DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
CHEMISTRY- 104
Dr. Ahmed Khamis Mohamed Salama
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
Course Syllabus
1. Atomic Theory
2. Chemical Bonding
3. Chemical Reactions
4. Solutions
5. Acids and Bases
6. Chemical Equilibrium
7. Thermochemistry
8. Chemical kinetics
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
Periodic Table of the elements
Group
1A 2A
3B 4B 5B 6B 7B
8
9
10 1B
2B
3A
4A
5A
6A
7A
8A
Period
1
2
3
4
5
6
7
1
2
H
He
3
4
5
6
7
8
9
10
Li Be
B
C
N
O
F
Ne
11 12
13
14
15
16
17
18
Na Mg
Al
Si
P
S
Cl
Ar
31
32
33
34
35
36
V Cr Mn Fe Co Ni Cu Zn Ga Ge As
Se
Br
Kr
19 20
21 22 23 24 25 26 27 28 29
K Ca
Sc Ti
37 38
39 40 41 42 43 44 45 46 47
51
52
53
54
Rb Sr
Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb
Te
I
Xe
71 72 73 74 75 76 77 78 79
85
86
55 56
Cs Ba
87 88
Fr Ra
*
**
*Lanthanoids
*
**Actinoids
**
30
48
80
49
50
81
82
83
84
Lu Hf Ta W Re Os Ir Pt Au Hg Tl
Pb
Bi
Po At Rn
103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118
Lr Rf Db Sg Bh Hs Mt Ds Rg Uub Uut Uuq Uup Uuh Uus Uuo
57 58 59 60 61 62 63 64 65
66
67
68
69
70
La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb
89 90 91 92 93 94 95 96 97
98
99 100 101 102
Ac Th Pa U Np Pu Am Cm Bk Cf
Es Fm Md No
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DR. AHMED KHAMIS SALAMA
1.
GENERAL CHEMISTRY
ATOMIC THEORY
The Classical Atomic Theory
The classical atomic theory was first put forward by the
Greek philosophers Leucippus and Democritus between
450 and 420 B.C. and postulated that: If a substance could be divided into smaller and
smaller portions of itself, eventually one should reach
the level of particles that could not be divided any
further.
These
extremely
small,
invisible,
and
indivisible particles were called atoms (from the
Greek word for indivisible).
 Atoms
of
various
shapes
combine
through
interlocking patterns to form the objects of the
world.
The theory maintained that:
 Hard and compact substances (such as diamond, iron,
and brass) contained atoms which were interlocked
in a very tight pattern.
 In liquid substances (such as water and wine), the
atoms were thought to be held together much more
loosely. Also, they thought that liquids made up of
round atoms since they would pour so easily.
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
Dalton's Atomic Theory (Modern atomic theory):
Dalton was the founder of modern atomic theory. Dalton's
atomic theory makes the following assumptions:
1. Elements are composed of small, indivisible particles
called atoms. All atoms of a given elements are identical in
mass, size, and shape, and show the same physical and
chemical characteristics.
According to this postulate, all atoms of an element are
identical in all respects. Thus, an atom of the element
carbon is the same whether it is found in New York,
Yokohama, or on the Moon.
2. Atoms of different elements have a different mass, size,
and shape, and show different physical and chemical
properties.
According to this postulate, oxygen atoms are very
different from carbon atoms. Their masses, sizes, and
shapes are different. As a result, the physical and chemical
properties of oxygen are different from those of carbon.
3. Atoms of two or more elements combine together to
form a compound. In any compound, the atoms of the
different elements in the compound are joined in a definite
whole-number ratio, such as 1:1, 2:1, 3:2, etc. The
smallest particle that still has the properties of the
compound is called a molecule.
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
According to this postulate, carbon and oxygen can react
to form carbon monoxide, In the reaction, one carbon
atom (C) can also combine with two atoms of an oxygen
molecule (O2) to form a molecule of carbon dioxide CO2.
Internal Structure of the Atom
It has been found that atoms are also made up of three
subatomic particles: protons, neutrons, and electrons.
PROTON: The proton is an elementary particle with a mass
of 1.67 × 10-24 g and has the smallest unit of positive
charge. According to the fundamental laws of electricity,
protons will repel each other, attract particles with
negative charges, and do not interact with particles that
carry no charge.
ELECTRON: The electron has the lowest mass, only 1/1836
that of a proton and has a negative charge which is equal
in magnitude to that of the proton. Thus, electrons repel
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
each other, attract protons, and do not interact with
electrically neutral particles.
NEUTRON: The neutron is a subatomic particle with a mass
almost equal to that of the proton but with no electrical
charge. Because of its electrically neutral nature, this
particle will neither attract nor repel positively charged
protons, negatively charged electrons, or other neutrons.
The Three Particles that Make up the Atom
Particle
Relative mass
Relative charge
Proton
1
+1
Neutron
1
0
Electron
1/1836
-1
The Atomic Nucleus according to Rutherford Model
(Solar system model):
Rutherford (The New Zealand scientist, Ernest Rutherford)
described the atom as having a central positive nucleus.
The entire mass of the atom is concentrated in its nucleus
and the rest of the atom was mostly empty space .
- The observation that, the mass of an electron is
negligible compared to the mass of a proton or a neutron,
indicated that the protons and the neutrons are located in
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
the nucleus, while the electrons are found in the outer
regions of the atom.
- The positive charge of the nucleus is determined by the
number of the protons it contains. As protons and
electrons have equal but opposite charges, it follows that
in an electrically neutral atom the number of protons must
be the same as the number of electrons.
Rutherford proposed that the electrons (located in the
outer regions of the atom) orbit the nucleus in the same
manner that the Earth and other planets orbit the sun.
Because of this analogy with our planetary system,
Rutherford's model is often referred to as the solar-
system model of the atom. The model makes no
assumptions about the distance of the electrons from the
nucleus.
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
Bohr Model of the Atom (electron shell model):
Bohr model of the atom, also known as the electron shell
model, assumes that the electrons orbit around the
nucleus on the surfaces of imaginary spherical shells
(levels).
These electron shells are concentric about the
nucleus in the same way as the successive layers of an
onion are packed together.
ELECTRON SHELLS
There are seven electron shells according to the energy
level. The seven electron shells are labelled with integers
n = 1, 2, 3, 4, 5, 6 and 7 starting from the shell closest to
the nucleus. This number (n) is known as the principal
quantum number.
Another convention for labelling the electron shell in by
means capital letters; K (n = 1), L (n = 2), M (n = 3), N (n
= 4), …… An electron in a shell with a relatively low value
of n is at a shorter distance from the nucleus than an
electron in a shell with a higher value of n.
Since the principle quantum number (n) is a measure of
the distance of an electron from the nucleus, it is also a
measure of the energy possessed by that electron.
Electrons in shells with low value n have a lower energy
than electrons in shells with higher value of n.
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
Bohr deduced that:
1. Electrons inside an atom possess different energies:
Electrons in the first orbit belong to the first energy level.
Electrons in the second orbit belong to the second energy
level and so on.
2. Each energy level of an atom could only accommodate a
certain number of electrons. The maximum number of
electrons that can populate a certain energy level is given
by the following formula.
For examples:
a) The maximum number of electron in the first energy
level (n = 1) is 2 (1)2 = 2 electrons
b) The maximum number of electron in the second
energy level (n = 2) is 2 (2)2 = 8 electrons
c) The maximum number of electron in the third
energy level (n = 3) is 2 (3)2 = 18 electrons
d) The maximum number of electron in the fourth
energy level (n = 4) is 2 (4)2 = 32 electrons
e) The maximum number of electron in the higher
energy levels (n = 5, 6, or 7) is 32 electrons.
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
Atomic Number and Nucleon Number
The nucleus of an atom always contains a whole number
of protons, exactly equal the number of electrons in the
neutral atom.
Atomic number is known as the number of protons in the
nucleus of an atom.
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
Nucleon (Mass) number is known as the sum of the
numbers of protons and neutrons.
Example 1:
What is the atomic number of the element uranium which
contains 92 electrons in each neutral atom?
Solution:
The number of protons must equal the number electrons in
a neutral atom. Thus, the nucleus of a uranium atom
contains 92 protons.
The atomic number of uranium is 92.
Example 2:
The nucleus of an atom of fluorine contain 9 protons and
10 neutrons. What is the atomic number and the nucleon
number of fluorine?
Solution:
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
Atomic number = Number of protons = 9
Nucleon number =
Number of protons + Number of neutrons =
9 + 10 = 19
Isotopes
Isotopes are known as atoms that have the same number
of protons but a different number of neutrons in the
nucleus. Thus, isotopes have the same atomic number but
a different nucleon number.
To distinguish between the isotopes of an element, the
following symbolic representation is often used:
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
The most common isotope of the element oxygen has 8
protons and 8 neutrons in the nucleus of one of its atoms.
The atomic number of this isotope of oxygen is, therefore
8 and the nucleon number is 16. The great majority
(99.759%) of oxygen atoms in the nature occur as this
isotope.
Example: Hydrogen has three isotopes.
The common isotope has a nucleus that contains one
proton only. The second one exist in every 5000 hydrogen
atoms has a nucleus that contains one proton and one
neutron. This latter isotope has twice the mass of an
ordinary hydrogen atom and is called heavy hydrogen or
deuterium (D). An even smaller number of hydrogen
atoms, 1 in 1017, has a nucleus with one proton and two
neutrons. This isotope is called super heavy hydrogen or
tritium (T).
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
Electron Configuration of the Elements:
The arrangement of electrons in an atom is called the electron
configuration. When electron fill the energy levels, it fills the lowest
energy level first.
A- According to Bohr Model of the Atom
Example: For a hydrogen atom, H, has an atomic number 1, the one
electron goes into the first energy level, the K shell (n = 1).
Example: For a lithium atom, Li, has an atomic number 3, two of the
three electrons go into the first energy level (K shell) while the third
electron goes into the second energy level (L shell). This electron in
the outer energy level is called the valence electron. The two
electrons in the first energy level are called the
core electrons.
Problem: Give the electron configuration for silicon (atomic number
14).
Silicon, Si, atomic number 14 and hence 14 electrons. The first shell
(K shell) can accommodate 2 electrons, and the second shell (L
shell) can hold 8 electrons. That leaves 4 electrons to be
accommodated in the third shell (M shell).
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
B- According to the Quantum Mechanical Model of the Atom
The Quantum Mechanical Model of the atom presents a more
accurate model of the atom. We will take a look at this model and
summarize the results based on these mathematical calculations
without carrying them out ourselves.
The Quantum Mechanical Model assumes that each shell is
subdivided into several number of sublevels (s, p, d and f ).
There is only one s-type orbital - There are three p-type orbitals,
There are five d-type orbitals - There are seven f-type orbitals
The first shell (K) contains only one orbital s,
the second shell (L) subdivided into two sublevels (s and p orbitals),
the third shell (M) subdivided into three sublevels (s, p and d
orbitals), while
the fourth shell (N) and the other shells (n = 5, 6 and 7) subdivided
into four sublevels (s, p, d and f orbitals)
a- Electrons in the Sublevels
Each orbital can contain a maximum of two electrons. Wolfgang
Pauli states that if two electrons occupy the same orbital they must
have opposite spin. This is known as the Pauli exclusion principle.
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
Summary: The Distribution of Electrons in each Principal Energy
Level
Energy
Level, n
Type of
Atomic
Orbital
Number of
Atomic
Orbitals
1
Maximum
Number of
Electrons per
Sublevel
2
Maximum
Total
Number of
Electrons
2
1
1s
2
2s
2p
1
3
2
6
8
3
3s
3p
3d
1
3
5
2
6
10
18
4, 5, 6,
7
4s
4p
4d
4f
1
3
5
7
2
6
10
14
32
Numbers on the last column is equivalent to the prediction using the
formula 2n2
b - Filling Order of the Sublevels
How do we go about remembering the sequence in which electrons
fill the sublevels?
The order in which electrons fill the sublevels is easy to remember if
you follow these steps:
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
1. Write the principal
energy levels and their
sublevels on separate
lines (as shown on the
diagram).
2. Draw arrows over the
sublevels
3. Join the diagonal lines
from end to end
4. The order is: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, ……etc
c - Electron Configuration Notations
There is a way to represent precisely the electron arrangement in
atoms. Let's take a look at the simplest atom, hydrogen.
A hydrogen atom has 1 electron. That electron will occupy the
lowest principal energy level, n = 1, and the only sublevel, s. We
denote the electron configuration of hydrogen as 1s1.

Helium has 2 electrons; the 2 electrons both occupy the s
sublevel in principal energy level 1.
o
Helium's electron configuration is 1s2
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DR. AHMED KHAMIS SALAMA

GENERAL CHEMISTRY
Lithium has 3 electrons; 2 of the 3 electrons occupy the s
sublevel in principal energy level 1. The 3rd electron must go
in the next available sublevel, 2s.
o

Lithium's electron configuration is 1s2 2s1
Beryllium has 4 electrons; 2 of the 3 electrons occupy the s
sublevel in principal energy level 1. The 3rd and 4th electrons
must go in the next available sublevel, 2s. Beryllium's electron
configuration is 1s2 2s2
d - Electron Configuration for Atoms of the First 20
Elements
When the electrons are arranged in their lowest energy state, the
atom is in the ground state. The following table summarizes the
ground state electron configuration of the first 20 elements on the
periodic table.
Name
Atomic Number
Electron
Configuration
Hydrogen
1
1s1
Helium
2
1s2
Lithium
3
1s2 2s1
Beryllium
4
1s2 2s2
Boron
5
1s2 2s22p1
Carbon
6
1s2 2s22p2
Nitrogen
7
1s2 2s22p3
Oxygen
8
1s2 2s22p4
Fluorine
9
1s2 2s22p5
Neon
10
1s2 2s22p6
Sodium
11
1s2 2s22p63s1
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
Magnesium
12
1s2 2s22p63s2
Aluminum
13
1s2 2s22p63s23p1
Silicon
14
1s2 2s22p63s23p2
Phosphorus
15
1s2 2s22p63s23p3
Sulfur
16
1s2 2s22p63s23p4
Chlorine
17
1s2 2s22p63s23p5
Argon
18
1s2 2s22p63s23p6
Potassium
19
1s2 2s22p63s23p64s1
Calcium
20
1s2 2s22p63s23p64s2
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
2. CHEMICAL BONDING
Atoms have high energy and therefore they very active and have
tendency to be more stable and carry less energy through
combination together to give molecules. Molecules are more stable
and have less energy than atoms.
Atoms join together through
chemical bonding.
Chemical bonding may be formed between atoms of two
nonmetals similar or different by sharing electrons to complete
the outermost shell of each one to the electronic configuration of
the nearest inert gas thus can acquire a stable, nobel-gas structure
(Lewis suggestion).
Chemical bonding may also be formed between a metal and
non-metal by transferring of one or more electron from the
outermost shell of the metal to the outermost shell of the nonmetal
to complete the outermost shell of each one to the electronic
configuration of the nearest inert gas (Kossel suggestion).
Chemical bonding may also be formed between two metals via
accumulation and rearrangement of these metals in a shape of
metallic crystal.
TYPES OF CHEMICAL BONDS:
The covalent bond:
Is a bond formed between atoms of two non-metals similar or
different by sharing electrons.
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
Why, for example, is the H2 molecule more stable than two isolated
hydrogen atoms ?
There are two somewhat different ways of explanation:
1. From a classical electrostatic point of view, locating two
electrons between the two protons of the H2 molecule
lowers the energy of the system.
energies
between
(electron-proton)
oppositely
exceed
the
The attractive
charged
particles
repulsive
energies
between particles of like charge (electron-electron,
proton-proton).
2. From a quantum mechanical point of view, the two
atomic orbitals of the hydrogen atoms overlap to form a
new bonding orbital. Putting two electrons of opposite
spin in this orbital lowers the energy of the system.
Single covalent bond: A bond formed by sharing of one pair of
electrons between two non-metallic atoms where each atom shares
one electron (H:H).
Double covalent bond: A bond formed by sharing of two pairs of
electrons between two non-metallic atoms where each atom shares
two electrons.
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
Examples, O2 , C2H4
Triple covalent bond: A bond formed by sharing of three pairs of
electrons between two non-metallic atoms where each atom shares
three electrons. Examples, N2 , C2H2
The ionic bond:
It is formed between a metal and non-metal by transferring of
one electron or more from the outermost shell of the metal to the
outermost shell of the non-metal.
Notes:

The outermost shell of an atom of a metal contains less
than 4 electrons (1 or 2 or 3).

Ions of metals are always positive because the metals
tend to lose the electrons of the outermost shell, so the
number of protons is greater than electrons.
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DR. AHMED KHAMIS SALAMA

GENERAL CHEMISTRY
The outermost shell of an atom of non-metal contains
more than 4 electrons (5 or 6 or 7).

Ions of non-metals are always negative because nonmetals tend to gain electrons, so electron number is
greater than proton number.
Formation of ionic bond:

Atom of metal loses electrons and changes into positive
ion. Atoms of non-metal gains the electrons lost by the
metal and changes into negative ion.

An electrostatic force of attraction is developed between
the positive ion and negative ion forming an ionic
compound.
Formation of NaCl:
Sodium atom (atomic number = 11):
Chlorine atom (atomic number = 17):
Note: ionic bonds are strong while covalent bonds are weak.
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
Nonpolar covalent bond:
A symmetrical distribution of electrons leads to a bond or molecule
with no positive or negative poles. When the covalent bond formed
between two identical non-metals such as Cl-Cl ,
Br-Br,….. So
there is no differences in the electronegativity value between them.
Thus, the covalent bond in this case is considered nonpolar covalent
bond.
Polar covalent bond:
As a result of an unsymmetrical distribution of electrons, the bond or
molecule contains a positive and a negative pole and is therefore a
dipole. When the covalent bond formed between atoms of two
different non-metals such as
the
two
atoms
is
H-Cl the electronegativity between
different
where
chlorine
atom
is
more
electronegative than hydrogen atom.
Therefore, the pair of electrons will attracts to the chlorine atom
more than the hydrogen atom.
So, chlorine atom will carries
partially negative charge (-δ) while hydrogen atom will carries
partially positive charge (+δ), H+δ – Cl-δ . Thus, the covalent bond
in this case is considered polar covalent bond.
Polarity of molecules:
A polar molecule is one that contains positive and negative poles.
There is a partial positive charge (positive pole) at one point in the
molecule and a partial negative charge (negative pole) at a different
point.
If a molecule is diatomic, it is easy to decide whether it is polar or
non-polar. A diatomic molecule has only one kind of bond; hence,
the polarity of the molecule is the same as the polarity of the bond.
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
Hydrogen and fluorine (H2, F2) are nonpolar because the bonded
atoms are identical and the bond is non-polar. Hydrogen fluoride,
HF, on the other hand, has a polar bond, so the molecule is polar.
The bonding electrons spend more time near the fluorine atom so
that there is a negative pole at the end and a negative pole at the
hydrogen end.
In a molecule containing more than two atoms, it is not so easy to
decide whether it is polar or non polar. In this case, not only bond
polarity but also molecular geometry determines the polarity of the
molecule.
The geometry of a diatomic molecule such as Cl2 or HCl can be
described very simply. Since two points define a straight line, the
molecule must be linear
With molecules containing three or more atoms, the geometry is not
so obvious. Here, the angles between bonds, called angle bonds,
must be considered.
For example, a molecule of the type YX2 ,
where Y represents the central atom and X an atom bonded to it,
can be
Linear, with a bond angle of 180°
Bent, with a bond angle less than 180°
The major features of molecular geometry can be predicted on the
basis of a quite simple principle-electron pair repulsion. The valence
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
electron pairs surrounding an atom repel one another, so the
orbitals containing those electron pairs are oriented to be as far
apart as possible.
Ideal geometries with two to six electron pairs around a central
atom:
Example
Orientation of
Predicted bond
electron pairs
angels
BeF2
Linear
180°
BF3
Triangular Planer
120°
CH4
Tetrahedron
109.5°
PCl5
Triangular Bipyramid
90°
120°
180°
SF6
Octahedron
90°
180°
In BeF2:
There are two polar Be-F bonds; in both bonds, the
electron density is concentrated around the more electronegative
fluorine atom. However, since the BeF2 molecule is linear, the two
Be-F dipoles are in opposite directions and cancel one another. The
molecule has no net dipole and hence is nonpolar.
In CCl4 : the bond dipoles cancel and the molecule is non-polar.
In CCl3 or in the bent H2O molecule: there is a net dipole, and the
molecule is polar.
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
Coordinate covalent bond:
The coordinate covalent bond formed between two atoms, the doner
atom and acceptor atom.
The doner atom has non-bonding
electrons and can sharing the covalent bond by its non-bonding
electrons while the acceptor atom accepts these non-bonding
electrons in its empty orbital.
Examples, :NH3 and
H2O:
The nitrogen atom in ammonia has one pair of non-bonding
electrons and can make coordinate covalent bond.
Metallic bond:
Atoms of the same metal are tend to be closed and rearranged in a
crystal shape.
Hydrogen bond:
Hydrogen atom join with an electronegative atom through covalent
bond, where the electronegative atom attracts the pair of sharing
electrons and carries partially negative charge while hydrogen atom
carries a partially positive charge. This bond formed between the
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
molecules of water and ammonia to increase the cohesion force
between molecules.
Van der-Waals force:
These are weak electrostatic forces appears due to the occurrence of
a positive charge and negative charge at the terminals of the
molecule causing attraction between these molecules with others of
the same matter.
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
3. CHEMICAL REACTIONS
When a chemical reaction occurs, we frequently observe at least one
of the following:
1. Change in color:
For example, when a solution of potassium
dichromate (orange) is added to a solution of sodium hydrogen
sulfite (colorless), the resulting solution is green.
2. Formation of a gas: For example, if hydrochloric acid is added
to a solution of sodium carbonate, one readily sees bubbles of gas
(carbon dioxide) escaping from the resulting solution. In addition,
the gas being formed often has an odor which can assist in its
detection.
3. Formation of a solid: For example, when a clear solution of
calcium chloride is added to a clear solution of potassium carbonate,
a white solid (calcium carbonate) is produced. The solid formed is
usually called a precipitate.
4. Release or absorption of heat: For example, when
hydrochloric acid is added to a solution of sodium hydroxide, heat is
released and the reaction vessel gets warm.
Note:
We cannot always assume that a chemical reaction has
occurred when heat is absorbed or released. Many substances
release or absorb heat when they are mixed together, even though
no reaction has taken place.
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
CLASSES OF CHEMICAL REACTIONS
There are four common classes of chemical reactions; combination,
decomposition, replacement and metathesis.
1. COMBINATION REACTIONS:
In a combination reaction, two or more simple substances combine
to produce a more complex substance.
2. DECOMPOSITION REACTIONS:
This is just the reverse of combination, a complex substance is
broken down into a number of simpler substances.
3. SINGLE REPLACEMENT REACTIONS:
During a replacement reaction one atom or ion in a compound is
replaced by another.
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GENERAL CHEMISTRY
The activity series: We can usually predict when a replacement
reaction is likely to occur by referring to a list called the activity
series.
The following table consists of metallic elements (plus
hydrogen) arranged in such a way that any element in the table will
displace ions of the elements below it from aqueous solutions of
their salts.
The activity series (or electrochemical or electromotive series)
Lithium
These
metals
Potassium
from water
displace
hydrogen
displace
hydrogen
Barium
Calcium
Sodium
Magnesium These
A specific metal will Aluminum
metals
from acids
displace any metal ion Zinc
that appears below it.
Iron
Nickel
Tin
Lead
Hydrogen
Copper
These metals do not react with
Mercury
acids (or water)
Silver
gold
Thus, as we have seen in the previous example, iron is able to
displace copper from an aqueous solution of copper(II) sulfate since
iron is above copper in the electrochemical series.
The reverse reaction does not occur
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Metals very high in the activity series:
The metals that are very high in the activity series can not only
displace hydrogen from an acid, but they can also displace hydrogen
from water. For example, sodium metal reacts violently with water
to give hydrogen gas and a solution of sodium hydroxide:
The halogen replacement series:
A series similar to the activity series exists for the halogens.
The halogen replacement series
A specific halogen will displace any Fluorine
halide ion that appears below it
Chlorine
Bromine
Iodine
An element in this series can replace the ions of any element below
it from aqueous solutions of its salts; thus, chlorine will displace
bromine from sodium bromide solution as discussed above, but the
reverse reaction does not occur since bromine is below chlorine in
this series.
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4. Double replacement reactions (metathesis reaction):
This may be best thought of as being a change of partners, the
cation from compound A joins with the anion of compound B, while
the cation from compound B joins with the anion from compound A.
The overall effect of such a reaction is that the positive ions
exchange their negative partners.
Example: The precipitation of iron(II) sulfide when aqueous
solutions of iron(II) sulfate and ammonium sulfide are mixed.
The metathesis reactions that involve the formation of covalently
bonded
water
molecules
are
given
NEUTRALIZATION REACTION.
the
special
name
of
A neutralization reaction
involves the reaction of an acid with a base to produce a salt plus
water.
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Arrhenius definitions of acids and bases:
An acid is a
substance that gives H+(aq) ions when it is dissolved in water. A base
is a substance that gives OH-(aq) ions when it is dissolved in water.
First let us consider the reaction between nitric acid (a strong acid)
and potassium hydroxide (a strong base).
The products are
potassium nitrate and water.
HNO3 (aq)
+
KOH(aq)
→
KNO3(aq) + H2O(ℓ)
5. Oxidation – Reduction reactions (redox reaction)
This type of reaction in aqueous solution involves an exchange of
electrons between two species. One species loses electrons and is
said to be oxidized.
The other species, which gains electrons, is
reduced.
To illustrate this situation, consider the redox reaction that takes
place when zinc pellets are added to HCl.
Zn + HCl
H2 + ZnCl
Total ionic equation
Zn + 2H+ + Cl-
H2 + Zn++ + Cl-
The net ionic equation for the reaction is:
Here, zinc atoms are oxidized to Zn2+ ions by losing electrons; the
half-reaction is oxidation:
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While H+ ions are reduced to H2 molecules by gaining electrons; the
half-reaction is reduction:
It should be clear that oxidation and reduction occur together, in the
same reaction, you can’t have one without the other. There is no net
change in the number of electrons in a redox reaction. Those given
off in the oxidation half-reaction are taken on by another species in
the reduction half-reaction.
The ion or molecule that accepts
electrons is called the oxidizing agents; by accepting electrons it
brings about the oxidation of another species.
Conversely, the
species that donates electrons is called the reducing agent; when
reaction occurs it reduces the other species.
To illustrate these concepts consider the reaction
The H+ ion is the oxidizing agent ; it brings about the oxidation of
zinc. By the same token, Zinc acts as a reducing agent, it furnishes
the electrons required to reduce H+ ions.
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Oxidation Number:
For monatomic ion (e.g., Na+, S2-), the oxidation number is, quite
simply, the charge of the ion (+1, -2).
Oxidation numbers in all kinds of species are assigned according to a
set of 4 rules:

The oxidation number of an element in an elementary
substance is zero.
For example, oxidation number of
chlorine in Cl2 or phosphorus in P4 is zero.

The oxidation number of an element in a monatomic ion
is equal to the charge of that ion. In the ionic compound
NaCl, sodium has an oxidation number of +1, chlorine
has an oxidation number of -1. The oxidation numbers
of aluminum and oxygen in Al2O3 (Al3+, O2- ions) are +3
and -2, respectively.

Certain elements have the same oxidation number in all
or almost all their compounds.
The group 1 metals
always exist as +1 ions in their compounds and hence
are assigned an oxidation number of +1. By the same
token, group 2 elements always have
oxidation
numbers of +2 in their compounds.

Fluorine always has an oxidation number of -1. Oxygen
is ordinarily assigned an oxidation number of -2 in its
compounds.
(An
exception
arises
in
compounds
containing the peroxide ion, O2-- , where the oxidation
number of oxygen is -1).
Hydrogen in its compounds
ordinarily has an oxidation number of +1. (The major
exception is in metal hydrides such as NaH and CaH2 ,
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
where hydrogen is present at the H- ion and hence is
assigned an oxidation number of -1).

The sum of the oxidation numbers in a neutral species is
0 ; in a polyatomic ion, it is equal to the charge of that
ion.
According oxidation number concept, oxidation is defined as an
increase in oxidation number and reduction as a decrease in
oxidation number.
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These definitions are of course compatible with the interpretation of
oxidation and reduction in terms of loss and gain of electrons. An
element that loses electrons must increase in oxidation number.
The gain of electrons always results in a decrease in oxidation
number.
Balancing half-equations (oxidation or reduction):
it is clear that mass and charge balance can be achieved by adding
an electron to the right:
mass balance is obtained by writing a coefficient of 2 for Cl- ; charge
is then balanced by adding two electrons to the left:
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DR. AHMED KHAMIS SALAMA
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To balance half-equations such these, proceed as follows:
1. Balance the atoms of the element being oxidized or
reduced.
2. Balance oxidation number by adding electrons.
For a
reduction half-equation, add electrons to the left; for an
oxidation half-equation, add electrons to the right.
3. Balance charge by adding H+ ions in acidic solution, OHions in basic solution.
4. Balance hydrogen by adding H2O molecules.
5. Check to make sure that oxygen is balanced. If it is, the
half-equation is almost certainly balanced correctly with
respect to both mass and charge.
Example: Balance the following half equation:
Since there is one atom of Mn on both sides, no adjustment is
required here.
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
Since Manganese is reduced from an oxidation number of +7 to +2,
five electrons must be added to the left.
There is a total charge of -6 on the left versus +2 on the right. To
balance, add 8 H+ to the left to give a charge of +2 on both sides
To balance the 8 H+ ions on the left, add 4 H2O molecules to the
right
Note that there are the same number of oxygen atoms, four, on both
sides, as there should be.
Example: Balance the following half equation:
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
Balancing Redox equations:
Follow a systematic, four-step procedure:
1. Split the equation into two half-equations, one for
reduction, the other for oxidation.
2. Balance one of the half-equations.
3. Balance the other half-equation.
4. Combine the two half-equations in such a way as to
eliminate electrons.
Balance the following redox equation:
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6. REACTIONS in which HEAT is ABSORBED or RELEASED
Reactions which accomplished by the release of energy in the form
of heat are called Exothermic Reactions, whereas reactions that
absorbs heat from the surroundings are called
Endothermic
Reactions.
If you wish to show that a reaction is exothermic, there is a simple
way of doing it. For example,
2 Na(s) + Cl2(g) → 2 NaCl(s)
+ heat
However, sometimes we wish to be more specific and say exactly
how much heat is released.
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DR. AHMED KHAMIS SALAMA
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CH4(g) + 2 O2(g) → CO2(g)
+ 2 H2O(g) + 890 KJ
Each of the four substances involved in this reaction contains a
certain quantity of energy and this energy content is called the
ENTHALPY (H) of the Compound. In this example, 890 KJ of energy
was released as heat. This means that one mole of carbon dioxide
gas plus two moles of water vapor must contain 890 KJ of energy
less than one mole of methane gas plus two mole of oxygen gas.
Because the products themselves posses 890 KJ of energy less than
the reactants, we say that the change in Enthalpy (∆H) of the
reaction is -890 KJ, (∆H = -890 KJ).
Our equation can be alternatively written as:
CH4(g) + 2 O2(g)
→
CO2(g)
+ 2 H2O(g)
( ∆H = - 890 KJ)
If heat is absorbed during the reaction, the enthalpy of the products
will be greater than that of the reactants. The change in Enthalpy
(∆H) will, therefore, have a positive sign.
An example of such a
reaction is as follows:
2 HgO(s)
→
2 Hg(ℓ) +
O2(g)
( ∆H = + 181.7 KJ)
The study of the energy changes that take place during chemical
reactions is called thermo chemistry.
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4. SOLUTIONS
A solution, which may be gaseous, liquid or solid is defined as a
homogeneous mixture that has a uniform composition throughout.
Solutions play a very important role in chemistry. Chemical reactions
can occur only when molecules or ions of different substances come in
contact with one another. If we mix solid substances together,
reaction is usually very slow, as the only contact will be at the surface
of the particles. In solution, however, a reactant is dispersed in the
form of free molecules or ions, and these can come into contact with
any molecules or ions of the other reactant. Reaction in solution, then,
will be very much faster.
Components of a solution
A solution is composed of two pure substances (or more), a solute
and a solvent. The component which is present in larger quantity is
usually called the solvent, while the other present in small quantity
is referred to as the solute. It is a mixture and not a compound,
because we can change the quantity of solute or solvent greatly and
still we have a homogeneous solution.
If the quantity of solute is small in solution, the solution is described
as dilute solution.
However if the quantity of solute is large in
solution, the solution is described as concentrated solution.
A saturated solution is one in which the solvent cannot dissolve a
further quantity of the solute.
Supersaturated solution is obtained in which the solvent has
dissolved
more
solute
than
is
necessary
for
saturation.
Supersaturated solutions are usually unstable or in a metastable
state. By shaking or dropping a crystal of the solid or by the action
of dust particles the excess amount of solute is precipitated down.
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TYPES OF SOLUTIONS:
1. Solid in liquid: In this case, the solid is the solute and the liquid is
the solvent. When water is the solvent then the system is known as
an aqueous solution. A solid that dissolves in a particular liquid is
said to be soluble in that liquid; one which does not dissolve in a
given liquid is said to be insoluble in that particular liquid.
2. Solid in solid: The solutions formed by dissolving one solid in
another. Solutions of this type are usually referred to called as
alloys.
3. Solid in gas: The solutions formed by dissolving one solids in gas
e.g. smoke, dust in air.
4. Liquid in liquid: Two liquids that will dissolve in one another are
said to be miscible. Usually the liquid present in the smaller quantity
is referred to as the solute. Immiscible liquids are those that will not
dissolve when mixed together. Water and ethyl alcohol are miscible
in any proportions, while water and gasoline are immiscible.
5. Liquid in solid: A solution of a liquid in a solid is referred to as an
amalgam. Dental amalgams, used as tooth fillings, consist of liquid
mercury combined with one or more solid metallic elements such as
silver, tin, or copper.
6. Liquid in gas: A solution of a liquid in gas such as mist.
7. Gas in liquid: Solutions formed by dissolving gases in liquids are
the basis of the soft-drink industry. (Carbonated drinks such as
Coca-Cola( consist of a flavored liquid containing dissolved carbon
dioxide gas.) In these solutions the amount of solute (i.e., gas) that
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
can be dissolved in a given volume of solvent decreases as the
temperature is raised. Increasing the pressure of a gas increases the
amount of gas that can be dissolved in a particular volume of
solution.
8. Gas in gas: The most common example of a solution consisting of
one gas dissolved in another gas is the air around us. The main
component of air, nitrogen, may be considered to be the solvent,
with the minor component, oxygen, being the solute. A number of
other gases, such as carbon dioxide and argon, are also present as
very minor components of this solution.
9. Gas in solid: The most common example of a solution consisting
of gas dissolved in solid is charcoal.
Water:
Water is the only solvent to be commonly used. In view of this, it is
worthwhile considering some of the properties that make water
unique.

Water is the only common liquid on this planet that occurs
naturally in each of the three phases of matter solid, liquid,
and gas.

Water in solid phase is less dense than the liquid phase, that
is, ice is less dense than water. As a result of this, ice will
always form on the top of a pond or lake, rather than at the
bottom. Consequently, lakes rarely freeze solid right to the
bottom.

One of water's most important properties is its ability to
dissolve a wide range of substances. Most covalent liquids are
only able to dissolve other covalent substances, but water can
dissolve
many
ionic
substances,
47
such
as
salt
(sodium
DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
chloride), as well as many covalent substances, such as sugar
(sucrose).

The large variety of ions present in seawater has resulted from
the leaching of rocks through thousands of millions of years.
Although we think of seawater as just containing sodium ions
(Na+) and chloride ions (Cl-), there are six other important
ions present: magnesium (Mg2+), calcium (Ca2+), potassium
(K+), hydrogen carbonate (HCO3-), bromide (Br-), and sulfate
(SO42-). Much of the world's supply of bromine and magnesium
is obtained from seawater.
What Happens When a Substance Dissolves?
We will discuss some of the important features that make water
such a good solvent for many ionic compounds. There is an unequal
sharing of bonding electron pairs in covalent bonds formed between
atoms of different electronegativity. In such bonds, the bonding
electrons are, on average, closer to the more electronegative atom.
This results in one atom having a slightly positive charge (δ+) and
the other atom having a slightly negative charge (δ-).
It is the polar nature of the water molecule that enables it to
dissolve many ionic compounds.
Let us consider what happens when a crystal of sodium chloride is
dissolved in water. In the solid, the sodium ions and chloride ions
are packed in negative charge (δ-). a regular arrangement.. When
the crystal is placed in the water, water molecules surround the ions
on the surface of the crystal. The slightly positive hydrogen atoms of
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DR. AHMED KHAMIS SALAMA
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some water molecules are attracted by the negative chloride ions
and the slightly negative oxygen atoms of other water molecules are
attracted by the positive sodium ions. These water molecules will
remove the ions into solution, exposing more ions to the "attack".
This process will continue until the entire crystal dissolves.
Sodium Chloride before dissolving (Crystal)
Sodium Chloride after dissolving (water molecule surround Ions of
NaCl)
Why, then, are some ionic compounds insoluble in water? It is the
strength of the ionic bonding in the crystal (compared to the
attraction of the water molecules for the ions) that determines
solubility. A compound with strong ionic bonds, whose ions are
comparatively weakly attracted by water molecules, will be
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
insoluble. Conversely, a compound with weak ionic bonds, whose
ions are strongly attracted by water molecules, will be very soluble.
Expression of Concentration
The concentration is a RATIO of one substance to another. Of all the
concentration units that are in general use, only four are likely to be
encountered in an introductory chemistry course: mass percent,
volume percent, mole fraction, and moles per liter (mol/L), which
called it molarity (M).
Mass Percent:
mass percent of solute= (mass solute/ total mass solution) x 100%
Thus we are really talking about the percentage by mass of the
solute in the solution.
Example:
In a solution prepared by dissolving 24 g of NaCl in 152 g of water,
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
Example:
A 20.0 g of Sodium Chloride NaCl is dissolved in sufficient quantity
of water, and the final mass of the solution is 200 gr. Find the mass
percent of NaCl.
Solution:
Mass of solute
 100 %
Mass of solute  Mass of solvent
Mass of (NaCl)
% NaCl 
 100 %
Mass of (NaCl)  Mass of Water
Mass of (NaCl)
% NaCl 
 100 %
Mass of Solution
20.0g
% NaCl 
 100 %  10%
200.0g
% NaCl 
Alloys consist of two or more metals that have been melted together
and then cooled back to the solid state (a solid solution). Then the
composition of alloys is usually quoted in mass percent.
Volume Percent
When two liquids are mixed to form a solution, it is often more
convenient to give the concentration in terms of volume percent
than mass percent since measuring the volume of a liquid is more
convenient than measuring its mass. We normally take the liquid
with the smaller volume as being the solute, and define volume
percent in the following way:
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
Example:
A 25.0 mL of alcohol is placed in a container and sufficient water is
added to bring the volume of the solution up to 225 mL. Find the
percent alcohol by volume.
Solution:
Volume of solute
 100 %
Volume of solute  Volume of solvent
Volume of Alcohol
%Alcohol 
 100 %
Volume of Alcohol  Volume of Water
Volume of Alcohol
% NaCl 
 100 %
Volume of Solution
25.0mL
% NaCl 
 100 %  11.11%
225.0mL
%Solute 
Parts per million (ppm) and parts per billion (ppb).
These terms are often used in the discussion of environmental
problems caused by the presence of minute quantities of toxic
chemicals. Parts per million and parts per billion are calculated by a
method similar to that used for mass percent
When the amount of solute is very small, as with trace impurities in
water, concentration is often expressed in parts per million (ppm):
ppm solute = (mass solute / total mass solution) x 106
Comparing the defining equations for mass percent and parts per
million, it should be clear that:
ppm = mass percent x 104
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DR. AHMED KHAMIS SALAMA
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Example:
In USA, drinking water cannot contain more than 5 x 10-4 mg of
mercury per gram of sample. In parts per million that would be :
Solution:
ppm Hg = (5 x 10-4 mg Hg / 1 x 103 mg) x 106 = 0.5
Mole Fraction
Mole fraction is the fraction of the total number of moles that is
accounted for by substance A.
Using XA to represent the mole
fraction of A
The mole fractions of all components of a solution (A, B, C,…) must
add to unity:
XA + XB + Xc + ….. = 1
The advantage of expressing a concentration as mole fraction is that
it provides more information concerning the actual ratios of particles
in the solution than the units previously discussed. The mole fraction
of the solute is the ratio of the amount of solute to the amount of
solution:
XA = nA / n total = moles A / total moles
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DR. AHMED KHAMIS SALAMA
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Example:
What are the mole fractions of CH3OH and H2O in a solution
prepared by dissolving 1.20 g of methyl alcohol in 16.8 g of water ?
Strategy: First convert grams to moles for both components. Then
calculate the mole fraction of CH3OH. Finally obtain the mole
fraction of H2O by subtraction from unity.
Solution:
n CH3OH = 1.20 g CH3OH / 32.04 g CH3OH = 0.0375 mol CH3OH
n H2O
= 16.8 g H2O / 18.02 g H2O = 0.932 mol H2O
X CH3OH = n CH3OH / (n CH3OH + n H2O)
= 0.0375 / (0.0375 + 0.932) = 0.0387
X H2O
= 1 - X CH3OH = 1 - 0.0387 = 0.9613
Molality:
Molality is the number of moles of solute per kilogram (1000 g) of
solvent.
Molality (m) = moles of solute / kilograms solvent
Example:
A solution used for intravenous feeding contains 4.8 g of glucose,
C6H12O6 , in 90.0 g of water. What is the molality of glucose ?
Strategy: First, calculate the number of moles of glucose (M =
180.16 g/mol), then the number of kilograms of water.
Solution:
n glucose = 4.8 g glucose / 180.16 g glucose = 0.0266 mol glucose
Kg water = 90 g water / 1000 = 0.090 kg water
Molality = 0.0266 mol glucose / 0.090 kg water = 0.296 m
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DR. AHMED KHAMIS SALAMA
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Molarity (Moles per Liter):
Molarity is the number of moles of solute per liter (1000 mL) of
solvent.
Molarity (M) = moles of solute / liters of solution
Example: For a solution containing 1.20 mol of substance A in 2.5 L
of solution,
Molarity (M) = 1.20 mol / 2.5 L = 0.480 mol / L = 0.480 M
A solution can be prepared to a specified molarity by weighing out
the calculated mass of solute and dissolving in enough solvent to
form the desired volume of solution.
Alternatively, you can start
with a more concentrated solution and dilute with water to give a
solution of the desired molarity.
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DR. AHMED KHAMIS SALAMA
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Keep a simple point in your mind:
Adding solvent cannot change the number of moles of solute , nA in
concentrated solution = nA in dilute solution.
Example: How would you prepare 1.00 L of 0.100 M CuSO4 starting
with 2.00 M CuSO4 ?
Or What volume of 2.00 M CuSO4 should be diluted with water to
give 1.00 L of 0.100 M CuSO4 ?
Solution:
Use the equation: [CuSO4]
concentrated
x Vconcentrated = [CuSO4]
diluted
x Vdiluted
Vconcentrated = 0.100 M x 1.00 L / 2.00 M = 0.0500 L = 50 mL.
Measure out 50.0 mL of 2.00 M CuSO4 and dilute with enough water
(about 950 mL) to form 1.00 L of 0.100 M solution.
Conversions between concentration units:
It is necessary to convert from one concentration unit to another.
When the original concentration is
Start with
Mass percent (%)
100 g solution
Molarity (M)
1.00 L solution
Molality (m)
1000 g solvent
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DR. AHMED KHAMIS SALAMA
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Mole fraction (X)
1 mol (solute + solvent)
Example:
Calculate the molarity of a concentrated solution of hydrochloric
acid that is 37.7% by mass HCl; the solution has a density of 1.19
g/mL
Solution:
Molarity of HCl = number of moles of HCl / solution volume, L
Molarity of HCl = mass of HCl, g / ( MW x solution volume, L )
HCl is 37.7 % by mass means, w/w, 37.7 gram HCl in 100 gram
solution
Weight of HCl = 37.7 g
MW = atomic weight of H + atomic weight of Cl
= 1.008 + 35.45 = 36.45
Solution volume of HCl = mass , g / density, (g/ml)
= 100 g / 1.19 = 84.03 mL
= 84.03 / 1000 = 0.084
Molarity = Wg / MW x VL
Molarity of HCl = 37.7 g / (36.45 x 0.084) = 12.3 M
This same approach can be used to convert between molarity and
molality.
Example:
A 1.13 M solution of KOH has a density of 1.05 g/mL. Calculate its
molality.
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
Solution:
Molecular weight of KOH = 39.10 + 16.00 + 1.008 = 56.11
Mass of KOH = 1.13 x 56.11 = 63.4 g
Density of sol. = 1.05 g/ml
Total mass of solution (mass of KOH + mass of 1L of water)=
1000 ml x 1.050 g/ml = 1050 g
Mass of 1L of water = 1050 - 63.4 = 986.6 g = 0.9866 kg
Molality = 63.4 / (56.11 x 0.9866) = 1.14 m
You will notice from this example that the molarity (1.13 M) and
molality (1.14 m) of the KOH solutions are very close to one another.
This is generally true for dilute water solutions; one liter of a dilute
aqueous solution contains approximately one kilogram of water.
For concentrated or non-aqueous solutions, molarity and molality
ordinarily differ considerably from each other.
Standard Solutions
Solutions which are prepared so that they have a precisely known
mass of solute in a precisely known volume of solution are called
standard solutions. These solutions are prepared in a special
container known as a volumetric flask.
Dilution
Occasionally you may find that your laboratory work calls for the use
of a dilute solution of a substance, but you discover that the only
solutions available are more concentrated than what you need. This
situation calls for a dilution to be carried out. In order to dilute a
concentrated solution you would require a volumetric flask and a
volumetric pipette. Pipettes are available in sizes from 0.50 mL to
100.0 mL. The 10.0 mL and 25.0 mL are the most commonly used
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
sizes. Again we have used the number of significant figures to
indicate the precision of each particular size of pipette.
Colloids
In a solution, one substance (the solute) is dispersed as separate
molecules or ions in another substance (the solvent). These
solutions are clear. For example, when sodium chloride (salt) is
dissolved in water no solid particles are visible. Both in everyday life
and in the laboratory we occasionally encounter a "solution" that is
not clear, but appears to be cloudy an excellent example of this is
milk. Even after standing for long periods, the matter responsible for
this cloudiness does not settle out; nor is it possible to remove it
from the solvent by the process of filtration. Such "solutions" are
referred to as colloidal solutions or colloids.
Principles of solubility:
The extent to which a solute dissolves in a particular solvent
depends upon several factors. The most important factors of these
are:
1. The nature of solvent and solute particles and the interactions
between them:
Like
dissolves
like.
It
means
that
two
substances
with
intermolecular forces of about the same type and magnitude are
likely to be very soluble in one another. For example, molecules of
the non-polar pentane and hexane are completely miscible with
each other and they held together by dispersion forces of about the
same magnitude. A pentane molecule experiences little or no change
in intermolecular forces when it goes into solution in hexane. Most
non-polar substances have very small water solubility. To dissolve
appreciable amounts of pentane in water, it would be necessary to
break the hydrogen bonds holding H2O molecules together. There is
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
no comparable attractive force between pentane and water to
supply the energy required to break into the water structure.
Of the relatively few organic compounds that dissolve readily in
water, most contain –OH groups such as methyl alcohol, ethyl
alcohol and ethylene glycol.
These compounds, as in water, the
principal intermolecular forces are hydrogen bonds. When these
compounds dissolve in water, it forms hydrogen bonds with H2O
molecules. The hydrogen bond as example, joining methyl alcohol
molecule to an water molecule are about as strong as those in the
pure substances. Not all organic compounds that contain –OH
groups are soluble in water. As molar mass increase, the polar –OH
group represents a smaller portion of the molecule while, the nonpolar hydrocarbon portion becomes larger. As a result, solubility
decreases with molar mass.
The solubility of gases in water is usually decreased by the addition
of other solutes, particularly electrolytes.
This phenomenon is
known as salting out. The salting out effect may be explained as
caused by the hydration of salt. A portion of the water combines
with the salt, and the water thus removed from the role of solvent is
no longer free to absorb gas.
2. The temperature at which the solution is formed.
If the solution process absorbs heat (endothermic process),
∆Hsolution > 0, an increase in temperature increases the solubility.
Conversely, if the solution process is exothermic, ∆Hsolution < 0, an
increase in temperature decreases the solubility.
As regard the
effect of temperature, we notice that most gases dissolve in water
with liberation of heat; hence according to Le Chatelier principle, the
solubility of gases decreases with rise in temperature.
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Dissolving a solid in a liquid is usually an endothermic process; heat
must be absorbed to break down the crystal lattice. So, the solubility
of solids usually increase as the temperature rises.
Cooling a saturated solution usually causes a solid to crystallize out
of solution, since the solubility is smaller at the lower temperature.
3. The pressure of a gaseous solute.
Pressure has a major effect on solubility only for gas-liquid systems.
According to Henry's law, The mass of gas dissolved by a given
volume of solvent at constant temperature, is proportional to the
pressure of the gas in equilibrium with the solution. Indeed, at low
to moderate pressures, gas solubility is directly proportional to
pressure.
The influence of partial pressure on gas solubility is used in making
carbonated beverages such as beer and many soft drinks.
These
beverages are bottled under pressures of CO2 as high as 4 atm.
When the bottle or can is opened, the pressure above the liquid
declines to 1 atm, and CO2 bubbles rapidly out of solution.
Colligative properties of solutions:
The properties of a solution differ considerably from those of the
pure solvent. Those solution properties that depend primarily upon
the concentration of solute particles rather than their nature are
called colligative properties. These properties depend only on the
number of molecules present their nature or magnitude playing no
part.
Thus, 1 mole of any gas at STP occupies a volume equal to
22.4 liters, gaseous volume being a colligative property. Other such
properties are vapor pressure lowering, osmotic pressure, boiling
point elevation and freezing point depression.
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The relation between colligative properties and solute
concentration:
(1)
In case of non electrolyte solutes:
Non electrolytes are
exist in solution as molecules.
Lowering of vapor pressure:
Whenever a substance is dissolved in a liquid, the vapour pressure
of the latter is lowered. The vapour pressure of dilute solution
follows Raoult's law according to which " the relative lowering of
the vapour pressure is equal to the mole fraction of the solute in
solution" .
Concentrated aqueous solutions evaporate more slowly than does
pure water. This reflects the fact that the vapor pressure of water
over the solution is less than that of pure water. Vapor pressure
lowering is a true colligative property; that is, it is independent of
the
nature
of
the
solute
but
directly
proportional
to
its
concentration. For example, the vapor pressure of water above
0.1 M solution of either glucose or sucrose at 0◦C is the same, about
0.008 mm Hg less than that of pure water. In 0.3 M solution, the
vapor pressure lowering is almost exactly three times as great,
0.025 mm Hg.
The vapor pressure of the solvent in solution is directly proportional
to the mole fraction of solvent if Raoult's law is obeyed.
Where:
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DR. AHMED KHAMIS SALAMA
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P1 is vapor pressure of the solvent in solution.
X1 is the mole fraction of solvent.
P1◦ is the vapor pressure of the pure solvent.
Note that since X1 in a solution must be less than 1, P1 must be less
than P1◦. This relationship is called Raoult's law
X1 = 1 – X2
where X2 is the mole fraction of solute.
Substituting 1 – X2 for X1 in Raoult's law
P1 = (1 – X2 ) P1◦
P1◦ - P1 = X2 P1◦
The quantity P1◦ - P1 is the vapor pressure lowering (∆P), it is the
difference between the solvent vapor pressure in the pure solvent
and in solution.
∆P = X2 P1◦
X2 is the mole fraction of solute
P1◦ is the vapor pressure of the pure solvent.
Water moves by evaporation and condensation from a region in
which its vapor pressure is high (pure water) to one in which its
vapor pressure is low (sugar solution).
Osmosis and osmotic pressure:
Osmosis is used to describe the spontaneous flow of solvent into a
solution; or from a more dilute to a more concentrated solution;
when the two liquids are separated from each other by a suitable
membrane.
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If a sugar solution is separated from water by a semi permeable
membrane (animal bladder, a slice of vegetable tissue or a piece of
parchment), the membrane allows water to pass through it, but not
sugar molecules. Here, as before, water moves from a region where
its vapor pressure is high (pure water) to a region where it is low
(sugar solution).
This process, taking place through a membrane
permeable only to the solvent, is called osmosis.
As a result of osmosis a pressure is developed which opposes the
tendency for the solvent to pass through the semi permeable
membrane into the solution.
This pressure is called the osmotic
pressure of the solution. It is defined as " the excess pressure which
must be applied to a solution to prevent the passage into it of
solvent when the two liquids are separated by a perfectly semi
permeable membrane".
If pressure (p) is less than osmotic pressure (π),
osmosis takes
place in the normal way, and water moves through the membrane
into the solution. By making the external pressure large enough, it
is possible to reverse this process.
When p > π, water molecules
move through the membrane from the solution to pure water. This
process, called reverse osmosis, is used to obtain fresh water from
seawater in arid regions of all the world, including Saudi Arabia.
Osmosis can be prevented by applying to the solution a pressure, p
that just balances the osmotic pressure, π.
Osmotic pressure, like vapor pressure lowering, is a colligative
property.
For any non electrolyte B, π is directly proportional to
molarity, [B].
The equation relating these two quantities is very similar to the ideal gas
law:
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DR. AHMED KHAMIS SALAMA
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Examples: When a cucumber is pickled, water moves out of the
cucumber by osmosis into the concentrated brine solution (the skin
of the cucumber acts as a semi permeable membrane). When a dried
prune is placed in water, the skin also acts as a semi permeable
membrane but this time the solution inside the prune is more
concentrated than the water, so that water flows into the prune,
making the prune less wrinkled.
Elevation of boiling point:
As defined the boiling point is that temperature at which the vapour
pressure of the liquid is equal to the external, usually atmospheric
pressure.
A direct consequence of the reduction of the vapour pressure by a
non volatile solute is that the boiling point of solution must be
higher than that for the pure solvent.
When a solution of a nonvolatile solute is heated, it does not begin
to boil until the temperature exceeds the boiling point of the
solvent.
The difference in temperature is called the boiling point
elevation, ∆Tb .
∆Tb = Tb - Tb°
where:
Tb and Tb° are the boiling points of the solution and the pure solvent,
respectively.
As boiling continues, pure solvent distils off, the concentration of
solute increases, and the boiling point continues to rise.
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DR. AHMED KHAMIS SALAMA
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Boiling point elevation is a direct result of vapor pressure lowering.
At any given temperature, a solution of a nonvolatile solute has a
vapor pressure lower than that of the pure solvent. Hence, a higher
temperature must be reached before the solution boils; that is ,
before its vapor pressure becomes equal to the external pressure.
Lowering of freezing point:
When a solution is cooled, it does not begin to freeze until a
temperature below the freezing point of the pure solvent is reached.
The freezing point lower in, ∆Tf , is defined to be a positive quantity:
∆Tf = Tf° - Tf
where:
Tf° , the freezing point of the solvent, lies above Tf , the freezing
point of the solution. As freezing takes place, pure solvent freezes
out, the concentration of solute increases, and the freezing point
continues to drop.
The freezing point lowering, like the boiling point elevation, is a
direct result of the lowering of the solvent vapor pressure by the
solute.
Notice that the freezing point of the solution is the
temperature at which the solvent in solution has the same vapor
pressure as the pure solid solvent.
This implies that it is pure
solvent (e.g., ice) that separates when the solution freezes.
(2)
In case of electrolyte solutes:
Electrolytes are exist in
solution as ions.
As noted earlier, colligative properties of solutions are directly
proportional to the concentration of solute particles. On this basis,
it is reasonable to suppose that, at a given concentration, an
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DR. AHMED KHAMIS SALAMA
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electrolyte should have a greater effect upon these properties than
does a nonelectrolyte.
When 1 mole of nonelectrolyte (sugar) dissolves in water, one mole
of solute molecules is obtained.
On the other hand, when 1 mole of electrolyte (NaCl) dissolves in
water, two mole of ions is obtained ( 1 mol of Na+ , 1 mol of Cl-).
When 1 mole of electrolyte (CaCl2) dissolves in water, three mole of
ions is obtained ( 1 mol of Ca2+ , 2 mol of Cl-).
The freezing points of electrolyte solutions, like their vapor
pressures, are lower than those of non electrolytes at the same
concentration.
Sodium chloride and calcium chloride are used to
lower the melting point of ice on highways, their aqueous solutions
can have freezing points as low as -21 and -55 °C, respectively.
Solutions of liquid in liquids:
On the basis of the mutual miscibility of components, solutions of
liquids in liquids can be classified into three classes:
1. Completely miscible liquids: Alcohol + water
2. Partially miscible liquids: phenol + water , chloroform + water
3. Practically (completely) immiscible liquids: benzene + water ,
nitrobenzene + water
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5. Acids and Bases
Arrhenius Model of Acids and Bases
The classical, or Arrhenius, model defined an acid as any substance
that liberates or yields hydrogen ions (H+) or protons in water. An
example would be hydrogen chloride, HCl, gas, which when put in
water ionizes to yield hydrogen ions, H+, and chloride ions. The
resulting water solution of ionized H+ and Cl- is known as
hydrochloric acid.
This process involving the breakdown of a substance into ions is
known as ionization.
An Arrhenius base is a substance that dissociates in water to
produce hydroxide ions, OH-. Two examples of strong, or almost
completely dissociated bases are potassium hydroxide, KOH, and
sodium hydroxide, NaOH or lye.
Brønsted-Lowry Acid-Base Model
The Arrhenius theory applies only when water is used as the solvent.
It restricts the term acid to substances yielding hydronium ions and
the term base to those yielding hydroxide ions.
Brønsted and Lowry independently proposed a much broader and
more useful concept of acids and bases. According to their model, a
Brønsted-Lowry acid is any substance capable of donating a
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
hydrogen ion or proton to another substance, and a Brønsted-
Lowry base is any substance capable of accepting a proton or
hydrogen from another substance. In other words, acids are proton
donors, and bases are proton acceptors.
According to this concept, any reaction involving the transfer of a
proton or H+ from one substance to another is an acid-base
reaction. Therefore, base is a proton acceptor and an acid is a
proton donor.
For example:
Notice that water can act as either an acid or as a base. For this
reason it is called amphoteric.
Some examples of acids and bases
Acid
HCl
=
H+
CH3COOH
=
H+
+ CH3COO_
NH4+
=
H+
+
NH3
H2CO3
=
H+
+
HCO3-
HCO3-
=
H+
+
CO3--
H2O
=
H+
+
OH-
H3O+
=
H+
+
H2O
69
Conjugate base
+
Cl-
DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
Hydronium ions
Consider HCl, a gas composed of polar covalent molecules. When
HCl gas is passed through water we achieve the classic substance,
hydrochloric acid, HCl(aq). The original gas does not have any of the
properties of the resulting solution. It is reasonable to assume that
molecules of HCl react with the water to produce ions. It is these
ions that ultimately give the water and HCl solution it's acidic
properties.
The reaction above consists of a breaking away of a proton, H+, from
the HCl molecules. A stable co-ordinate bond is formed when a
proton, H+, shares a pair of electrons with an oxygen atom of the
highly polar water molecule. A hydrated proton, called the
hydronium ion, H3O+, is formed.
Concentrated vs. Dilute; Strong vs. Weak:
These terms are often the most misused in chemistry. Concentrated
and dilute refer to the concentration of an acidic or basic substance
in a solvent. eg. 16 M HCl is more concentrated than a 0.5 M
solution of the same acid.
Strong and weak refer to the ability of an acid or base to dissociate.
A strong acid will dissociate completely in water to form hydronium
ions. i.e. 100% of it will form H3O+. A weak acid or base will only
dissociate to a certain percentage. Often a very small percentage
only.
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Strong and Weak Bases
NH3(aq) is a poor conductor of electricity when compared with
NaOH(aq). This means that the degree of dissociation of NH3 in
water is relatively small when compared with that of the NaOH.
A base which is only slightly dissociated in aqueous solution is called
a weak base; one which is highly dissociated is called a strong base.
All the strong bases happen to be inorganic, that is, the NaOH, KOH,
RbOH group. Even Ca(OH)2 and Ba(OH)2 are considered to be strong
bases. All of the rest are too insoluble to provide a significant [OH-]
in water. The double arrow convention should be used when dealing
with a weak base. A single arrow is to be used when showing the
dissociation of a strong base since for all practical purposes,
dissociation is 100% complete. i.e., a water solution made from
NaOH(s) will have no molecules of NaOH in it. The NaOH will be
completely ionized into Na+ and OH-.
The pH Scale
Every aqueous solution is either acidic, basic or neutral. There is a
quantitative relationship between the concentration of hydronium
and hydroxide ions in the solution.
The pH scale is a numerical scale which, for most applications
extends from 0 through to 14. The numbers on the scale represent
the relative acidity of solutions and can be converted into actual
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
hydronium ion concentrations. The brackets as usual denote molar
concentrations.
The pH scale is based on the self-ionization of pure water. Two
water molecules will sometimes combine into hydronium and
hydroxide ions.
Pure water is considered to neutral and the hydronium ion
concentration is 1.0 x 10-7 mol/L which is equal to the hydroxide ion
concentration.
The equilibrium law for this reaction at 25oC should be:
You will please note that at neutrality the molarity of the hydronium
ion is 10-7. The 7 plays a part in the pH scale by indicating neutrality.
The scale reaches a maximum at 14. Please note again that the
hydronium and hydroxide concentrations multiply out to 10-14 M.
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The pH scale was derived around this relationship:
So the pH is the -log of the [hydronium ion].
pH
[H3O+]
[OH-]
pOH
1
10-1
10-13
13
2
10-2
10-12
12
3
10-3
10-11
11
4
10-4
10-10
10
5
10-5
10-9
9
6
10-6
10-8
8
7
10-7
10-7
7
8
10-8
10-6
6
9
10-9
10-5
5
10
10-10
10-4
4
11
10-11
10-3
3
12
10-12
10-2
2
13
10-13
10-1
1
The pH of a solution may be determined by the use of an electronic
instrument known as a pH meter, or through the use of chemical
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DR. AHMED KHAMIS SALAMA
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indicators. Acid-base indicators are dyes which undergo slight
changes in molecular structure and color when the pH value of the
solution changes.
Specific colors correspond to specific pH values. Some examples are:
litmus, phenolphthalein, bromothymol blue, etc. There is a list of
acid-base indicators in the databook.
Sample Problems
What is the pH of an HCl solution which has a [H3O+] = 1.0 x 10-3?
pH = -log[H3O+] = -log[1.0 x 10-3] = -(-3) = 3
What is the pH of an acetic acid solution whose [H3O+]=2.5 x 10-4?
What is the hydronium concentration of nitric acid if the pH=4.0?
[H3O+] = 10-pH = 10-(4) = 1.0 x 10-4 mol/L
What is the [H3O+] of HCl if the pH = 2.57?
What is the pH of 0.010 mol/L hydrochloric acid?
The pOH Scale
The pOH scale is the corollary of the pH scale
ie. pH + pOH = 14 You'll remember from math class that when you
multiply two numbers you only add their logs.
[H3O+]*[OH-] = (1.0 x 10-7 )*(1.0 x 10-7) = 1.0 x 10-14
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or
-log[H3O+] + -log[OH-] = 7 + 7 = 14
Thus a solution that has a pH = 7 must also have a pOH = 7.
Problems
What is the pOH of a 0.010 mol/L NaOH solution?
What is it's pH?
2- What are the hydronium ion and hydroxide ion concentration of a
solution prepared by adding 1 mL of 1.0 mol/L HCl to 9 mL of water?
Assume that volumes are additive and that the 1 mol/L HCl
dissociates completely.
3- What are the hydronium ion and hydroxide ion concentrations of
a solution made by adding 1 mL of 0.1 mol/L NaOH to 9 mL of
water?
4- Find the pH, pOH, [OH-] of a 0.00010 mol/L HCl solution.
5- Find the pH of a 0.00325 mol/L NaOH solution.
6- What is the hydronium ion concentration of a solution that has a
pH = 2.6?
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Dissociation of water:
From conductivity measurements, water has been shown to be very
weakly ionized and at 25°C the concentration of hydrogen ions is
only 10-7 gram equivalents per liter.
The equilibrium constant for the dissociation of water is given by:
or
Now the concentration of water to all intents and purposes is
constant, so we can write:
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DR. AHMED KHAMIS SALAMA
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The ionic product of water at 25°C is, therefore,
At neutrality the pH of pure water at 25°C is 7
At other temperatures, the pH at neutrality is not 7 since K w varies
with temperature. Even a small change in temp from 37 to 40°C
causes an 8% increase in hydrogen and hydroxyl ions so that a
slight rise or fall in temp may produce a profound biological change
in a living system sensitive to hydrogen ion concentration.
Temperature [°C]
pH of neutrality
0
7.97
25
7.00
37
6.80
40
6.77
75
6.39
100
6.16
Measurement of pH:
The most convenient and reliable method for measuring pH is by the
use of a pH meter. This instrument measures the EMF of a
concentration cell using a reference electrode (Calomel reference
electrode) and a glass electrode reverseble to hydrogen ions.
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DR. AHMED KHAMIS SALAMA
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The usual type of cell employed is shown below:
Glass electrode is very sensitive to pH change. The glass electrode
rapidly responds to hydrogen ion concentration and can be used in a
wide variety of media.
The electrode must be always be thoroughly washed after use and
stored in distilled water.
Standard pH solutions:
The pH meter is calibrated before use by means of a standard
solution. In the United Kingdom, potassium hydrogen phthalate is
the recommended standard; at 15°C a 0.05 M solution has a pH of
4.000. The pH at other temperatures (0 – 60°C) can be obtained
from the equation:
pH = 4.000 + ½ ( t-15 / 100)2
The meter should be calibrated with a solution whose pH is close to
that under test and several convenient standards are given below.
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DR. AHMED KHAMIS SALAMA
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Primary standards for the calibration of a pH meter
Dissociation of acids and bases
Strong acids: They are completely dissociated to hydrogen ions and
the conjugate base occurs, so that the hydrogen ion concentration is
the same as that of the acid. The pH can therefore, be very easily
calculated:
In a solution of 0.01 N HCl, H concentration = HCl concentration = 0.01 N
= 10-2 N
pH = -log [10-2] = 2
Strong basis:
They are also completely dissociated .
In a solution of 0.01 N NaOH, OH concentration =
0.01 N = 10-2 N
pOH = -log [OH-]
pOH = -log [10-2] = 2
pH = pKw – pOH
pH = 14 – pOH = 14- 2 = 12
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NaOH concentration =
DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
Weak acids: Weak acids are only slightly ionized in solution and a
true ewquilibrium is established between the acid and the conjugate
base.
If HA represents a weak acid,
HA = H+ + A-
According to the law of mass action, Ka the acid dissociation
constant is defined as:
Taking negative logarithms,
This formula is known as the Henderson-Hasselbalch equation and is
valid over the pH range 4-10 where the hydrogen and hydroxyl ions
do not contribute significantly to the total ionic concentration. In
addition, the ratio of conjugate base to acid should be less than 0.1.
pKa is the negative logarithm of the acid dissociation constant of a
waek acid
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DR. AHMED KHAMIS SALAMA
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Another way of defining pKa is the pH at which the concentrations
of the acid and its conjugate base are equal.
Buffer Solutions
A buffer solution is one that resists pH change on the addition of
acid or alkali. Buffer consisted of weak acid + its salt (acetic acid +
sodium acetate) or weak base + its salt (ammonium hydroxide +
ammonium chlorid). Such solutions are used in many biochemical
experiments where the pH needs to be accurately controlled.
From the Henderson-Hasselbalch equation, the pH of a buffer
solution depends on two factors; one is the pKa value and the other
the ratio of salt to acid.
Let us take as an example acetate buffers consisting of a mixture of
acetic acid and sodium acetate:
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DR. AHMED KHAMIS SALAMA
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Since acetic acid is only weakly dissociated, the concentration of
acetic acid is almost the same as the amount put in the mixture;
likewise the concentration of acetate ion can be considered to be the
same as the concentration of sodium acetate placed in the mixture
since the salt is completely dissociated.
Example:
What is the pH of a mixture of 5 ml of 0.1 M sodium acetate and 4 ml
of 0.1 M acetic acid. (pKa CH3COOH = 4.76)
Solution
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Example:
How is the pH changed on adding 1 ml of 0.1 N HCl to the above
mixture.
Solution
Addition of HCl provides H+ which combines with the acetate ion to
give acetic acid. This reduces the amount of acetate ion present and
increases the quantity of undissociated acetic acid, leading to an
alteration in the salt/acid ratio and hence to a change in pH.
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DR. AHMED KHAMIS SALAMA
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Concentration of CH3COO- = (5/10)
x 0.1 M
- (1/10) x 0.1 M
= 0.04 M
Concentration of CH3COOH = (4/10) x 0.1 M + (1/10) x 0.1 M =
0.05 M
pH = pKa + log10 [salt] / [acid]
pH = 4.76 + log10 [0.04] / [0.05]
pH = 4.76 + (- 0.097)
pH = 4.66
The pH of the solution has been reduced from 4.86 to 4.66, a change
of only 0.2 of a unit, whereas if the HCl had been added to distilled
water, the pH would be 2. The solution has, therefore, acted as a
buffer by resisting pH change on the addition of acid.
Titration curves
When a strong base is mixed with a solution of acid and the pH
recorded, a plot of the base added against pH recorded can be
obtained and this is known as a titration curve.
Strong acid and a strong base:
There is little change in pH value on adding base until complete
neutralization when only a slight excess of base causes a large
increase in pH.
In effect, the strong acid is acting as a buffer
solution in resisting change in pH.
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Example: Suppose 10 ml o.1 N HCl is titrated with 0.1N NaOH
1. Initial pH value:
[H+] = 0.1 = 1 x 10-1 g mole/L
pH = 1
2. After the addition of 5 ml of 0.1N NaOH:
On adding 5 ml of strong base, 5 ml of the HCl solution is neutralized
leaving 5 ml of 0.1N HCl in a total volume of 15 ml.
Normality of HCl = (5/15) x 0.1 = 3.33 x 10-2
pH = - log10 (3.33 x 10-2)
pH = - [log10 3.33 + (-2)]
pH = - ( 0.523-2) = 1.48
3. After the addition of 9.9 ml of 0.1N NaOH:
On adding 9.9 ml of strong base, 9.9 ml of the HCl solution
is
neutralized leaving 0.1 ml of 0.1 N HCl in a total volume of 19.9 ml.
Normality of HCl = (0.1/19.9) x 0.1 = 5.03 x 10-4
pH = - log10 (5.03 x 10-4)
pH = - [log10 5.03 + (-4)]
pH = - (0.702 -4) = 3.30
4. After the addition of 10.1 ml of 0.1 N NaOH:
On adding 10.1 ml of strong base, all the HCl is neutralized leaving
0.1 ml of 0.1 N NaOH in a total volume of 20.1 ml.
Normality of NaOH = (0.1 / 20.1) = 4.98 x 10-4
[OH-] = 4.98 x 10-4
pOH = 3.3
pH = 14-3.3 = 10.70
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
Determination of pKa
pKa values can be obtained from titration data by the following
methods:
1. By definition, the pKa value is equal to the pH at which the
acid is half titrated. The pKa can, therefore, be obtained from
a knowledge of the end point of the titration.
2. The ratio of salt / acid can be calculated from the
experimental data and a graph prepared of log10 salt / acid
against pH. The intercept on the axis is the pKa value.
Lewis Acids and Bases
Lewis had suggested in 1916 that two atoms are held together in a
chemical bond by sharing a pair of electrons. When each atom
contributed one electron to the bond it was called a covalent bond.
When both electrons come from one of the atoms it was called a
dative covalent bond or coordinate bond. The distinction is not clearcut as the diagram at the right shows; although the ammonia
molecule donates a pair of electrons to the hydrogen ion, the
identity of the electrons is lost in the ammonium ion that is formed.
Nevertheless, Lewis suggested that an electron-pair donor be
classified as a base and an electron-pair acceptor be classified as
acid.
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
MO diagram depicting the formation of a dative covalent bond
between two atoms
The modern definition of a Lewis acid is an atomic or molecular
species that has an empty atomic or molecular orbital of low energy
(LUMO) that can accommodate a pair of electrons, as illustrated in
the molecular orbital diagram at the right.
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GENERAL CHEMISTRY
6. CHEMICAL EQUILIBRIUM
Reversible reactions are those which can proceed either in the
forward or in the backward direction according to the relative
amounts of the substances present.
The two oppositely directed arrows refer to a reversible reaction.
If we mix 1 mole of H2 (2.0 gram) and 1 mole of I2 (254 gram) we
should expect to obtain 2 moles of HI (256 gram). This will be true
if the reaction was irreversible. Actually we obtain about 76% only
of this quantity. This does not at all indicate that the reaction stops
at this limit but that only at this stage the rate of combination of
hydrogen and iodine is equal to the rate of decomposition of
hydrogen iodide and the system is said to be in equilibrium.
Chemical equilibrium is reached when the rate of the forward
reaction equals the rate of the reverse reaction. This does not mean
that the quantities of reactants and products are equal.
LAW OF MASS ACTION:
The influence on concentration of the rate of a chemical reaction is
expressed by the law of mass action. According to this law, the rate
of a chemical reaction is directly proportional to the product of
active masses of the reacting materials. By active mass is meant the
molar concentration of the substance; i.e., the number of moles of
substance per volume (liter) and is designated by two square
brackets.
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In general, a reversible reaction may be written as:
where v1 and v2 are the rates (or speed) of the forward and
backward reaction, respectively.
If A and B represents the
concentration of the reactants in moles per liter, then according to
the law of mass action.
where k1: is the specific rate constant for the forward reaction
where k2 : is the specific rate constant for the backward reaction.
At equilibrium, there is no further apparent change and the rate of
the forward reaction becomes equal to that of the backward one,
hence,
and
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GENERAL CHEMISTRY
Where [A], [B], [G],
[H] are the equilibrium concentrations. The
constant Kc is the equilibrium constant with respect to molar
concentration.
At equilibrium we can determine the equilibrium constant, Kc.
For the general reaction:
The equilibrium constant can be expressed as:
in case of ideal diluted solutions.
Uppercase letters represent the various compounds in any reaction
and the lowercase letters represent the coefficients for each
compound.
For example:
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To determine Kc, substitute in the concentrations of each compound
at equilibrium.
For example, using
[N2] = 3.0 x 10-2 M
[H2] = 3.7 x 10-2 M
[NH3] = 1.6 x 10-2 M
Calculte the equilibrium constant for the reaction?
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You can also calculate the concentration of one compound if you
know the concentrations of the others and Kc
To determine if a reaction is at equilibrium, you use a term called Qc.
It is calculated in the same way as the kc only using the current
concentrations of the compounds. At equilibrium the Qc equals the
Kc. When Qc is larger than Kc the reaction has too many products
and the reaction will move to the left to reach equilibrium.
When
Qc is smaller than Kc the reaction has too many reactants and will
move to the right to reach equilibrium.
For the reaction
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Is the reaction at equilibrium, if not in which direction will the
reaction proceed?
Qc is larger than Kc, therefore the reaction will move to the left
(making more reactants).
Relationship between Kc and Kp
Consider the reversible reaction
for which the equilibrium constant Kc is expressed as:
The molar concentration may be expressed as being equal to n/v ,
M = n/V
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By applying the ideal gas equation
For each gas it follows that:
Substituting these values of active mass or molar concentration into
the expression
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Or
Where: Δn = number of resulting molecules – number of reacting
molecules
Kp = equilibrium constant in case of ideal gases
Kc = equilibrium constant in case of ideal solutions
R = general gas constant (0.082 atm.L. mol-1 .deg-1)
T = Absolute temperature, K.
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Three cases may be considered:
i) Δn is 0, e.g.,
H2(g) + I2 (g) = 2 HI(g)
Δn = 2 -1 -1 = 0
Hence,
Kp = Kc
ii) Δn is positive, e.g.,
PCl5(g) = PCl3 (g) + Cl2(g)
Δn = 1 +1 -1 = +1
Hence,
Kp >
Kc
iii) Δn is negative, e.g.,
N2(g) + 3H2 (g) = 2 NH3(g)
Δn = 2 -1 -3 = -2
Hence,
Kp < Kc
Problem:
For the reaction
N2 + 3H2
found to be 0.500
= 2 NH3
at 400°C
Kc was
Calculate Kp value
Solution:
Δn = number of resulting molecules – number of reacting molecules
Δn = 2 – (1+3) = -2
T = 400°C + 273 = 673 k
R = 0.082 atm.L. mol-1 .deg-1
Since
Kp = Kc (RT)Δn
Kp = 0.500 (0.082 x 673)-2 = 1.64 x 10-4
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Determination of equilibrium constant:
Many chemical and physical measurements can be used in the
determination of equilibrium constants. The method used depends
on the nature of the reversible reaction under investigation.
The following are some representative examples.
1) Equilibrium constant for the homogeneous liquid reaction of
acetic acid and ethyl alcohol,
CH3COOH + C2H5OH
= CH3COOC2H5
+ H2O
To determine Kc for this reaction, different quantities from the acid
and alcohol are placed in sealed glass tubes then left in a thermostat
at a certain fixed temperature. After reasonable sufficient time, the
remaining amount of acetic acid is quantitatively determined by
titration with sodium hydroxide solution of appropriate known
concentration. From the acid and alcohol it is possible to calculate
the concentration of the reaction constituents at equilibrium, hence
Kc can be calculated.
2) Equilibrium constant for the gaseous reaction
N2(g) + 3H2(g)
=
2NH3(g)
The equilibrium constant for this reaction can be determined by
passing a mixture of N2 gases and hydrogen at a 1:3 ratio in an iron
coil immersed in a thermostat at a certain temperature. In order to
accelerate the reaction, the inside of the coil may be covered with a
catalyst.
The outlet gases from the coil are then rapidly cooled,
collected and then analysed quantitatively for nitrogen, hydrogen
and ammonia, hence the value of Kp can be determined.
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Le Chatelier Principle:
This principle describes the effect of temperature, volume, pressure,
concentration, etc. on the position of equilibrium in a system. It
stated that the equilibrium is shifted in the direction that opposes
such effect.
Accordingly, increase of temperature will favor
endothermic reactions, increase of pressure will shift the equilibrium
in the direction accompanied with a fewer number of molecules.
As an example for chemical equilibrium,
N2 + O2
= 2 NO
- 43,200 Cal
The combination of N2 and O2 to form NO is an endothermic process.
According to Le Chatelier principle, increase of temperature will
favour the formation of NO. The reaction on the other hand, is not
accompanied with a change in volume, pressure, temperature,
therefore, has no effect on the position of equilibrium. The excess
amounts of reactants added consumed in the formation of more
nitric oxide to maintain the position of equilibrium.
As an example of a physical equilibrium,
Ice = liquid water -heat
By raising the temperature the equilibrium will shift in the direction
of ice to melt. By increasing the pressure the equilibrium will shift
in the direction accompanied with a decrease in volume (ice
converts to liquid water).
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Application of the law of mass action
A) Homogeneous Gaseous reaction
1. Reactions occurring without change in the number of molecules:
Consider the formation of hydrogen iodide from hydrogen and iodine
Δn = number of resulting molecules – number of reacting molecules
Δn for this reaction = zero
Therefore,
Kp = Kc
Suppose we start with (a) moles of H2 and (b) moles of I2. When
equilibrium is established, let (x) mole be the quantity reacted of
hydrogen and iodine to form (2x) moles of HI. The remaining
quantity of H2 is then (a-x) mole, and of iodine (b-x) mole.
If V is the volume of the system, it follows that:
By applying the law of mass action :
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Or
It will be noted that the volume V of the system does not appear in
the last equation.
Therefore, the volume and consequently the
pressure should not alter the position of equilibrium in such system.
The above conclusion may be arrived at by considering Kp.
According to Dalton's law of partial pressure,
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GENERAL CHEMISTRY
In the above example
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By applying the law of mass action :
It follows that
Which is the same value for Kc previously derived. So, the pressure
has no effect on the position of equilibrium.
Problem:
A mixture of 7.9 cc H2 and 33.1 cc I2 vapors was heated at 444 °C
until equilibrium was reached. If the equilibrium constant for the
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reaction is 36.68, Calculate the number of cc of H2 present when
equilibrium is attained.
Solution:
H2 + I2
= 2 HI
Let X cc of H2 and I2 be changed to HI
Then the volume of H2 remaining at equilibrium = (a-x) = (7.9 – X)
cc
The volume of I2 remaining at equilibrium = ( b-x) = (33.1 – X) cc
The volume of HI formed at equilibrium = (2 X) cc
For the reaction
4x2
Kp = ----------(a-x) (b-x)
4x2
36.68 = --------------(7.9-x) (33.1-x)
4x2 = 36.68 (7.9-x)(33.1-x)
X = 7.8 or 38.2
38.2 can not be the solution of the problem, therefore, x = 7.8
Hence, the number of cc of H2 present at equilibrium = (7.9 -7.8) =
0.1 cc
2. Reactions occurring with a change in the number of molecules:
Consider the dissociation of nitrogen tetra oxide to nitrogen dioxide
If x represents the degree of dissociation (i.e., fraction of mole that
dissociates at equilibrium).
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1-x will represents the fraction of undissociated N2O4 and 2x the
number of moles of NO2 formed. If the total volume is V and the
total pressure is P, therefore
By applying the law of mass action for the equation
N2O4(g) = 2 NO2(g)
Total number of moles present at equilibrium = 1-x + 2x = 1+x
The partial pressures will be
PNO2
= P. 2x / (1+x)
PN2O4
= P. (1-x) / (1+x)
Substituting in the expression
(PNO2)2 / (PN2O4)
= Kp
It follows that
(P. 2x / 1+x)2 / (P. 1-x / 1+x)
= Kp
P2. 4x2 . [1/ (1+x)2 ]/ (P. 1-x / 1+x)
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GENERAL CHEMISTRY
4x2 P . [1/ (1+x)2 ]. [1/1-x] . [1+x]
= Kp
The last
p
equation shows that the pressure
should affect the
position of equilibrium but not the equilibrium constant Kp. Thus if
P increases, x should decrease so that
Kp remains constant at
constant temperature.
B) Heterogeneous reaction
1) Equilibrium involving gases and solids
When the reaction involves one or more solids in equilibrium with a
gas or a liquid, the reaction is a heterogeneous one.
As an example consider the dissociation of calcium carbonate
CaCO3 = CaO(s) + CO2 (g)
By applying the law of mass action, it follows that
Kp = PCaO . PCO2 / PCaCO3
But since PCaO and PCaCO3 are constant at any one temperature
constant Kp is used where
Kp = PCO2
So at any particular temperature the equilibrium constant is
determined solely by the equilibrium pressure of carbon dioxide.
This equilibrium pressure of carbon dioxide is generally calculated x
from the dissociation pressure of calcium carbonate at that
particular temperature.
As in the case of homogeneous equilibrium, the influence of
pressure on heterogeneous equilibrium can be predicted by means
of Le Chatelier principle.
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Ionic equilibrium
and applications of law of mass action
Electrolytic solutions are divided into two main groups:
a) Strong electrolytes:
This group includes strong acid like (HNO3,
HCl, H2SO4).
Strong
alkalies like (KOH, NaOH) and salt solutions. By the word (strong)
we mean that the solutions of these substances are completely
ionized.
Thus if we have a 0.1N solution of HCl the whole
concentration of the acid will be present as H+ and
Cl- each of
which will be 0.1 g ion. Such state is represented as:
b) Weak electrolytes:
Which includes weak acids like acetic acid
CH3COOH, oxalic acid
H2C2O4, weak alkalies like NH4OH. By the word (weak) we mean
that such solutions is not completely ionized. For example, if we
have a 0.1N solution of acetic acid, then only part of this
concentration will ionize to give H+ and CH3COO- while the rest will
remain in the undissociated from CH3COOH. A state of equilibrium
will then be established between the formed ions and the
undissociated acid as follows:
In this case it is clear that the concentration of each of CH3COOand H+ will not be equal 0.1
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Ostwald law of dilution:
Consider 1 g equivalent of a weak acid HA dissolved in V liters of
solution. If α is the degree of dissociation of the acid. We have at
equilibrium.
If in general the acid concentration is C g. mole. Then we have
By applying the law of mass action we obtain
Where ka is the equilibrium constant, which in this case is known as
the and [A-] we obtain
The above relation is known as Ostwald law of dilution.
It indicates that if C increases, α should decrease.
Since Ka is a
constant. For weak electrolytes the degree of dissociation is usually
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very small compared to unity and the Ostwald law may be simpled
as
A relation similar to that given above can be obtained by applying
the law of mass action to the ionization of a weak base BOH, i.e.
Where kb is the ionization constant for the weak base BOH.
Neglecting α with respect to unity.
The following table shows the ionization constant for some weak
acids (ka) and weak bases (kb).
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Acid
Ka
Base
Kb
HCOOH
1.8 x 10 – 4
NH4OH
1.7 x 10 - 5
CH3COOH
1.8 x 10 – 5
CH3NH2
5.0 x 10 - 4
C6H5.COOH
6.3 x 10 – 5
C2H5NH2
4.1 x 10 - 10
Example:
Calculate the degree of ionization and hydrogen ion concentration
for 0.005 M solution of acetic acid knowing that its ionization
constant is 1.8 x 10-5
Solution:
Ka = α2 C
α2 = Ka / C
α2 = 1.8 x 10 -5 / 0.005
α = 6 x 10-2
Since
[H +] = C α = 6 x 10-2 x 0.005 = 3.0 x 10-4
Example:
Calculate the percent ionization of a 1.00 M solution of HCN acid,
knowing that ka is 4.8 x 10-10
Solution:
Ka = α2 C
α2 = Ka / C
α2 = 4.8 x 10 -10 / 1.00
α = 2.2 x 10-5
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Ionization of polybasic acids
When a polybasic acid is dissolved in water, the various hydrogen
atoms undergo ionization to different extents. For a dibasic acid
H2A, the primary and secondary ionization can be represented as
follows:
If the dibasic acid is a weak electrolyte the law of mass action may
be applied and therefore:
[H+] [ HA- ] / [H2A] = k1
[H+] [A- -] / [HA-] = k2
K1 and k2 are known as the primary and secondary dissociation
constants respectively. Each stage of the dissociation process has
its ionization constant, and the magnitude of these constants give a
measure of the extent to which each ionization has proceeded at
any given concentration.
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A tribasic acid (e.g. orthophosphoric acid) will similarly give three
dissociation constants K1, K2, and K3 according to:
Common ion effect
Consider a weak electrolyte such as acetic acid.
In solution the
following equilibrium exists:
Consider the addition of a strong electrolyte having ion in common
to this solution, e.g. sodium acetate.
The new equilibrium will be
The acetate ion is now called the common ion. The result is that the
equilibrium of acetic acid is shifted to the left, i.e. the degree of
ionization is decreased.
Another examples:
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Rule:
GENERAL CHEMISTRY
Addition of a strong electrolyte
to a weak electrolyte
containing a common ion, results in decrease of concentration of the
common and uncommon ion for the latter.
Electrolyte: The common ion effect provides a valuable method for
controlling the concentration of the ions furnished by a weak
electrolyte.
Solubility Product Constant
In general, when ionic compounds dissolve in water, they go into
solution as ions. When the solution becomes saturated with ions,
that is, unable to hold any more, the excess solid settles to the
bottom of the container and an equilibrium is established between
the undissolved solid and the dissolved ions.
Consider a sparingly (slightly) soluble electrolyte e.g. silver chloride.
When it is shaken up with water, some of it (an exceedingly small
quantity) passes in solution to form a saturated solution of the salt.
The following equilibrium exists between the insoluble AgCl and ions
in solution
Applying law of mass action (or Ostwald law of dilution) to such an
equilibrium then
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Since the silver chloride is present in the solid state its active mass
may be taken as unity. Therefore, the above equation reduces to:
This new constant S is known as the "solubility product" of silver
chloride and can be written as Ksp .
Like all equilibrium constants, the Ksp is temperature dependant, but
at a given temperature it remains relatively constant.
It is also like any equilibrium expression, each ion concentration in
the expression is raised to the power of its coefficient in the
solubility equation.
In general, the solubility product (Ksp), is the equilibrium constant
for the solubility equilibrium of a slightly soluble ionic compounds.
Slightly soluble substances such as AgCl, PbSO4, etc. posses small
solubility product and can therefore be easily precipitated in
solution. On the other hand, substances possessing higher solubility
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have high solubility products and cannot be precipitated easily from
solution.
In general any substances cannot be precipitated except if the
product of its ions in solution exceeds the solubility product of its
sparingly soluble compound.
For any electrolyte having the general formula AxBy in contact with
its saturated solution, the equilibrium between the solid and ions in
solution is
and the solubility products is given by
The solubility product of a sparingly soluble substance may be
derived from a knowledge of the solubility of the substance in pure
water. For example if the solubility of silver chloride in water is "S"
gram mole per liter, then the concentration of both silver and
chloride ions is "s" gram ion per liter, therefore,
Solubility product = [Ag+] [Cl-] = s X s = solubility (s)2
Or
solubility (s) = solubility product (S)½
In case of a salt giving more than two ions in solution, e.g. Ag2CrO4
S = [Ag+]2 [CrO4--] = (2s)2 (s) = 4s3
Solubility (s) = (s/4)1/3
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Example:
The solubility of silver chloride is 0.0015 gram per liter, calculate the
solubility product knowing that the MW of AgCl is 143.5
Solution:
Solubility in gram mole/ Liter = 0.0015 / 143.5 = 1.05 x 10-5.
Therefore in a liter of solution
the concentration of silver ions = 1.05 x 10-5 gram ions
and since
S = [ Ag+] [ Cl-]
S = (1.05 x 10-5 ) (1.05 x 10-5) = 1.1 x 10-10
Problem:
What is the molar solubility of barium fluoride in a solution that is
0.15 M NaF at 25 °C
Solution:
Since the solution is already 0.15 M in F-1 ions, we must make an
addition to our equilibrium concentrations.
BaF2(s)
Ba++(aq)
"x"
+
2F-(aq)
"2x +0.15"
(at equilibrium)
Ksp = [Ba++] [ F-]2
Because BaF2 is only slightly soluble, you might expect "2x" to be
negligible compared to 0.15. In that case
(2x +0.15 ) ≈ (0.15) and substituting into the Ksp expression, we
get
1.0 x 10-6 = (x)(0.15)2
solving for x, we get
x= 4.44 x 10-5 M
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7. THERMOCHEMISTRY
Introduction:
This chapter deals with energy and heat, two terms used widely by
both the general public and Scientists. Scientifically, these terms
have quite different meanings.
capacity to do work.
Energy can be defined as the
Heat is a particular form of energy that is
transferred from a body at a high temperature to one at a lower
temperature when they are brought into contact with each other.
Two centuries ago, heat was believed to be a material fluid
("caloric"); we still use the phrase "heat flow" to refer to heat
transfer or to heat effects in general.
Thermochemistry refers to the study of the heat flow that
accompanies chemical reactions. Our discussion of this subject will
focus upon

The basic principles of heat flow.

The
experimental
measurement
of
the
magnitude
and
direction of heat flow, known as calorimetry.

The concept of enthalpy (heat content) and enthalpy change,
ΔH.

The calculation of ΔH for reactions, using thermochemical
equations and enthalpies of formation.

Heat effects in the breaking and formation of covalent bonds.

The relation between heat and other forms of energy, as
expressed by the first law of thermodynamics.
Principles of heat flow:
In any discussion of heat flow, it is important to distinguish between
system and surroundings. The system is that part of the universe
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upon which attention is focused. In a very simple case it might be a
sample of water in contact with a hot plate.
The surroundings,
which exchange energy with the system, comprise in principle the
rest of the universe.
For all practical purposes, however, they
include only those materials in close contact with the system. If we
heat some of water in a beaker, the surroundings would consist of
the hot plate, the beaker holding the water sample, and the air
around it.
When a chemical reaction takes place, we consider the substances
involved,
reactants
and
products,
to
be
the
system.
The
surroundings include the vessel in which the reaction takes place
(test tube, beaker, and so on) and the air or other material in
thermal contact with the reaction system.
State properties:
The state of a system is described by giving its composition,
temperature, and pressure.
For example, if the system which
consists of 50.0 g of H2O(L) at 50.0 °C and at 1 atm is heated, its
state changes, perhaps to one described as 50.0 g of H2O(L) at 80.0
°C and 1 atm.
Certain quantities, called state properties, depend only upon the
state of the system, not upon the way the system reached that
state. Put it another way, if X is a state property, then
ΔX = X final – X initial
That is, the change in X is the difference between its values in final
and initial states. Most of the quantities that you are familiar with
are state properties; volume is a common example.
You may be
surprised to learn, however, that heat flow is not a state property;
its magnitude depends upon how a process is carried out.
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Direction and sign of heat flow:
If the hot plate is turned on, there is a flow of heat from the
surroundings into the system, 50.0 g of water.
This situation is
described by stating that the heat flow, q , for the system is a
positive quantity.
The (q) value is (+) when heat flows into the system from the
surroundings.
Usually, when heat flows into a system, its temperature rises. In
this case, the temperature of the 50.0 g water sample might
increase from 50.0 to
80.0 °C. When the hot plate is shut off, the
hot water gives off heat to the surrounding air. In this case, q for
the system is a negative quantity.
The (q) value is (-) when heat flows out of the system into the
surroundings.
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Here, as in usually the case, the temperature of the system drops
when heat flows out of it into the surroundings. The 50.0 g water
sample might cool from 80.0°C back to 50.0°C.
This same reasoning can be applied to a reaction where the system
consists of the reaction mixture (products and reactants). We can
distinguish between:
-
an endothermic process (q>0), in which heat flows from the
surroundings into the reaction system.
An example is the melting of ice:
The melting of ice absorbs heat from the surroundings, which might
be the water in a glass of iced tea. The temperature of the
surroundings drops, perhaps from 25 to 0 °C, as they give up heat to
the system.
-
an exothermic process (q<0), in which heat flows from the the
reaction system into the surroundings.
An example is the combustion of methane gas:
This reaction evolves heat to the surroundings, which might be the
air around a Bunsen burner in the laboratory or a potato being
baked in a gas oven. In either case, the effect of the heat transfer is
to raise the temperature of the surroundings.
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Magnitude of heat flow:
In any process, we are interested not only in the direction of heat
flow but also in its magnitude. The magnitude of q is ordinarily cited
in joules (J) or kilo-joules (kJ).
1 kJ = 103 J
The SI unit of heat, the joule, is named for James Joule, who carried
out very precise thermometric measurements that established the
first law of thermodynamics.
Most of the remainder of this chapter is devoted to a discussion of
the magnitude of the heat flow in chemical reactions or phase
changes. Here, however, we will focus upon a simpler process in
which the only effect of the heat flow is to change the temperature
of a system. In general, the relationship between the magnitude of
the heat flow, q , and the temperature change, Δt , is given by the
equation:
The quantity C appearing in this equation is known as
the heat capacity of the system. It represents the amount of heat
required to raise the temperature of the system 1°C and has the
units J/°C.
For a pure substance of mass m, the expression for q can be written
as
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The quantity , C, is known as the specific heat; it is defined as the
amount of heat required to raise the temperature of one gram of a
substance 1°C.
Specific heat, like density or melting point, is an intensive property
which can be used to identify a substance. Water has an unusually
large specific heat, 4.18 J/g.°C. This explains why swimming is not
a popular pastime in northern Minnesota in May.
Even the air
temperature rises to 90°F, the water temperature will remain below
60°F. Metals have a relatively low specific heat. When you warm
water in a stainless steel saucepan, nearly all of the heat is absorbed
by the water, very little by the steel.
Example:
How much heat is given off by a 50.0 g sample of copper when it
cools from 80.0 to 50.0°C , Knowing that the specific heat of copper
is 0.382 J/g.°C?
Solution:
The temperature change is
Δt = t final – t initial
= 50.0°C - 80.0°C
= - 30.0°C
q = m x C x Δt
= 50.0 g x 0.382 J/g.°C x (-30.0 °C)
= - 573 J
The negative sign indicates that heat flows from copper to the
surroundings.
Specific heats of a few common substances
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Substance
C (J/g.°C)
Substance
C (J/g.°C)
Br2(L)
0.474
Cu(s)
0.382
Cl2(g)
0.478
Fe(s)
0.446
C2H5OH(L)
2.43
H2O(g)
1.87
C6H6(L)
1.72
H2O(L)
4.18
CO2(g)
0.843
NaCl(s)
0.866
Heat flow in an reaction is measured using a calorimeter device.
Enthalpy:
We have referred several times to " the heat flow for the reaction
system, " symbolized as q
reaction.
At this point, you may well fined
this concept a bit nebulous and wonder if it could be made more
concrete by relating q
products.
reaction
to some property of reactants and
This can indeed be done; the situation is particularly
simple for reactions taking place at constant pressure. Under that
condition, the heat flow for the reaction system is equal to the
difference in enthalpy (H) between products and reactants. That is,
Enthalpy is a type of chemical energy, sometimes referred to as "
heat content. " Reactions that occur in the laboratory in an open
container or in the world around us take place at a constant
pressure, that of the atmosphere. For such reactions, the equation
just written is valid, making enthalpy a very useful quantity.
Reactions which accomplished by the release of energy in the form
of heat are called Exothermic Reactions, whereas reactions that
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absorbs heat from the surroundings are called
Endothermic
Reactions.
If you wish to show that a reaction is exothermic, there is a simple
way of doing it.
For example,
2 Na(s) + Cl2(g)
→
2 NaCl(s)
+ heat
However, sometimes we wish to be more specific and say exactly
how much heat is released.
CH4(g) + 2 O2(g)
→
CO2(g)
+ 2 H2O(g) + 890 KJ
Each of the four substances involved in this reaction contains a
certain quantity of energy and this energy content is called the
Enthalpy (H) of the Compound. In this example, 890 KJ of energy
was released as heat. This means that one mole of carbon dioxide
gas plus two moles of water vapor must contain 890 KJ of energy
less than one mole of methane gas plus two mole of oxygen gas.
Because the products themselves posses 890 KJ of energy less than
the reactants, we say that the change in Enthalpy (∆H) of the
reaction is -890 KJ, (∆H = -890 KJ).
Our equation can be alternatively written as:
CH4(g) + 2 O2(g)
→
CO2(g)
+ 2 H2O(g)
∆H = - 890 KJ
If heat is absorbed during the reaction, the enthalpy of the products
will be greater than that of the reactants. The change in Enthalpy
(∆H) will, therefore, have a positive sign.
reaction is as follows:
123
An example of such a
DR. AHMED KHAMIS SALAMA
2 HgO(s)
→
2 Hg(ℓ) +
GENERAL CHEMISTRY
O2(g)
∆H = + 181.7 KJ
The study of the energy changes that take place during chemical
reactions is called thermochemistry.
The following figure shows the enthalpy relations for an exothermic
reaction such as
CH4 (g) + 2 O2 (g)
→
CO2 (g) + 2 H2 (ℓ)
ΔH < 0
Here, the products, 1 mol of CO2 (g) and 2 mol of H2O(ℓ), have a lower
enthalpy than the reactants, 1 mol of CH4 (g) and 2 mol of O2(g). The
decrease in enthalpy is the source of the heat evolved to the
surroundings.
In an exothermic reaction, the products have a lower enthalpy than
the reactants; thus, ΔH is negative, and heat is given off to the
surroundings.
The following figure shows the situation for an endothermic process
such as
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DR. AHMED KHAMIS SALAMA
H2O (s)
H2O (ℓ)
GENERAL CHEMISTRY
ΔH > 0
Since liquid water, H2O(ℓ), has a higher enthalpy than ice, heat must
be transferred from the surroundings to melt the ice.
In an endothermic reaction, the products have a higher enthalpy
than the reactants so ΔH is positive, and heat is absorbed from the
surroundings.
In general, the following relations apply for reactions taking place at
constant pressure.
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The enthalpy of a substance, like its volume, is a state property. A
sample of one gram of liquid water at 25.0 °C and 1 atm has a fixed
enthalpy, H. In practice, no attempt is made to determine absolute
values of enthalpy.
Instead, scientists deal with changes in
enthalpy, which are readily determined. For the process
1.00 g H2O (ℓ, 25.0°C, 1atm)
1.00 g H2O (ℓ, 26.0°C, 1atm)
ΔH is 4.18 J because the specific heat of water is 4.18 J/g.°C.
THERMOCHEMICAL EQUATIONS:
A chemical equation that shows the enthalpy relation between
products and reactants is called a thermochemical equation. This
type of equation contains, at the right of the balanced chemical
equation, the appropriate value and sign for ΔH.
The thermochemical equation for the formation of HCl from the
elements is found to be
In other words, 185 KJ of heat is evolved when two moles of HCl are
formed from H2 and Cl2.
Rules of thermochemistry:
To make effective use of thermochemical equations, three basic
rules of thermochemistry are applied.
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Rule 1. The magnitude of ΔH is directly proportional to the amount
of reactant or product. The amount of heat that must be absorbed
to boil a sample of water is directly proportional to its mass.
In
another case, the more gasoline you burn in your car's engine, the
more energy you produce.
This rule allows you to find ΔH corresponding to any desired amount
of reactant or product.
Example:
Consider the thermochemical equation for the formation of two
moles of HCl from the elements.
Calculate ΔH when
a) 1.00 mol of HCl is formed
b) 1.00 g of Cl2 reacts.
Solution:
- 185 KJ
a) ΔH = 1.00 mol HCl x ------------- = -92.5 KJ
2 mol HCl
This means the thermochemical equation for the formation of one
mole of HCl would be
½ H2(g) + ½ Cl2(g)
b) ΔH = 1.00 g Cl2
HCl(g)
ΔH = - 92.5 KJ
1 mol Cl2
- 185 KJ
x --------------- x ------------70.90 g Cl2
1 mol Cl2
= - 2.61 KJ
This relation between ΔH and amounts of substances is equally
useful in dealing with chemical reactions or phase changes.
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Example:
Using a coffee-cup calorimeter, it is found that when an ice cube
weighing 24.6 g melts, it absorbs 8.19 KJ of heat. Calculate ΔH for
the phase change represented by the thermochemical equation
H2O (s)
H2O (L)
ΔH = ?
Solution:
18.02 g H2O
8.19 KJ
ΔH = 1 mol H2O x ----------------- x ------------1 mol H2O
24.6 g H2O
= 6.00 KJ
This calculation shows that 6.00 KJ of heat must be absorbed to
melt one mole of ice:
H2O (s)
H2O (L)
ΔH = + 6.00 KJ
ΔH per one mole of ice = 6.00 KJ
ΔH per gram of ice = 6.00 KJ / 18.02 g
= 0.333 KJ/g
Rule 2. ΔH for a reaction is equal in magnitude but opposite in sign
to ΔH for the reverse reaction. Another way to state this rule is to
say that the amount of heat evolved in a reaction is exactly equal to
the amount of heat absorbed in the reverse reaction. This again is a
common-sense rule. If 6.00 KJ of heat is absorbed when a mole of
ice melts,
H2O(s)
H2O(L)
ΔH = + 6.00 KJ
Then 6.00 KJ of heat should be evolved when a mole of liquid water
freezes.
H2O(L)
H2O(s)
ΔH = - 6.00 KJ
Example:
Given
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H2(g) + ½O2(g)
H2O(L)
ΔH = - 285.8 KJ
Calculate ΔH for the equation:
2 H2O(L)
2 H2(g) + O2(g)
Solution:
Applying rule 1:
H2(g) + ½O2(g)
2 H2(g) + O2(g)
H2O(L)
ΔH = - 285.8 KJ
2 H2O(L)
ΔH = 2 (- 285.8 KJ) = -571.6
2 H2O(L)
ΔH = -571.6 KJ
KJ
2 H2(g) + O2(g)
Applying rule 2:
2 H2O(L)
2 H2(g) + O2(g)
ΔH = +571.6 KJ
Rule 3. The value of ΔH for a reaction is the same whether it
occurs in one step or in a series of steps.
ΔH = ΔH1 + ΔH2 + ΔH3 + ΔH4 + ………
This relationship is referred to as Hess' law
Example:
Given
Calculate ΔH for the reaction
Solution:
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To get one mole of CO on the right side, reverse Equation (2) and
divide the coefficients by two.
Applying Rule 1 and rule 2 in
succession.
Now add Equation (1) and simplify:
Summarizing the rules of thermochemistry:
1. ΔH is directly proportional to the amount of reactant or product.
2. ΔH changes sign when a reaction is reversed.
3. ΔH for a reaction has the same value regardless of the number of
steps.
Enthalpies of formation:
Meaning of ΔHf°:
The standard molar enthalpy of formation of a compound, ΔHf° , is
equal to the enthalpy change when one mole of the compound is
formed at a constant pressure of 1 atm and a fixed temperature,
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ordinarily 25°C, from the elements in their stable states at that
pressure and temperature.
From the equations
It follows that:
ΔHf° AgCl (s) = -127.1 KJ/mol
ΔHf° NO2 (s) = +33.2 KJ/mol
Enthalpies of formation, ΔHf° (KJ/mol), of compounds at 25°C, 1atm
AgBr(s)
-100.4
BaCl2 (s)
-858.6
AgCl(s)
-127.1
BaCO3(s)
-1216.3
AgI(s)
-61.8
C2H2(g)
+226.7
AgNO3(s)
-124.4
C2H6(g)
-84.7
Ag2O(s)
-31.0
C2H4(g)
+52.3
Al2O3 (s)
-1675.7
ZnS(s)
-206.0
Notice that, with a few exceptions, enthalpies of formation are
negative quantities. This means that the formation of a compound
from the elements is ordinarily exothermic.
Conversely, when a
compound decomposes to the elements, heat usually must be
absorbed.
The enthalpy of formation of an element in its stable state at 25°C
and 1atm is taken to be zero.
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DR. AHMED KHAMIS SALAMA
That is,
GENERAL CHEMISTRY
ΔHf° Br2(L) = ΔHf° O2(g) = zero
Example:
Calculate ΔH° for the combustion of one mole of propane, C3H8,
according to the equation
Solution:
ΔH° = heat of formation of products – heat of formation of reactants
Heat of formation of an element (at 25 °C, 1 atm) is equal zero
H°f, Br2 = 0
H°f, O2 = 0
Expressing ΔH° in terms of enthalpies of formation,
Taking the enthalpy of formation of O2(g) to be zero and substituting
values for the other substances tables.
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8. CHEMICAL KINETICS
This chapter deals with the measurements of chemical rates and the
factors that affecting its velocity. In case of reactions in solution
(liquid state) the rate depends on the effective concentration of
reactants while for gaseous and solid state reaction it depends on
the pressure and surface area, respectively.
Kinetic classification of chemical reactions:
1. Molecularity:
The chemical reactions are classified in this method as the following:

Uni-molecular reactions
if there is only one reactant material.

Bi-molecular reactions
if there is two reactant materials.

Tri-molecular reactions
if there is three reactant materials.
2. Order of the reaction (n):
The order of the reaction is depending on the number of effective
concentrations that participating in the rate determining step.
If only one concentration affecting the rate of reaction it is named
as first order (1st.) reaction, when two concentrations it is called
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second-order reaction (2nd), if three, it is called third-order (3rd).
The last type is very rare to have three molecules collide at the same
time, but the reaction may be complicated and takes place in several
steps.
The rate of the reaction is a positive quantity that expresses how the
concentration of a reactant or product changes with time.
To illustrate what this means, consider the reaction
As you can see the concentration of N2O5 decreases with time; the
concentration of NO2 and O2 increase. Because these species have
different
concentrations
in
the
balanced
equation,
their
concentrations do not change at the same rate. When one mole of
N2O5 decomposes, two moles of NO2 and one-half mole of O2 are
formed. This means that:
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GENERAL CHEMISTRY
Where Δ [ ] refers to the change in concentration in moles per liter.
The minus sign in front of the N2O5 term takes account of the fact
that [N2O5] decreases as the reaction takes place; the numbers in
the denominator of the term on the right (2, ½) are the coefficients
of these species in the balanced equation.
The rate of reaction can now be defined by dividing by the change in
time, Δt:
More generally for the reaction
where A, B, C, and D represent substances in the gas phase (g) or in
aqueous solution (aq), and a, b, c, d are their coefficients in the
balanced equation:
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To illustrate the use of this expression, suppose that for the
formation of ammonia:
Molecular nitrogen is disappearing at the rate of 0.10 mol/L per
minute, i.e., Δ[N2]/ Δt = -0.10 mol/L.min. From the coefficients of
the balanced equation, we see that the concentration of H2 must be
decreasing three times as fast:
Δ[H2]/ Δt = -0.30 mol/L.min. By the same token, the concentration
of NH3 must be increasing at the rate of 2 x 0.10 mol/L. min:
Δ[NH3]/ Δt = 0.20 mol/L.min. It follows that:
By defining rate this way, it is independent of which species we
focus upon, N2 , H2 , or NH3.
Notice that reaction rate has the units of concentration divided by
time. We will always express concentration in moles per liter. Time,
on the other hand, can be expressed in seconds, minutes, hours.
A rate of 0.10 mol/L. min corresponds to:
0.10 mol/L. min x 1 min / 60 sec = 1.7 x 10 -3 mol/L. s = 6.0
mol/L. h
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Measurement of rate
For the reaction N2O5 (g)
2NO2 (g) + ½ O2 (g)
The rate could be determined by measuring

The absorption of visible light by the NO2 formed; this species
has a reddish-brown color, whereas N2O5 and O2 are colorless.

The change in pressure that results from the increase in the
number of moles of gas
(1mole reactant
2½ mole product).
Reaction rate and concentration
Ordinarily, reaction rate is directly related to reactant concentration.
The higher the concentration of starting materials, the more rapidly
a reaction takes place.
Pure hydrogen peroxide, in which the
concentration of H2O2 molecules is about 40 mol/L, is an extremely
dangerous substance to work with.
In the presence of trace
impurities, it decomposes explosively at a rate too rapid to measure.
H2O2 (L)
H2O (g) + ½ O2 (g)
The hydrogen peroxide you buy in a drugstore is a dilute aqueous
solution in which [H2O2] ≈ 1 M. At this relatively low concentration,
decomposition is so slow that the solution is stable for several
months.
The dependence of reaction rate upon concentration is readily
explained.
Ordinarily, reactions occur as the result of collisions
between reactant molecules.
The higher the concentration of
molecules, the greater the number of collisions in unit time and
hence the faster the reaction.
As reactants are consumed, their
concentrations drop, collisions occur less frequently, and reaction
rate decreases. This explains the common observation that reaction
rate drops off with time, eventually going to zero when all the
reactants are consumed.
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Rate expression and rate constant:
In the following decomposition reaction, the rate increases as
concentration increases.
N2O5 (g)
2NO2 (g) + ½ O2 (g)
The plot of rate versus concentration is a straight line through the
origin, which means that rate must be directly proportional to
concentration.
Rate expression
α
[N2O5]
= k [N2O5]
Where, k is the rate constant.
Rate depends upon concentration but rate constant does not.
For the decomposition of N2O5 , a plot of rate versus concentration of N2O5 is a
straight line. The line, if extrapolated, passes through the origin. This means that
rate is directly proportional to concentration; that is, rate = k[N 2O5].
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Order of reaction involving a single reactant:
The rate expression for the process
A
products has
Rate = k[A]m
the general form
The power to which the concentration of reactant A is raised in the
rate expression is called, the order of the reaction, m. If m is 0, the
reaction is said to be "zero order". If m=1, the reaction is "first
order". If m=2, the reaction is "second order" and so on.
The order of a reaction must be determined experimentally; it is
cannot be deduced from the coefficients in the balanced equation.
This must be true because there is only one reaction order, but there
are many different ways in which the equation for the reaction can
be balanced. For example, although we wrote
N2O5
(g)
2NO2
(g)
+ ½ O2
(g)
to describe the
decomposition of N2O5 , it could have been written as
2N2O5 (g)
4NO2 (g) + O2 (g)
The reaction is still first-order no matter how the equation is
written.
One way to find the order of a reaction is to measure the initial rate
(i.e., the rate at t=0) as a function of the concentration of reactant.
Suppose, for example, that we make up two different reaction
mixtures differing only in the concentration of reactant A. We now
measure the rates at the beginning of reaction, before the
concentration of A has decreased appreciably.
This gives two
different initial rates (rate1 , rate2) corresponding to two different
starting concentrations of A,
[A]1 and [A]2.
expression,
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From the rate
DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
Rate1 = k[A]1m
Rate2 = k[A]2m
Dividing the second rate by the first,
Since all the quantities in this equation are known except m, the
reaction order can be calculated.
Example:
The initial rate of decomposition of acetaldehyde, CH3CHO, at 600°C
was measured at a series of concentrations with the following
results:
CH3CHO(g)
CH4 (g) + CO(g)
[CH3CHO]
0.10 M
0.20 M
0.30 M
0.40 M
Rate (mol/L.s)
0.085
0.34
0.76
1.4
Using these data, determine the reaction order, that is, determine
the value of m in the equation rate = k[CH3CHO]m
Strategy: Choose the first two concentrations, 0.10 M and 0.20 M.
Calculate the ratio of the rates, the ratio of the concentrations, and
finally the order of reaction, using the general relation derived
above.
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Solution:
The reaction is second-order.
Once the order of the reaction is known, the rate constant is readily
calculated.
Consider, for example, the decomposition of acetaldehyde, where
we have shown that the rate expression is
rate = k[CH3CHO]2
The data in the above example show that the rate at 600°C is 0.085
mol/L.s when the concentration is 0.10 mol/L.
It follows that:
Rate = K [CH3CHO] m
rate
0.085 mol/L.s
K= ------------- = -------------------- = 8.5 L/ mol.s
[CH3CHO]2
(0.10 mol/L)2
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The same value of k would be obtained using any other data pair.
Having established the value of k and the reaction order, the rate is
readily
calculated
at
any
concentration.
Again,
using
the
decomposition of acetaldehyde as an example, we have established
that
Rate = 8.5 L/mol.s [CH3CHO]2
If the concentration of acetaldehyde were 0.50 M,
Rate = 8.5 L /mol.s) (0.50 mol /L)2 = 2.1 mol / L.s
Order of reaction with more than one reactant:
Many reactions (Indeed, most reactions) involve more than one
reactant. For a reaction between two species A and B,
aA+ bB
products
The general form of the rate expression is
Rate = k[A]m x [B]n
Here m is referred to as " the order of the reaction with respect to
A." Similarly, n is " the order of the reaction with respect to B."
The overall order of the reaction is the sum of the exponents, m + n.
If m=1, n=2, then the reaction is first-order in A, second-order in B,
and third-order overall.
When more than one reactant is involved, the order can be
determined by holding the initial concentration of one reactant
constant while varying that of the other reactant. From rates
measured under these conditions, it is possible to deduce the order
of the reaction with respect to the reactant whose initial
concentration is varied.
To see how this is done, consider the reaction between A and B
referred to above. Suppose we run two different experiments in
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which the initial concentrations of A differ ([A]1 , [A]2) but that of B
is held constant at [B]. Then:
rate1 = k[A]1m x [B]n
rate2
= k[A]2m x [B]n
Dividing the second equation by the first:
Rate2
k[A]2m x [B]n
[A]2m
-------- = ------------------- = ------- = [[A]2 / [A]1]m
Rate1
k[A]1m x [B]n
[A]1m
Knowing the two rates and the ratio of the two concentrations, we
can readily find the value of m.
Example:
Consider the reaction at 55°C
(CH3)3CBr(aq) + OH-(aq)
(CH3)3COH(aq) + Br-(aq)
A series of experiments is carried out with the following results:
[(CH3)3CBr]
0.50
1.0
1.5
1.0
1.0
[OH-]
0.050
0.050
0.050
0.10
0.20
Rate
0.0050
0.010
0.015
0.010
0.010
(mol/L.s)
Find the order of the reaction with respect to both (CH3)3CBr and
OH-.
Strategy: To find the order of the reaction with respect to (CH3)3CBr,
choose two experiments, perhaps 1 and 3, where [OH-] is constant.
A similar approach can be used to find n ; compare experiments 2
and 5, where [(CH3)3CBr] is constant.
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Solution:
Rate3
[CH3)3CBr]3
------- =
Rate1
[--------------- ]m
[CH3)3CBr]1
0.015 / 0.005 = (1.5 / 0.05)m
3
31
3m
=
3m
=
m= 1
Rate5
[OH-]5
------- =
Rate2
0.010
[---------]
n
[OH-]2
---------0.010
0.20
=
[---------]n
1 = 4n
0.050
0.010 / 0.010 = (0.20 / 0.050)n
1
=
4n
40
=
4n
Any number raised to the power zero equals one. 40 = 1
In this case, n = 0 ; the rate is independent of the concentration of
OH-.
n=0
The overall order = m+n = 1 +0 = 1
Reactant concentration and time
The rate expression
Rate = k[N2O5]
Shows how the rate of decomposition of N2O5 changes with
concentration.
From a practical standpoint, however, it is more
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GENERAL CHEMISTRY
important to know the relation between concentration and time
rather than between concentration and rate. Suppose for example,
you are studying the decomposition of N2O5. Most likely, you would
want to know how much N2O5 is left after 5 min, 1h, or several days.
An equation relating rate to concentration does not answer that
purpose.
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Pseudo-order reaction:
When one of the reactants present in large excess amount compared
to other one the reaction is then called pseudo-order reaction.
For example, hydrolysis of ethyl acetate in presence of large excess
of water and acid catalysed, the reaction is pseudo-first order:
H+
CH3COOC2H5 + H2O
(low amount)
(excess)
CH3COOH
+
C2H5OH
The rate of reaction is only dependent on [CH3COOC2H5].
First-order reactions
For the reaction
A
products
The rate equation will be written as follows:
Rate = K[A]1
Or
dx/dt = k(a-x)
where:
(a) is the initial concentration,
(x) is the product concentration,
(a-x) is the concentration at time t,
(k) is the first-order rate constant,
It can be shown by using calculus that the relationship between
concentration and time is
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Since ln a/b = ln a – ln b, the first-order equation can be written in
the form
ln (a) - ln (a-x) = kt
ln (a-x) = ln (a) - kt
The rate constant for a first-order reaction can be determined from the slope of a
plot of ln (a-x) versus time. From the graph of data for the reaction
N2O5 (g)
2NO2 (g) + ½ O2 (g)
at 67°C, it appears that the first-order rate
constant is about 0.35/min.
Solving for ln (a-x),
ln (a-x) = ln (a) – kt
Comparing this equation to the general equation of a straight line,
Y = b + mx
(b is y-intercept, m is slope)
It is clear that a plot of ln (a-x) versus t should be a straight line
with a
y-intercept of ln (a) and a slope of –k. This is indeed the
case, as you can see from the above figure where we have plotted ln
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
[N2O5] versus time for the decomposition of N2O5. Drawing the best
straight line through the points and taking the slope based upon the
two points on the y- and x- axes,
Slope = y2 – y1 / x2 - x1
Slope = -3.5 – (-1.8) / 4.9 – 0 = - 0.35
It follows that the rate constant is 0.35 / min; the integrated firstorder equation for the decomposition of N2O5 is
Ln a / (a-x) = Kt
ln [N2O5]0 / [N2O5] = k t
ln [N2O5]0 / [N2O5] = (0.35/min) t
Using the equation
Ln a / (a-x) = Kt
The first order representation will be:
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GENERAL CHEMISTRY
Example:
N2O5 (g)
2NO2 (g) + ½ O2 (g)
For the decomposition of N2O5 at 67°C, where k= 0.35/min ,
calculate:
a) the concentration after 4.0 min, starting at 0.160 M.
b) the time required for the conc. to drop from 0.160 to 0.100 M.
c) the time required for half a sample of N2O5 to decompose
N2O5 (g)
2NO2 (g) + ½ O2 (g)
Solution:
a)
ln (a) / (a-x) = kt
ln [N2O5]0 / [N2O5] = kt
ln 0.160 M / [N2O5] = (0.35 /min) t
ln 0.160 M / [N2O5] = (0.35 /min) (4.0 min)
ln 0.160 M / [N2O5] = 1.4
Taking inverse logarithms,
0.160 M / [N2O5] = e1.4
0.160 M / [N2O5] = 4.0
[N2O5] = 0.160 M / 4.0 = 0.040 M
b)
ln (a) / (a-x) = kt
t = (1/ k) ln (a) / (a-x)
t = 1 / (0.35/min) ln 0.160 M / 0.100 M
t = ln 1.60 / (0.35 / min) = 0.47 min / 0.35 = 1.3 min
c) When half of the sample has decomposed
[N2O5] = [N2O5]0 / 2
[N2O5]0 = 2 [N2O5]
[N2O5]0 / [N2O5]
= 2
Using the equation in (b) t = (1/ k) ln (a) / (a-x)
t = (1/ k) ln 2
t = 0.693 / k
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DR. AHMED KHAMIS SALAMA
when k = 0.35 / min, we have
GENERAL CHEMISTRY
t = 0.693 / 0.35 = 2.0 min
The time required for one half of a reactant to decompose via a
first–order reaction has a fixed value, independent of concentration.
This quantity, called the half-life, is given by the expression
t½ = ln 2 / k
t½= 2.303 log 2 / K
t½ = 0.693 / k
Example:
Plutonium-240, produced in nuclear reactors, has a half-life of 6.58
x 103 years. Calculate:
a) the first-order rate constant for the decay of plutonium-240
b) the fraction of a sample that will remain after 100 years.
Solution:
a)
t½ = 0.693 / k
k = 0.693 / t½
k = 0.693 / 6.58 x 103 yr = 1.05 x 10-4 yr
b)
ln (a) / (a-x) = kt
ln (a) / (a-x) = (1.05 x 10-4 yr) (100 yr)
Taking the inverse log, (a) / (a-x) = 1.01
Hence (a-x) / (a) = 1 / 1.01 = 0.99, that is 99% remains.
Zero-order reactions
This type of reactions is independent on reactant concentration but
may depend on physical changes in properties such as viscosity,
refractive index, catalytic activity, ……..etc.
For the reaction
A
products
Rate = K[A]0 = K
Or
dx/dt = k
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
Zero-order reactions are relatively rare. Most of them take place at
solid surfaces, where the rate is independent of concentration in the
gas phase.
A typical example is the thermal decomposition of
hydrogen iodide on gold:
When the gold surface is completely covered with HI molecules,
increasing the concentration of HI(g) has no effect on reaction rate.
dx/dt = k
if the initial concentration (at time t=0) is a,
if the concentration after time t (remaining) is (a-x),
By integration:
a – (a-x) = kt
t½ = a – (a-x) / k
at t½ , the remaining amount = a/2
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GENERAL CHEMISTRY
t½= (a - a /2) / k
t½= (½ a) / k
t½= 1/ k (½ a)
t½= a/ 2k
The half-life period:
Second-order reactions:
Consider the following reaction is a general form of a second order
type:
A
+
B
C
t=0
a …….
b………
zero
t=t
a-x …….
b -x………
x
For this reaction we have two cases;
1. CA = CB : the rate equation will be
dx /dt = k (CA)2
2. CA ≠ CB : the rate equation will be
dx /dt = k CA CB
Here we will consider the simplest case when CA = CB or when two
concentrations of A participating in the reaction and equation is
used as follows:
dx/dt = k(a-x)2
dx/(a-x)2 = kt
By integration:
1/(a-x) = kt + C
At
t=0,
x=0,
and C=1/a
1/(a-x) = kt + 1/a
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
The half-life period:
By applying for x = a/2 in the equation,
x / a(a-x) = kt
(a/2) / a (a- (a/2)) = kt½
The plot shows how to determine the rate constant k from the slope
using equation
1/ (a-x) = kt + C
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DR. AHMED KHAMIS SALAMA
In case of using equation
GENERAL CHEMISTRY
x / a(a-x) = kt
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
Summary:
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
Graphical method to determine the reaction order:
In this method the data obtained were plotted according to 1 st order
equation
ln a/(a-x) = kt .
If a straight line is obtained, the reaction is then obey first order
reaction (n=1).
If not, apply in the second order equation x/
(a-x) = kt .
When a straight line is obtained this means that the reaction is 2 nd
order.
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
Example:
The following data were obtained for the gas-phase decomposition
of HI
Time (hr)
0
2
4
6
[HI]
1.0
0.5
0.33
0.25
Is this reaction zero-, 1st -, or 2nd – order in HI ?
Solution:
Not from the previous explanation that:
1. if the reaction is zero-order, a plot of (a-x) versus t should be
linear.
2. if the reaction is 1st -order, a plot of ln(a-x) versus t should be
linear.
3. if the reaction is 2nd -order, a plot of 1/(a-x) versus t should be
linear.
Using this data, prepare these plots and determine which one is a
straight line.
T(hr)
[HI]
ln[HI]
1/[HI]
0
1.0
0
1
2
0.5
-0.69
2
4
0.33
-1.10
3
6
0.25
-1.39
4
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
Zero order reactions:
[HI]
Not linear
0
2
4
6
0
2
4
6
0
2
4
6
1st order reactions:
Not linear
Ln[HI]
2nd order reactions:
Linear
1/[HI]
The reaction is 2nd – order.
Effect of temperature on the reaction rate:
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
Since the reaction occurs via collision of the reacting molecules, this
effective collision mostly increases on increasing temperature. This
accelerate the reaction depending on the activation energy (Ea).
The activation energy has the units of Cal/mol or Joul/mol.
Arrhenius equation was applied to correlate the rate constant (k)
with the temperature as following:
K = A exp (-Ea / RT)
Where T is the absolute temperature, R is the gas constant, Ea is the
activation energy and A is the frequency factor.
Taking logarithm of both side;
ln k = ln A - Ea / RT
The plot of ln k versus reciprocal of T it showed a straight line of
slope equal to Ea / R as shown from the figure.
Substituting for the value of the gas constant R in the slope we got:
Ea = slope x 1.987 Cal/mol
Ea = slope x 8.314 J/mol
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
Complete the following:
1. The extremely small, invisible, and indivisible particles were
called ……………..
2. Iron atoms are …………….. in mass, size, and shape, and show the
same physical and chemical characteristics.
3. Atoms of two or more elements combine together to form a
………………...
4. The smallest particle that still has the properties of the compound
is called a …………….
5. According to Rutherford, the entire mass of the atom is
concentrated in its …………. and the rest of the atom was mostly
……………….
6. Isotopes are known as atoms that have the ………. number of
protons but a ……….. number of neutrons in the nucleus.
7. The electron configuration of sodium that has atomic number, 11
is ………………………………
8. ………………… bond is taking place between atoms of two
nonmetals by ……………… to complete the outermost shell of each
one to the electronic configuration of the nearest inert gas.
9. ………………bond is formed between a metal and non-metal by
………………….. from the ………………. to…………………
10. The oxidation number of carbon atom may be differ according
the compound formula, where it is …………… in H2CO3 , …………….
in CO, ……………. in CO3 -- and ……………. in NaHCO3.
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DR. AHMED KHAMIS SALAMA
GENERAL CHEMISTRY
1st Quiz
Majmaah University
Faculty of Science
General Chemistry (CHEM-101)
Student Name:
Put circle around the correct answer of the following questions:
1. The extremely small, invisible, and indivisible particles were called
(a) molecules
(b) atoms
(c) compounds
element
2. Atoms of two or more elements combine together to form
(a) molecules
(b) particles
(c) compounds
mixtures
(d)
(d)
3. The electron configuration of chlorine atom (atomic weight = 17) is:
(a) 1s2 1p6 2s2 2p6 3s1
(b) 1s2 2s2 2p6 2d7
2
2
6
2
5
(c) 1s 2s 2p 3s 3p
(d) 1s2 2s2 2p6 3s1 3p6
4. Which of the following correctly describes the composition of Ca2+ ions in
calcium (atomic number=20, and mass number=40)?
(a) 18 protons, 20 electrons.
(c) 20 protons, 18 electrons
(b) 20 protons, 20 electrons.
(d) 20 protons, 22 electrons
5. Which of the following sublevels do not exist?
(a) 1d
(b) 2s
(c) 3d
6. The chemical bonding in CH4 is :
(a) polar covalent bond
(c) non-polar covalent bond
(d) 6d
(b) ionic bond
(d) metallic bond
7. The chemical bonding in NaCl is :
(a) covalent bond
(c) coordinate covalent bond
(b) ionic bond
(d) metallic bond
8. Covalent bond is taking place between:
(a) atoms of two non-metals
(c) atoms of two metals
(b) atoms of metal and non-metal
(d) All answers are wrong
9. Ionic bond is taking place between:
(a) atoms of two non-metals
(c) atoms of two metals
(b) atoms of metal and non-metal
(d) All answers are wrong
10. Triple covalent bond formed by sharing of :
(a) three pairs of electrons between two metallic atoms
(b) two pairs of electrons between two non-metallic atoms
(c) three pairs of electrons between two non-metallic atoms
(d) one pair of electrons between two non-metallic atoms
With my best wishes
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