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- Cl-

When 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

1s22s22p6 1s22s22p63s23p6

NaCl formation

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

bond): O ═ O

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:

C...

d20

12

30

1053210136

10343031

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 (-252C) 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 (-88.7C),

respectively 194.7 K (-78.5C).

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 dipole-

dipole 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·s-2

1 atm 1.01325·105 Pa

1 atm 760 torr (mmHg)

1 bar 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: p V k where: p – is the pressure (Pa);

V – is the volume (m3);

k – is a constant.

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 hyperbola,

corresponding to a particular temperature.

Boyle’s law may be written as the relationship:

1 1p V p V

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:

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:

0

0

VVT T

k

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:

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:

0

0

ppT T

k

With the addition of Avogadro’s law we obtain the

ideal gas law

0 0

0

p Vp VT T

These three laws were combined to give the combined 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:

Mp V 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:

n1 2p p p ... p

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.

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.

Liquid solutions are:

Gaseous solutions: are gas mixtures, like air.

Gases, regardless of their chemical nature, are miscible

in any proportion.

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.

Solvents

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.

The dissolving process of potassium nitrate in water:

Solvation (hydrating) represents the process of attaching

solvent molecules to the separated ions from the crystal

lattice.

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:

100% s

d

m

mc [%]

Where: 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:

100.)%( s

d

V

Vvolc [%]

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:

sd

dM VM

mc

[molL-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:

solvd

dm mM

mc

[molkg-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:

BA

AA nn

nx

Similarly, the mole fraction of compound B is:

BA

BB nn

nx

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:

This way of expressing concentration is commonly

used in analytical chemistry.

s

d

V

mT [g mL-1] or [g cm-3]

Transformation relations between different ways of

expressing concentration:

Way of expressing

concentration

Percentagec%

MolaritycM

Molalitycm

Percentage c%

MolaritycM

Molalitycm

s

dM

Mcc

100

% s

solvdm m

mMcc

100%

d

sM M

cc100

%

s

ssolvmM m

mcc

solvd

sm mM

mcc

100%

solvs

sMm m

mcc

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.

Solubilit

y

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.

0

0

p pp

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 = the vapor pressure of the pure solventp = 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:

21

1

nn

n

p

pp

o

o

Considering x1 as the molar fraction of the solute and x2

the molar fraction of the solvent, the relation can be written:

11 xp

p

o

from which results: 11 xp

p

o

Considering the fact that x1 + x2 =1, we obtain:

p = x2po

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 = kp

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:

pkT

The increase of the boiling point is:

Tf - the boiling temperature of the solution;

Tf0 - the boiling temperature of the solvent

The decrease of the vapor pressure is:

p0 is the vapor pressure of the solvent;

p – is the vapor pressure of the solution.

0f fT T T

0p p p

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:

meb cKT

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:

Mm

mKT

solv

deb

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:

mcrss cKTTT o

where: Ts is the freezing temperature of the solution;

Ts0 – the freezing temperature of the solvent;

Kcr – the cryoscopic constant.

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:

Mm

mKT

solv

dcr

The cryoscopic constant represents the drop of the

freezing point produced by dissolving one mole of

substance in 1 kg of solvent.

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

membrane in the sugar solution.

The membrane is permeable only

for the small water molecules but

not for the large sugar molecules.membrane

sugar solution

water

The movement of the solvent through a semi-

permeable 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 was

formulated by van’t Hoff:

= cRT

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 be emphasized through the next

manifestations:

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 lead

dichromate.

Pb(NO3)2 + Na2Cr2O7 2NaNO3 + PbCr2O7

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.

Sometimes the modifying of colour can be the sign of a

physical process, not necessary chemical.

3NH4SCN + FeNH4(SO4)2 Fe(SCN)3 + 2(NH4)2SO4

d) Appearance of flame

This is another sign that a

chemical reaction takes

place. An example is the

ignition reactions of hydro-

carbons. The flame that

appears at the Bunsen bulb

is the sign of the oxidation

reaction of methane with

oxygen from air.

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.

N2 + 3H2 = 2NH3

Fe + S = FeS

Ca + Cl2 = CaCl2

SO3 + H2O = H2SO4

a) Combination reactions (synthesis) are reactions in

which two substances interact to form a single compound.

There are many examples for this:

CaCO3 = CaO + CO2

4NH4NO3 = 3N2 + N2O4 + 8H2O

Fe2(SO4)3 = Fe2O3 + 3SO3

b) Decomposition reactions are transformations in

which from one substance, two or more substances are

formed:

Fe + CuSO4 = Cu + FeSO4

Mg + 2H2O = Mg(OH)2 + H2

Zn + 2HCl = ZnCl2 + H2

Cl2 + 2KI = 2KCl + I2

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:

Pb(NO3)2 + 2KI = PbI2 + 2KNO3

AgNO3 + KCl = AgCl + KNO3

H2SO4 + BaCl2 = BaSO4 + 2HCl

CaCl2 + K2CO3 = CaCO3 + 2KCl

d) Double displacement or coupling substitutions are

transformations in which two elements or groups of

elements are exchanged between two chemical

combinations:

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) or sulfur

compounds (thiophene)

C6H5-CH3 + 9O2 = 7CO2 + 4H2O

CH4 + 2O2 = CO2 +2H2O

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:

HCCH + 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 chemical equations, the stoichiometric

coefficients 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

________________________________________

20.23 2 18x 0.18gH O

2 23

Similarly, we calculate the mass of the resulted NaOH:

2·23g Na………………………………2·40g NaOH

0.23g Na………………………………x g NaOH

_________________________________________

NaOH0.23 2 40

x 0.4g2 23

The resulted hydrogen mass:

2·23g Na………………………………2g H2

0.23g Na………………………………x g H2

_____________________________________

2

0.23 2x 0.01gH

2 23

We can calculate directly the volume of H2 that

results from the reaction in normal conditions of

temperature and pressure:

2·23g Na………………………………22.4L H2 (cn)

0.23g Na………………………………x L H2 (cn)

___________________________________________

2(cn)0.23 22.4

x 0.12LH2 23

CHAPTER 7

CHEMICAL EQUILIBRIUM

CHAPTER 7 CHEMICAL EQUILIBRIUM

Reversible reactions

Reactions that may proceed in both directions are called

reversible reactions.

H2 + I2 2HI Example:

The reversible equation is represented using arrows in

both ways instead of the equality sign.

The law of mass action

We consider the reversible reaction:

CH3-COOH + C2H5-OH CH3COOC2H5 + H2O

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.

alcoholacid

wateresterc cc

ccK

Kc is the equilibrium constant.

For a general reversible reaction:

aA + bB mM + nN

the expression of the law of mass action is:

bB

aA

nN

mM

c cc

ccK

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.

3. Increasing the pressure will shift the equilibrium so that

molecules with smaller volume are being formed.

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;

Electrolytic dissociation of water

The water molecules dissociates according to the reaction:

H2O + H2O H3O+ + HO-

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:

2

2

3

OH

OHOH

c

ccK

WOHOHOH KcccK 32

2

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:

7103

WOHOHKcc mol L-1

Kw is called the ionic product of water

In order to express the concentration of the hydrogen

ions in aqueous solutions, the notion of pH was introduced

by Sörensen (1909):

OH

cpH3

lg

OH

apH3

lg

The relation was modified by Bates by replacing the

concentration of the hydronium ions with their activity:

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:

HOcpOH lg

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 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

The dissociation degree is defined as the ratio

between the number of dissociated molecules and the total

number of dissolved molecules:

molecules ofnumber total

molecules ddissociate ofnumber

We consider an acid that dissociates according to the

equation:

HA + H2O H3O+ + A-

The equilibrium constant is given by the relation:

O][H[HA]

][A]O[H

2

3

K

Acidity constant

For diluted solutions, the concentration of water can be

considered constant and it is included in K. We obtain the

acidity constant:

[HA]

][A]O[H][ 3

2

aKOHK

Very weak acids: the first acidity constant lower than 10-7.

Acid Constant K1

HClO (hypochlorous acid) 3.2·10-8

H3BO3 (boric acid) 5.8·10-10

Weak acids: the first acidity constant between 10-7 and 10-2.

Acid Constant K1

H3PO4 (phosphoric acid) 7.5·10-3

CH3COOH (acetic acid) 1.8·10-5

H2CO3 (carbonic acid) 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-

the base constant is given by the relation: [BOH]

][HO][B bK

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:

2

1041010

2

104 14147142

ccx =1.1210-7

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 + 1,1210-7 = 2.1210-7

Thus, the pH of a 10-7 mol L-1 solution of HCl, will be:

pH = -lg[H3O+] = -lg 2.1210-7 = 6.67