ and chlorine ions Cl- , which are
held together by the electrostatic force of their opposite
charges. NaCl is an ionic compound.
1s22s22p63s1
1s22s22p63s23p5
NaCl formation
1s22s22p6
1s22s22p63s23p6
NaCl formation may be illustrated showing the outer
electrons only (Lewis symbol):
In a similar way, a calcium atom may lose two electrons
to two chlorine atoms forming a calcium ion Ca2+ and two
chloride ions Cl-, that is calcium chloride CaCl2 :
In sodium chloride, the ionic bonds are not only
between a pair of sodium ion Na+ an chlorine ion Cl-, but
also between all the ions. These electrostatic interactions
have as a result the formation of NaCl crystal.
We write the formula of sodium chloride as NaCl, but
this is the empirical formula. The sodium chloride crystal
contains huge and equal numbers of Na+ and Cl- ions pocket
together in a way that maximizes the electrostatic forces of
the oppositely charged ions.
Figures 3.2 and 3.3 show the crystal lattice of NaCl
and LiBr.
Sodium chloride crystal
Lithium bromide crystal
Covalent bonds
The covalent bond is a type of chemical bond
formed by sharing pairs of electrons between atoms.
When two electronegative atoms react together,
ionic bonds are not formed because both atoms have a
tendency to gain electrons. In such cases, an stable
electronic configuration may be obtained only by sharing
electrons. First, consider how chlorine atoms Cl react to
form chlorine molecules Cl2 :
Each chlorine atom shares one of its electrons with
the other atom. The electron is shared equally between both
atoms, and each atom in the molecule has in its outer shell 8
electrons – a stable electronic configuration corresponding
to that of Ar.
In a similar way a molecule of carbon tetrachloride
CCl4 is made up of carbon and four chloride atoms:
The carbon atom shares all its four electrons and the
chlorine atoms share one electron each. The carbon atom
forms 4 covalent bonds with 4 chlorine atoms. In this way,
both the carbon and all four chlorine atoms attain a stable
electronic structure.
The sharing of a single pair of electrons results in a
single covalent bond, often represented by a dash sign, so
chlorine molecule may be written as follow:
Cl — Cl
carbon tetrachloride
For oxygen molecule O2, there are two pairs of
electrons shared between the O atoms (double covalent
O═O
bond):
In nitrogen molecule (N2) each nitrogen atom shares
three electrons. The sharing of three pairs of electrons
between two atoms leads to a triple covalent bond
N≡N
Coordinate bond
A molecule of ammonia NH3 is made up of one nitrogen
and three hydrogen atoms:
The nitrogen atom forms three bonds and the
hydrogen atoms one bond each. In this case, one pair of
electrons
is not involved in bond formation and this is
called a lone pair of electrons.
It is possible to have a shared electron pair in which
the pair of electrons comes just from one electron and not
from both. Such bond is called coordinate covalent bond.
Even though the ammonia molecule has a stable
configuration, it can react with hydrogen H+ by donating the
lone pair of electrons, forming the ammonium ion NH4+:
Partial ionic character of covalent bonds
In the chlorine molecule Cl – Cl the pair of electrons
of the covalent bond is shared equally between both
chlorine atom. Because there is not a charge separation
toward one of the Cl atoms, Cl2 molecule is nonpolar.
On the contrary, in HCl molecule, there is a shift of
electrons
toward
the
chlorine
atom
which
is
more
electronegative than hydrogen one. Such molecules, in
which a charge separation exists is called polar molecule or
dipole molecule
The polar molecule of hydrochloric acid
The magnitude of the effect described above is
denoted through the dipole moment μ. The dipole moment
is the product of the magnitude of the charges (δ) and the
distance separating them (d):
μ=δ·d
The symbol δ suggests small magnitude of charge, less
than the charge of an electron ( 1.602 · 10-19 C ).
The magnitude of 3.34 · 10-30 Cm means Debye ( D ):
1D = 3.34 · 10-30 C m
The hydrochloric acid molecule has a dipole moment
μ=1.03 D and the distance between H and Cl atoms is
136 pm ( 136 · 10-12 m ). A charge δ will be:

1.03  3.34  10 30
 20
 

2
.
53

10
C
12
d
136  10
The charge δ is about 16% of the electron charge
(1.602 · 10-19 C ). We can say therefore that the covalent
H – Cl bond has about 16% ionic character.
Metallic bond
Metals tend to have high melting and boiling points
suggesting strong bonds between the atoms.
Sodium has the electronic structure 1s22s22p63s1. When
sodium atoms come together, the electron in the 3s atomic
orbital of one sodium atom shares space with the corresponding
electron of a neighbouring atom to form a molecular orbital in
the same way that a covalent bond is formed.
The difference, however, is that each sodium atom is
touched by eight other sodium atoms, and the sharing occurs
between the each atom and 3s orbitals of all the eight other
atoms. And each oh these eight is in turn touched by eight
sodium atoms and so on.
All of the 3s orbitals of all the atoms overlaps to give a
vast number of molecular orbitals which extend over the whole
piece of metal. There have to be huge numbers of molecular
orbitals because any orbital can only hold two electrons.
The electron can move freely within these molecular
orbitals and so each electron becomes detached from its parent
atom. The electrons are said to be delocalised. The metal is
held together by the strong forces of attraction between the
positive nuclei and the delocalised electron. This may be
described as “ an array of positive ions in a sea of electrons “.
Metallic bond
The “ free “ electrons of the metal are responsible for the
characteristic metallic properties: ability to conduct electricity
and heat, malleability (ability to be flattened into sheets),
ductility (ability to be drawn into wires) and lustrous
appearance.
Intermolecular bonds
Van der Waals forces
Intermolecular forces are attractions between one molecule
and neighboring molecules. All molecules are under the influence
of intermolecular attractions, although in
some cases those
attractions are very weak. These intermolecular interactions are
known as van der Waals forces. Even in a gas like hydrogen
(H2), if you slow the molecules down by cooling the gas, the
attractions are large enough for the molecules to stick together in
order to form a liquid and then a solid.
In hydrogen’ s case the attractions are so weak that the
molecules have to be cooled to 21 K (-252C) before the
attractions, are enough to consider the hydrogen as a liquid.
Helium’ s intermolecular attractions are even weaker – the
molecules won’ t stick together to form a liquid until the
temperature drops to 4 K ( -269 C).
Attractions are electrical in nature. In a symmetrical
molecule like hydrogen, however, these doesn’t seem to be any
electrical distortion to produce positive or negative parts. But
that’ s only true in average. In the next figure the symmetrical
molecule of is represented.
H2 symmetrical molecule
The even shading shows that on average there is no
electrical distraction. But the electrons are mobile and at
any one instant they might find them selves towards one
out if the molecule. This end of the molecule becomes
slightly negative (charge -). The other end will be
temporarily short of electrons and so becomes slightly
positive (+ ) as we can see in the next figure. An instant
later the electrons may well have moved up to the other
end, reversing the polarity of the temporary dipole of
molecule.
Temporary dipole of H2
This
phenomena
even
happens
in
monoatomic
molecules of rare gases, like helium, which consists of a simple
atom. If both the helium electrons happen to be on one side of
the atom at the same time, the nucleus is no longer properly
covered by electrons for that instant.
Temporary dipole of He
The
question
is
how
temporary
dipoles
give
intermolecular bonds?
Imagine a molecule which has a temporary polarity being
approached by one which happens to be non- polar just at that
moment.
Induced dipole
As the right hand molecule approaches, its electrons
will tend to be attracted by the slightly positive end of the left
hand one. This sets up on induced dipole in the approaching
molecule, which is orientated in such a way that the + end of
one is attached to the - end of the other.
Dipole-dipole attraction
An instant later the electrons in the left hand molecule
may well have up the other end. In doing so, they will repel the
electrons in the right hand one.
The polarity of both molecules reverses, but there is still
attraction between - end and + end. As long as the molecules
stay close to each other, the polarities will continue to fluctuate
in synchronization so that the attraction is always maintained.
This phenomena can occur over huge numbers of
molecules. The following diagram shows how a whole lattice of
molecules could be held together in a solid.
Molecular distribution in a solid
The interactions between temporary dipoles and induced
dipoles are known as van der Waals dispersion forces .
Now, let us consider a molecule like HCl. Such a
molecule has a permanent dipole because chlorine is more
electronegative than hydrogen. These permanent dipoles will
cause the HCl molecules to attract each other rather than if
they had to rely only on dispersion forces.
It’s important to realize that all molecules experience
dispersion forces. Dipole-dipole interactions are not an
alternative to dispersion forces. They occur in addition to
them. Molecules which have permanent dipoles will have
boiling points higher than molecules which have only
temporary fluctuating dipoles. Surprisingly, dipole-dipole
attractions are fairly minor compared with dispersion forces,
and their effect can be seen if we compare two molecules
with the same number of electrons and the same size.
For example, the boiling points of ethane (CH3-CH3)
and
fluoromethane
(CH3F)
are:
184.5
K
(-88.7C),
respectively 194.7 K (-78.5C).
The molecule of ethane is symmetric while that of
fluoromethane has permanent dipole.
Hydrogen bond
If we plot the boiling points the hydride of the elements of
groups 15, 16 end 17 we find that the boiling point of the first
elements in each group is abnormally high.
In the cases of NH3, H2O and HF there must be some
additional intermolecular forces of attraction, requiring
significantly more heat energy to break. These relatively
powerful intermolecular forces are described as hydrogen
bonds.
We can observe that in each of these molecules:
 the hydrogen is attached directly to one of the most
electronegative elements, causing the hydrogen to acquire a
significant amount of positive charge;
 each of the elements to which the hydrogen atom is
attached is not only significant, but also has one “active“ lone
pair of elements.
Lone pairs of the 2nd level have the elements contained
in a relatively small volume of space which therefore has a high
density of negative charge. Lone pair at higher levels are more
diffuse and not so attractive to positive particles.
Let’s consider two water molecules coming close together:
The slightly + charge of hydrogen is strongly
attracted to the lone pair end as a result a coordinate bond
is formed. This is a hydrogen bond .
Hydrogen bond is significantly stronger than a dipoledipole interaction, but has about a tenth of the strength of an
average covalent bond. In liquid water, hydrogen bonds are
constantly broken and reformed.
In solid water each water molecule can form
hydrogen bond surrounding water molecules as we can see
in the next figure.
This is why the boiling pint of water is higher than of
ammonia or hydrogen fluoride. In the case of ammonia, the
amount of hydrogen bonding is limited by the fact that each
nitrogen atom has only one lone pair. As well, in hydrogen
fluoride, the number of hydrogen atoms is not enough to
form a three- dimensional structure.
CHAPTER 4
GAS LAWS
GAS LAWS
In a gas the molecules are in a permanent and
chaotic motion. Each particle travels in random directions at
high speed until it reaches another one, when it is deflected,
or until it collides with the wall of the vessel. This movement
is called Brownian motion and the gas phase is a completely
disordered state.
The thermodynamic state of a gas is characterized by
its pressure, its volume, and its temperature.
The relationship between the pressure, volume,
temperature and amount of gas are called gas laws.
Pressure is measured as force per unit area.
The SI unit for pressure is Pa (Pascal).
However, several other units are commonly used.
The table below shows the conversion between these units:
Units of Pressure
1 Pa
1 N·m-2=1 kg·m-1·s2
1 atm
1 atm
1 bar
1.01325·105 Pa
760 torr (mmHg)
105 Pa
Volume is related between all gases by Avogadro’s
hypothesis, which states:
Equal volumes of gases, at the same temperature
and pressure contain equal numbers of molecules.
From this, one can derive the molar volume of a gas,
that is the volume occupied by one mole of gas under
certain conditions. This values, at 1atm and 0°C is:
VM = 22.41 L·mole-1
Temperature is a measure of how much energy the
particles have in a gas.
1. Boyle’s law
This law was discovered by Robert Boyle (1662) and
describes the relationship between the gas pressure and
volume.
The volume occupied by a given amount of gas is
inversely proportional to the pressure at constant
temperature:
where:
p V  k
p – is the pressure (Pa);
V – is the volume (m3);
k – is a constant.
Boyle’s law may be written as the relationship:
p  V  p1  V1
where p1 and V1 are the pressure and the volume in another
state, at the same temperature.
If we represent this relationship we obtain a set of
curves
with
a
shape
called
equilateral
corresponding to a particular temperature.
hyperbola,
The explanation of Boyle’s law is based on the fact
that the pressure exerted by a gas arises from the impact of
its molecules to the walls of the vessel.
If the volume is halved, the density of molecules is
doubled. In a given interval of time twice as many molecules
strike the walls and so, the pressure is doubled in accord
with Boyle’s law.
This law is universal in the sense that it applies to all
gases without reference to their chemical composition.
2. Charles’s law
The volume of a given amount of gas, at constant
pressure, increases proportionally to the temperature:
V V
0
k

T T0
where: V – is the gas volume (m3);
T – is the temperature (K);
k – is a constant.
For two different states, at the same gas pressure, the
relationship becomes:
V V0

T T0
If we represent the relationship, we obtain a set of
straight lines for each pressure considered.
The point of intersection between the straight lines
and the temperature axis is the same: -273.15°C. It is the
temperature at which the volume of a gas would become
zero , called absolute zero temperature (0 K).
3. Gay Lussac’s law
The pressure of a given amount of gas at constant
volume increases proportionally to the temperature:
p p0
 k
T T0
where: p – is the pressure at temperature T;
T – is the temperature (K);
k – is a constant.
For two different states, at the same gas volume, the
relationship becomes:
p p0

T T0
These three laws were combined to give the combined gas law:
p  V p0  V0

T
T0
With the addition of Avogadro’s law we obtain the
ideal gas law
p V  n RT
where: n is the amount of substance expressed in mole;
R – universal gas constant (8.314 J·mole-1·K-1).
For a mole of gas, the relationship becomes:
p  VM  R  T
VM is the molar volume.
4. Dalton’s law of partial pressures
Studies of gaseous mixtures showed that each
component behaves independently of the others.
The total pressure exerted by a gaseous mixture is
equal to the sum of the partial pressures of each
component:
p  p1  p2  ...  pn
The partial pressure of a gas is the pressure that the
gas would exert if it were alone in the container.
CHAPTER 5
SOLUTIONS
SOLUTIONS
Solutions are homogeneous mixtures formed by two
or more components.
For one solution, we distinguish the component that
dissolves, called solvent, and the compound that is
dissolved, called solute.
The notion of solution is not limited to a certain state
of aggregation of the substances. There can be liquid, solid
or gaseous solutions.
Liquid solutions are:
 a gas dissolved in a liquid – carbon dioxide in water;
 liquid dissolved in liquid – ethanol in water;
 solid dissolved in liquid – sodium chloride in water,
naphthalene in benzene.
Solid solutions: the most important are metal alloys, but in
this category are included only the alloys which are
homogenous mixtures.
Gaseous solutions: are gas mixtures, like air.
Gases, regardless of their chemical nature, are miscible
in any proportion.
Solvents
The most common dissolving agent is water; it can
dissolve many solid, liquid or gaseous substances.
Other usual dissolving agents are: ethanol, ethyl
ether, toluene, chloride derivatives and others.
Substances are dissolved in solvents differently. For
example, fats are not dissolved in water, but are well
dissolved in petrol; iodine is barely dissolved in water, but is
well dissolved in alcohol.
Dissolving is a consequence of molecular movement.
When a solid substance is introduced in water, its
particles (molecules or ions) interact with the water
molecules, are separated from the solid and diffuse inside
the solution.
The higher the number of particles separated in the
time unit, the faster the dissolving process.
The finely divided substances, having a higher
surface area in contact with the solvent, are dissolved faster
than massive substances. Also, agitation and temperature
intensifies the dissolving process.
The thermal effect of dissolving
The dissolving of the substances is accompanied by
a thermal effect: either heat absorption or heat
release.
For example, dissolving one mole of potassium
nitrate in a large quantity of water requires 36 kJ absorbed
from the environment.
The dissolving process of an ionic substance, like
potassium nitrate, consists of two successive processes:
 the separation of K+ and NO3- ions from the crystal
lattice, process that requires energy from the exterior,
 the solvation of the ions (hydrating when the solvent is
water), that takes place with heat release.
Solvation (hydrating) represents the process of attaching
solvent molecules to the separated ions from the
crystal lattice.
The dissolving process of potassium nitrate in water:
Because the energy absorbed for the extraction of the
ions from the crystal lattice is higher than the energy
released during the solvation of the ions, the dissolution of
potassium nitrate is an endothermic process, meaning that
dissolving potassium nitrate in water the solutions will cool
down. In case of other ionic substances, like copper sulfate,
the dissolution is an exothermic process.
Concentration of the solutions
Concentration expresses the quantitative relation
between the components of the solution. There are several
ways of expressing the concentration of solutions:
1. Mass percentage: represents the mass of
substance (g) dissolved in 100 g of solution. The relation of
mass percent is:
Where:
md
c% 
 100
ms
[%]
md is the mass of the dissolved substance;
ms – the mass of the solution.
2. Volume percentage: represents the volume of
substance (m3) dissolved in 100m3 of solution:
Vd
c%(vol.) 
 100 [%]
Vs
Where:
Vd is the volume of the dissolved substance;
Vs – the volume of the solution.
This way of expressing the concentration of solutions
is used especially in the case of liquids dissolved in other
liquids. 80% (v) ethanol contains 80 volumes of pure ethanol
and 20 volumes of water. 80° alcohol means 80% (v).
3. Molarity: represents the number of moles of
substance dissolved in 1L of solution:
md
cM 
M d  Vs
[mol L-1]
Where: Md is the molecular mass of the dissolved substance;
Vs – the volume of the solution (L).
4. Molality: represents the number of moles of
substance dissolved in 1kg of solvent:
md
cm 
M d  msolv
[mol kg-1]
Unlike molarity, which depends on temperature, the
molality is independent on temperature.
5. Molar fraction (mole fraction): is the number of
moles of solute divided by the total number of moles of a
solution.
For a solution that contains nA moles of compound A
and nB moles of compound B the mole fraction of compound
A in the solution is:
nA
xA 
n A  nB
Similarly, the mole fraction of compound B is:
nB
xB 
n A  nB
From these relations results that the sum of the mole
fractions of compounds A and B is 1. Similar relations result
in the case of solutions that have several compounds.
6. Titer: represents the mass of dissolved substance
(expressed in g) that is found in 1mL of solution:
md
T
Vs
[g mL-1] or [g cm-3]
This way of expressing concentration is commonly
used in analytical chemistry.
Transformation relations between different ways of
expressing concentration:
Way of
expressing
concentration
Percentage
c%
Molarity
cM
Molality
cm
Percentage
c%
Molarity
cM
c%  c M
c M  c%
100M d
s
s
100M d
ms
ms
c m  c%
c  cM
100M d  msolv m
 s  msolv
Molality
cm
100M d  msolv
c%  c m
ms
c M  cm
msolv   s
ms
Solubility
Introducing sodium chloride in a certain amount of
water, in small portions and under stirring, it can be seen
that at a certain moment, the quantities of NaCl that are
added don’t dissolve anymore, they remain in solid state.
The solution that at a certain temperature contains the
maximum proportion of dissolved substance is called
saturated solution. For example, at 20°C, 35.8g NaCl is the
maximum quantity of NaCl that can be dissolved in 100g of
water. The maximum concentration of the substance in the
saturated solution represents the solubility.
The solubility of sodium chloride is 35.8 g/100 g of
water at the considered temperature.
Solubility depends on the nature of the substances.
Substances that at 20°C have the solubility of more than 1 g
solute per 100 g solvent are considered soluble.
Substances with solubility under this value are
considered slightly soluble.
Soluble substances in water are: NaCl, KNO3, AgNO3,
KBr, NaOH, sodium acetate, sulfuric acid, sugar, etc.
Slightly soluble substances in water are: AgBr,
PbSO4, Fe(OH)3, CaCO3, BaSO4.
The solubility of substances depends on temperature.
The variation of solubility with temperature is
represented by the solubility curves.
The solubility of salts generally increases with
increasing temperature. For some solid substances like
Ce(SO4)3 or Ca(OH)2 the solubility decreases with
increasing temperature.
The solubility of liquids increases with increasing
temperature.
The solubility of gases decreases with increasing
temperature. The solubility of gases is also influenced by the
pressure of the gas above the solution. The higher the
pressure of the gas, the higher the solubility.
Solubility curves for some solid substances
The vapor pressure of solutions
Vapor pressure is the pressure of a vapor in
equilibrium with its non-vapor phases.
The transition of a liquid substance in gaseous state
(evaporation) takes place even before reaching boiling point.
At the liquid – air interface the molecules of the substance
are stopped from leaving the liquid due to the intermolecular
forces which are orientated towards the mass of the liquid.
But, if the kinetic energy of the molecule becomes
very large, this molecule can “escape” from the solution and
passes in the gaseous state.
This phenomenon is reversible, and at the interface
there is a dynamic equilibrium, when the number of
molecules that passes from liquid to air is equal to the
number of molecules that passes from air to liquid. This
means that at equilibrium, the gaseous state is saturated
with the molecules of the liquid substance.
The vapor pressure is an indication of a liquid's
evaporation rate. A substance with a high vapor pressure at
normal temperatures is often referred to as volatile.
The temperature at which the vapor pressure of a
liquid becomes equal to atmospheric pressure (or in case of
closed spaces – the pressure above the liquid) is the boiling
temperature.
Vapor pressure of mixtures
If a non-volatile substance is dissolved in a solvent,
the vapor pressure of the solution is smaller than the one
of the pure solvent, at the same temperature. The relative
drop of vapor pressure is given by:
p0  p
p0
p0 = the vapor pressure of the pure solvent
p = the vapor pressure of the solvent above the solution.
Raoult’s law (1877): the relative drop of the vapor
pressure of a diluted solution is equal to the molar fraction
of the solute in solution:
po  p
n1

po
n1  n2
Considering x1 as the molar fraction of the solute and x2
the molar fraction of the solvent, the relation can be written:
p
1
 x1
po
from which results:
p
 1  x1
po
Considering the fact that x1 + x2 =1, we obtain:
p = x2po
The vapor pressure of a solvent in a solution is
directly proportional to its molar fraction.
The solutions that respect the Raoult’s law are called
ideal solutions. Diluted solutions are approaching the state
of ideal solution.
If a gaseous substance is dissolved in a liquid solvent,
the molecules of gas are dispersed in the mass of the
solvent. They can reach the liquid – gas interface, and if their
kinetic energy is sufficiently high, they pass in the gaseous
state.
Equilibrium is reached at a certain concentration of the
gas in solution, when the number of the gas molecules
that pass from the solution in gaseous state is equal to
the number of gas molecules that pass the opposite way. At
equilibrium, the solution is saturated in gas.
The variation of the solubility of a gas with the
pressure is expressed by Henry’s law: the molar fraction of a
gas dissolved in a solvent is proportional to the pressure of
the gas in equilibrium with the solution:
x = k p
Increasing the boiling point of the solutions
According to Raoult’s law, by dissolving a non-volatile
substance in a solvent, the vapor pressure of the solvent
above the solution is smaller than the one above the pure
solvent.
Thus, the boiling temperature of the solution will be
higher than the one of the solvent.
The increase of the boiling point of the solution
compared to the solvent is proportional to the decrease of
the vapor pressure of the solution compared to the solvent:
T  k  p
The increase of the boiling point is: T  Tf  Tf0
Tf - the boiling temperature of the solution;
Tf0 - the boiling temperature of the solvent
The decrease of the vapor pressure is: p  p0  p
p0 is the vapor pressure of the solvent;
p – is the vapor pressure of the solution.
The variation of the boiling point of the solution
depends also on the concentration of the dissolved substance.
Vapor pressure and temperature for different
concentrations of the solute expressed in molality
The increase of the boiling point can be expressed by
the relation:
T  K eb  cm
where Keb is the ebullioscopic constant,
cm – the molal concentration of the solute.
The ebullioscopic constant, Keb represents the
increase of the boiling point when one mole of substance
is dissolved in 1 kg of solvent.
For diluted solutions, the ebullioscopic constant
does not depend on the nature of the dissolved
substance, as it is a characteristic of the solvent. This
means that solving the same quantity of substance in a
solvent, the increase of the boiling point of the solution
will be the same.
Ebullioscopic constant for different solvents.
Solvent
H2O
chloroform
ethanol
benzene
diethyl ether
Keb
0.52
3.88
1.15
2.57
2.11
Replacing the molarity with its expression, it results:
md
T  K eb 
msolv  M
where: md is the mass of solute (kg);
msolv – the mass of solvent (kg);
M – the molar mass (kg·mol-1).
This relation is used for determining the molecular
mass of the substances.
The research method, based on the experimental
determination of the increase of the solutions boiling point, is
called ebullioscopy.
Decreasing the freezing point of solutions
Another consequence of Raoult’s law is the drop of
the freezing point of solutions. The decrease of the
freezing point is proportional with the molal concentration of
the dissolved substance:
T  Ts  T  K cr  c m
o
s
where:
Ts is the freezing temperature of the solution;
Ts0 – the freezing temperature of the solvent;
Kcr – the cryoscopic constant.
The cryoscopic constant represents the drop of the
freezing point produced by dissolving one mole of
substance in 1 kg of solvent.
Solvent
H2O
Camphor
Naphthalene
Benzene
Cyclohexane
Kcr
1.8
40.0
7.0
5.12
20.2
The research method based on the experimental
determination of the decrease of the freezing point of
solutions is called cryoscopy. The relation used in
cryoscopy for determining the molecular masses of
substances is:
md
T  K cr 
msolv  M
Osmosis and osmotic pressure
If we carefully pour water on a copper sulfate solution
(blue), we will see at the beginning a clear separation
between the blue-coloured copper sulfate solution and the
colorless water. Because of the Brownian movement the
Cu2+ and SO42- ions are dislocated from the solution in the
water layer and the water in the copper sulfate solution, so
that, after a while, a homogenization of the copper sulfate
concentration is produced.
The effective movement of the chemical species,
ionic or molecular, under the influence of the difference
of
concentration is called diffusion. At equal concentrations
the diffusion stops.
The diffusion of some chemical species can be
prevented using membranes. There are semi-permeable
membranes that allow certain molecules or ions to pass
through, but prevent the passage of other molecules.
The osmosis can be evidenced by the following
experience:
At the beginning, the liquid from the
funnel is at the same level with the
liquid in the vessel. In time, the
liquid ascends in the gradual tube
to a certain level. This happens
because water diffuses through the
sugar
solution
membrane in the sugar solution.
The membrane is permeable only
for the small water molecules but
water
not for the large sugar molecules.
membrane
The movement of the solvent through a semipermeable membrane from the diluted solution into the
concentrated solution is called osmosis.
The increase of the level stops when the hydrostatic
pressure h is sufficiently high to prevent the passage of
water. The pressure necessary to stop the diffusion of water
is the osmotic pressure. It can be measured by the height
of the liquid column.
The
general
osmotic
pressure
expression
formulated by van’t Hoff:
 = cRT
was
where:
π is the osmotic pressure (N·m-2);
c – concentration (mole·m-3);
R – universal constant of gases;
T – thermodynamic temperature (K).
The van’t Hoff’s equation is similar to the general
equation of ideal gases.
CHAPTER 6
CHEMICAL REACTIONS
The chemical reaction represents the phenomenon
through which one or more substances are transformed in
other substances, without affecting the nature of the
constituent atoms of the transformed substances.
In
the
environment
several
reactions
can
be
observed, although most of them have a slow rate. Some
examples in this way are rusting of the steel pieces,
alcoholic fermentation, green turning of leaves due to the
forming of chlorophyll, the ignition of fuels. Chemical
reactions
can
manifestations:
be
emphasized
through
the
next
a) Evolution of gas bubbles
If we introduce a piece
of zinc in a hydrochloric
acid solution, we may
observe the hydrogen
evolution reaction.
Zn + 2HCl  ZnCl2 + H2
A more violent reaction
occurs between sodium
and water. The reaction
product is also hydrogen.
2Na + 2H2O  2NaOH + H2
b) Forming of precipitates
By
pouring
sodium
dichromate solution in a
lead nitrate solution we
observe the appearance
of
a
yellow-coloured
precipitate consisting of
slightly
soluble
dichromate.
Pb(NO3)2 + Na2Cr2O7  2NaNO3 + PbCr2O7
lead
c) Changing of colour
Substances absorb light of different wave lengths, so
they appear differently coloured. Changing the nature of a
substance through a chemical reaction can sometimes lead
to color modifications. So, if in a colourless solution of
ammonium thiocyanate we pour an iron (III) and ammonia
sulphate solution we observe the colouring of the solution in
deep red because of the forming of the iron (III) tiocyanate.
3NH4SCN + FeNH4(SO4)2  Fe(SCN)3 + 2(NH4)2SO4
Sometimes the modifying of colour can be the sign of a
physical process, not necessary chemical.
d) Appearance of flame
This is another sign that a
chemical
reaction
takes
place. An example is the
ignition reactions of hydrocarbons.
The
flame
that
appears at the Bunsen bulb
is the sign of the oxidation
reaction
of
methane
oxygen from air.
with
e) Modification of physical properties of solutions
This is another proof of a chemical reaction. Such
kind of property is conductivity.
If in a vessel with hydrochloric acid solution we add a
sodium hydroxide solution, with the help of a conductivity
meter one can measure the decreasing of the solution’s
conductivity until the complete neutralization of the acid.
Thermal effects
The chemical reactions take place through the
breaking of chemical bonds and the forming of new
ones. Therefore, chemical reactions are accompanied by
important thermal effects (heat release or absorption).
Exothermic reactions = reactions that take place with heat
release.
Endothermic reactions = reactions that take place with
heat absorption.
Chemical reactions are represented using chemical
equations.
Reactants = substances initially involved in a chemical
reaction. They are written in the left term of the equation.
Reaction products = substances formed in a chemical
reaction. They are written in the right term of the equation
Because in a chemical reaction, the nature of atoms of
the substances is not changed, the chemical equations are
equalized so that the number of atoms of a certain element
from the left term is equal to the one from the right term.
Let’s
consider
the
chemical
reaction
between
hydrogen and chlorine, when hydrochloric acid is formed:
H2 + Cl2 = 2HCl
For the hydrochloric acid we chose the coefficient 2 so
that the number of chlorine atoms, as well as the number of
hydrogen atoms is not modified.
The primary signification of this chemical reaction is
that a hydrogen molecule interacts with a chlorine molecule
in order to form two molecules of hydrochloric acid.
During this transformation, the covalent bonds: H – H
and Cl – Cl are broken, and a new bond is formed: H – Cl.
The chemical equations have the same properties as
mathematical
equations.
Thus,
the
equation
can
be
multiplied with Avogadro’s number, and we obtain:
NA  H2 + NA  Cl2 = 2 NA  HCl
The second signification of the chemical equation
is: that 1 mole of hydrogen reacts with 1 mole of chlorine
to obtain 2 moles of hydrochloric acid.
In some situations, in order not to create confusion,
chemical formulas of the reactants and the reaction products
are followed by the symbol of the aggregation state written
between brackets:
2Na (s) + 2H2O (l) = 2NaOH (aq) + H2 (g)
The next symbols are used: s – solid, l – liquid, g –
gas, aq – aqueous solution.
Classification of chemical reactions
It is very difficult to choose unique and well defined
criteria for the chemical reactions classification. One
criterion can be the way the reactants interact in order to
form the reaction products. Based on these criteria, we can
distinguish:
 combination reactions (synthesis),
 decomposition reactions,
 single displacement reactions,
 double displacement reactions.
a) Combination reactions (synthesis) are reactions in
which two substances interact to form a single compound.
There are many examples for this:
N2 + 3H2 = 2NH3
Fe + S = FeS
Ca + Cl2 = CaCl2
SO3 + H2O = H2SO4
b) Decomposition reactions are transformations in
which from one substance, two or more substances are
formed:
CaCO3 = CaO + CO2
4NH4NO3 = 3N2 + N2O4 + 8H2O
Fe2(SO4)3 = Fe2O3 + 3SO3
c) Single displacement or substitution reactions are
transformations in which one element or one group of
elements from a combination is replaced with another
element or group of elements:
Fe + CuSO4 = Cu + FeSO4
Mg + 2H2O = Mg(OH)2 + H2
Zn + 2HCl = ZnCl2 + H2
Cl2 + 2KI = 2KCl + I2
d) Double displacement or coupling substitutions are
transformations in which two elements or groups of
elements are exchanged between two chemical
combinations:
Pb(NO3)2 + 2KI = PbI2 + 2KNO3
AgNO3 + KCl = AgCl + KNO3
H2SO4 + BaCl2 = BaSO4 + 2HCl
CaCl2 + K2CO3 = CaCO3 + 2KCl
A special case of double substitution reactions is the
reaction between acids and bases:
H2SO4 + 2NaOH = Na2SO4 + 2H2O
Based on the nature of the reactants or products there
are:
- combustion reactions
- hydrolysis reaction
- precipitation and complexation reactions
a) Combustion reactions: oxygen reacts with a carbon
compound containing hydrogen and/or other element like O,
S, N. Example: the combustion of hydrocarbons (toluene,
methane,
acetylene),
alcohols
(methanol)
compounds (thiophene)
C6H5-CH3 + 9O2 = 7CO2 + 4H2O
CH4 + 2O2 = CO2 +2H2O
or
sulfur
2C2H2 + 5O2 = 4CO2 + 2H2O
2CH3OH + 3O2 = 2CO2 + 2H2O
C4H4S + 6O2 = 4CO2 + 2H2O + SO2
The burning of carbon can also be considered a
combustion reaction: C + O2 = CO2
b) Hydrolysis reaction: the reactant is water; this reactions
are frequent in inorganic chemistry as well as in organic
chemistry:
Al2(SO4)3 + 6H2O = 2Al(OH)3 + 3H2SO4
R-CN + H2O = R-CONH2
c) The precipitation and complexation reactions: the
classification criteria is the nature of the reaction products:
Pb(NO3)2 + K2SO4 = PbSO4 + 2KNO3
CoCl3 + 6NH3 = [Co(NH3)6]Cl3
In organic chemistry, the chemical reactions imply
usually the breaking and formation of covalent bonds.
There are three fundamental types of reactions:
substitution, addition and elimination.
Generally,
the
organic
molecule
that
suffers
a
transformation is called substrate, and the reactant used in it
is called reagent.
The substitution is the reaction in which an atom or a
group of atoms attached to a carbon atom is replaced with
another atom or group of atoms:
CH3-CH2-Cl + NaOH = CH3-CH2-OH + NaCl
The addition reaction is the transformation that leads
to the increasing of the number of atoms or groups of atoms
attached to the carbon atoms of the substrate:
HCCH + HCN => H2C=CH-CN
The elimination is the reverse of the addition and it
leads to the decreases of the number of atoms or groups of
atoms attached to the carbon atoms:
CH3-CH2-OH => H2C=CH2 + H2O
The breaking of the covalent C – C bonds can be
interpreted as an elimination reaction.
Stoichiometry
It is the part of chemistry that has as aim the
establishment of the quantitative relations between the
reactants and reaction products.
The
name
stoichiometry
derives
from
Greek
stoicheon that means element and metron that means
measurement. So, stoichiometry is the science of the
elements measuring.
As it was seen before, the atomic mass unit (uam)
was introduced, that represents the 12th part of the mass
of the C:
1 uam = 1.6605·10-27 kg
Based on the atomic mass unit the relative atomic
masses of all elements have been determined.
Knowing the atomic masses one can calculate the
(relative) molecular masses, as the sum of the relative
masses of all the atoms in the molecule.
For example, the molecular mass of water is
MH2O = 2·1+16=18
MH2SO4 = 2·1+32+4·16=98
MNaCl = 23+35.5=58.5
MCuSO4 = 63.5+32+4·16=159.5.
In
the
coefficients
chemical
equations,
the
stoichiometric
indicate the ratio between the number of
molecules of the reactants and reaction products.
The mole was initially defined as the mass of
substance, expressed in grams, equal to the molecular
mass of the substance.
Thus, 1 mole of H2SO4 is the quantity of substance
that contains 98 g H2SO4.
The definition of the mole, as a fundamental unit in the
International System of Units, is the following:
The mole is the quantity of substance of a system
that contains 6.022·1023 (the Avogadro’s number NA)
elementary particles.
Avogadro’s number refers to different elementary
particles that can be: molecule, atoms, ions or electrons.
Stoichiometric calculation
Stoichiometric calculation is based on the law of
conservation of mass:
In a chemical reaction, the mass of the reactants is
equal to the mass of the reaction products.
Let us consider the reaction between metallic sodium
and water that occurs according to the chemical equation:
2Na + 2H2O = 2NaOH + H2
Atomic masses: Na – 23, H – 1, O – 16.
In a vessel filled with sufficiently enough water we
introduce 0.23 g sodium. Calculate the quantity (mass) of
water that has reacted, as well as the quantities (masses) of
sodium hydroxide and hydrogen that have resulted.
The quantity of water that has reacted with sodium:
2·23g Na………………………………2·18g H2O
0.23g Na………………………………x g H2O
________________________________________
x
0.23  2  18
 0.18gH2O
2  23
Similarly, we calculate the mass of the resulted NaOH:
2·23g Na………………………………2·40g NaOH
0.23g Na………………………………x g NaOH
_________________________________________
x
0.23  2  40
 0.4gNaOH
2  23
The resulted hydrogen mass:
2·23g Na………………………………2g H2
0.23g Na………………………………x g H2
_____________________________________
x
0.23  2
 0.01gH2
2  23
We can calculate directly the volume of H2 that
results
from
the
reaction
in
normal
conditions
temperature and pressure:
2·23g Na………………………………22.4L H2 (cn)
0.23g Na………………………………x L H2 (cn)
___________________________________________
0.23  22.4
x
 0.12LH2 (cn)
2  23
of
CHAPTER 7
CHEMICAL EQUILIBRIUM
CHAPTER 7 CHEMICAL EQUILIBRIUM
Reversible reactions
Reactions that may proceed in both directions are called
reversible reactions.
Example:
H2 + I2
2HI
The reversible equation is represented using arrows in
both ways instead of the equality sign.
The law of mass action
The ratio between the product of the reaction
products concentrations and the product of the
reactants concentrations, all taken to the power of
their stoichiometric coefficients, is constant.
We consider the reversible reaction:
CH3-COOH + C2H5-OH
cester  cwater
Kc 
cacid  calcohol
CH3COOC2H5 + H2O
Kc is the equilibrium constant.
For a general reversible reaction:
aA + bB
mM + nN
the expression of the law of mass action is:
c Mm  c Nn
Kc  a b
c A  cB
Le Chatelier’s principle
If a dynamic equilibrium is disturbed by changing
the conditions (concentrations, temperature and
pressure) the position of equilibrium moves to
counteract the change.
Consequences of Le Chatelier’s principle:
1. Increasing the concentration of one of the
components will shift the equilibrium in the direction in
which this component reacts;
2. Increasing the temperature of the system will shift the
equilibrium in the direction of endothermic reaction, so
that the heat will be absorbed;
3. Increasing the pressure will shift the equilibrium so that
molecules with smaller volume are being formed.
Electrolytic dissociation of water
The water molecules dissociates according to the reaction:
H3O+ + HO-
H2O + H2O
The
equilibrium
is
shifted
far
to
the
left.
Experimentally, it was determined that at 25°C, only one
molecule of water, out of 556,000,000 is dissociated, which
means the dissociation degree of water is α = 18·10-10.
The equilibrium constant for the dissociation reaction
of water is:
K
c H O   cOH 
3
c H2 2O
K  c H2 2O  c H O   cOH   K W
3
Kw is called the ionic product of water
The
ionic
product
of
water
depends
on
the
temperature. At 25°C, the value of KW is 10-14 mol L-1.
In pure water the concentration of the H3O+ ions is
equal to that of the HO-, which means that at 25°C:
c H O  cOH   KW  10 7 mol L-1
3
In order to express the concentration of the hydrogen
ions in aqueous solutions, the notion of pH was introduced
by Sörensen (1909):
pH   lg c H O 
3
The relation was modified by Bates by replacing the
concentration of the hydronium ions with their activity:
pH   lg a H O 
3
In the case of diluted solutions, the activity can be
considered equal to the concentration
Similar to the pH notion the term of pOH was
introduced, that is a measure of the concentration of the
hydroxyl ions:
pOH   lg c HO
It is easily demonstrated that, at the temperature of 25°C:
pH + pOH = -lg KW = 14
Acid – base equilibrium
Acids are substances that, in aqueous solutions,
release hydrogen ions H+. For example, the hydrochloric
acid dissociates in H+ and Cl- ions:
HCl
H+ + Cl-
Bases are substances that, in aqueous solutions,
produce hydroxyl ions, like the case of sodium hydroxide,
that dissociates in Na+ and HO- ions:
NaOH
Na+ + HO-
In aqueous solutions, acids like HCl, H2SO4 or HNO3
are completely dissociated, the dissociation degree is 1.
They are called strong acids. Similarly, bases like KOH or
NaOH are completely dissociated in solution, reason for
which they are called strong bases.
Partially dissociated acids in aqueous solution, like
CH3COOH, HCN or H2S, are called weak acids, and
partially dissociated bases in solution, like NH3 or organic
amines, are called weak bases. The dissociation degree for
weak acids and bases is less than 1.
The dissociation degree is defined as the ratio
between the number of dissociated molecules and the total
number of dissolved molecules:
number of dissociate d molecules

total number of molecules
The reaction between an acid and a base is called
neutralization reaction and it leads to the formation of a
salt and water. For example, the reaction between nitric
acid and potassium hydroxide can be represented by the
equation:
HNO3 + KOH = KNO3 + H2O
Acidity constant
We consider an acid that dissociates according to the
equation:
HA + H2O
H3O+ + A-
The equilibrium constant is given by the relation:
[H 3O  ]  [A  ]
K
[HA]  [H 2O]
For diluted solutions, the concentration of water can be
considered constant and it is included in K. We obtain the
acidity constant:
[H 3 O  ]  [A  ]
K  [ H 2 O]  K a 
[HA]
Very weak acids: the first acidity constant lower than 10-7.
Acid
HClO (hypochlorous acid)
H3BO3 (boric acid)
Constant K1
3.2·10-8
5.8·10-10
Weak acids: the first acidity constant between 10-7 and 10-2.
Acid
H3PO4 (phosphoric acid)
CH3COOH (acetic acid)
H2CO3 (carbonic acid)
Constant K1
7.5·10-3
1.8·10-5
0.45·10-6
Strong acids are completely dissociated in aqueous solution;
one can not distinguish between their acidity constants.
Base constant
For a base that, in aqueous solution, dissociates
according to the reaction:
BOH
B+ + HO-
[B  ]  [HO  ]
the base constant is given by the relation: K b 
[BOH]
Weak bases, like ammonia, aniline, have the basicity
constant below 10-3:
Base
Constant Kb
Ammonia NH3
1.7·10-5
Aniline C6H5 – NH2
3.8·10-10
Strong bases, like sodium hydroxide, calcium hydroxide,
are completely dissociated in water, like in case of strong acids.
Calculation of pH for acid and base solutions
Monoprotic acids are completely dissociated in
aqueous solutions so the hydronium ions concentration is
equal to the concentration of the acid. For example, for a 10-3
mol L-1 solution of HCl, the concentration of the hydronium
ions is [H3O+] = 10-3 mol L-1. The pH of the solution is:
pH = -lg[H3O+] = -lg 10-3 = 3
For a strong base, for example 10-3 mol L-1 KOH the
concentration of hydroxyl ions is [HO-] = 10-3 mol L-1. It results
that the pOH of the solution is: pOH = -lg[HO-] = -lg 10-3 = 3
Considering the relation between pOH and pH one obtains:
pH = 14 – pOH = 14 – 3 = 11
For concentrations higher than 10-3 mol L-1 the pH is
calculated using the Bates relation because the activity
differs from the concentration
At very low acid concentrations for the calculation of
pH it is necessary to consider the hydronium ions coming
from the dissociation of both acid as well as water
molecule.
For example, the pH of a 10-7 mol L-1 solution of HCl is
not 7 because the hydronium ions result not only from the
dissociation of the acid, but from the dissociation of water as
well:
HCl + H2O
H3O+ + Cl-
H2O + H2O
H3O+ + HO-
Considering that the concentration of hydrochloric acid in
the solution is c and the concentration of hydronium,
respectively hydroxyl ions is x, the total hydronium ions
concentration will be c+x. The ionic product of water, at the
temperature of 25°C, will be:
(x + c)x = 10-14
One obtains a second degree equation:
x2 + cx - 10-14 = 0
Solving the equation one obtains:
 c  c 2  4  10 14  10 7  10 14  4  10 14
x

=1.1210-7
2
2
The solution with “minus” in front of the square root has no
meaning, since it is negative.
The concentration of the hydronium ions will be:
[H3O+] = 10-7 + 1,1210-7 = 2.1210-7
Thus, the pH of a 10-7 mol L-1 solution of HCl, will be:
pH = -lg[H3O+] = -lg 2.1210-7 = 6.67