SECTION B
1. Determine the period, group and block for the following elements based on their
electronic configutation.
A : 1s2 2s2
B : 1s2 2s2 2p6 3s2 3p6 3d10 4s1
C : 1s2 2s2 2p6 3s2 3p4
D : 1s2 2s2 2p6 3s2 3p6 3d10 4s2
E : 1s2 2s2 2p6 3s1
F : 1s2 2s2 2p6 3s2 3p6 4s2 3d6
2. Based in isotope notation for the following elements
P Q27 31 R35
13 15 17
Determine the element with the smallest atomic radius. Explain your answer.
3. The proton number for several elements is given as follow :
Element Proton number
J 17
H 16
I 19
4. Choose the larger species in each pair :
a) Na or Na+
b) Co2+ or Co3+
c) Cl or Cl-
5. Rank the following elements by increasing atomic radius:
carbon, aluminium, oxygen, potassium.
6. Define electronegativity.
7. Rank the following elements by increasing electronegativity:
sulphur, oxygen, neon, aluminium.
8. Define ionization energy .
9. Why does fluorine have a higher ionization energy than iodine?
10 Why do elements in the same family generally have similar properties?
11 Indicate whether the following properties increase or decrease from left to right across
the periodic table.
(a) atomic radius (excluding noble gases)
(b) first ionization energy
(c) electronegativity
12 What trend in atomic radius occurs down a group on the periodic table?
What causes this trend?
What trend in ionization energy occurs across a period on the periodic table? What
13 causes this trend?
Underline the atom in each pair that has the largest atomic radius.
97 | C h a p t e r 6 C h e m i s t r y 1 D K 0 1 4
14 a. Al or B
b. Na or Al
c. S or O
d. O or F
e. Br or Cl
f. Mg or Ca
Revised by:
Kasbah Bin Andi Dahran @ Asmad
Zanarina Binti Sapiai
98 | C h a p t e r 6 C h e m i s t r y 1 D K 0 1 4
CHAPTER 7: CHEMICAL BONDING
GENERAL OVERVIEW
a)Lewis dot
symbol
b)Octet rule
7.1 Valence c) How atom obtain i. Ionic or
electron and octet configation. electrovalent bond
Lewis structure
d) Using Lewis dot ii. covalent bond
symbol
e) Lewis structure
f) Formal charge i. incomplete
octet
g) Exception to octet ii. expanded
rule octet
7.0 CHEMICAL BONDING a) Valence Shell iii. odd number
Electron Pair electron
Repultion theory i. linear
(VSEPR)
ii. trigonal planar
7.2 Molecular Basic iii. tetrahedral
shape Molecular iv. trigonal bipyramidal
7.3 Shape
Intermolecular
v. octahedral
Forces
c) shape of i. Bond
molecule angle
a) Polarity and i. van der Dipole-dipole
dipole moment Waals interaction
forces
b) Intermolecular London
forces ii. hydrogen forces
bonding
99 | C h a p t e r 7 C h e m i s t r y 1 D K 0 1 4
7.1 VALENCE ELECTRON AND LEWIS STRUCTURE
Gilbert N. Lewis
American physical chemist
known for:
✓ the discovery of the
covalent bond and his
concept of electron pairs
✓ Lewis structures
✓ Other contributions to
valence bond theory have
shaped modern theories
of chemical bonding
(October 23, 1875 – March 23, 1946)
LEWIS SYMBOL FOR ATOMS
▪ When atoms react to form chemical bonds, only the electrons in the outermost or
valence shell are involved. Thus, only the electronic configuration of the valence shell
is important in the bond formation.
▪ In order to place emphasis on the electron in the outermost shell (valence electrons),
we use special symbols known as Lewis Structure or also called Lewis dot
structures.
▪ Lewis symbol for an element consists of the element’s symbol and dot or cross.
▪ Table 7.1 shows the Lewis Structures of Period 2 and Period 3 elements. Each ‘dot’
represents one electron in the valence shell of the atom.
▪
EXAMPLE 1:
Lewis structure for N
Answer:
EXAMPLE 2:
Draw a Lewis symbol for each of the
following atoms:
(a) K
(b) Si
(c) I
(d) Ba
(e) Ar
100 | C h a p t e r 7 C h e m i s t r y 1 D K 0 1 4
Answer
a)
b)
c)
d)
e)
OCTET RULE
▪ Octet rule, atoms tend to lose, gain or share electrons until they are surrounded
by eight electrons.
▪ In the Lewis symbol, an octet is shown as four pairs of valence electrons arranged
around the element symbol as in the Lewis symbol of noble gas.
▪ Atoms can achieve an octet configuration by:
i) losing valence electrons or gaining electrons to form positive ion and negative
ion
iii) sharing electrons to form covalent bonds.
➢ There are 3 type of stability of ion
i. Noble gas configuration
➢ The term given to the electron configuration of noble gases
➢ Atoms may lose or gain enough electron so as to forms stable ion with octet
(or duplet) configuration ns2 np6
➢ Cations with noble gas configurations
➢ Anions with noble gas configurations
➢
➢ Elements in Groups 1,2, 16 and 17 attain noble gas configuration. Their ions
101 | C h a p t e r 7 C h e m i s t r y 1 D K 0 1 4
are said to be isoelectronic with the nearest noble gas (Group 18).
ii. Pseudo-noble gas configuration
➢ Ions with stable electronic configurations in which all of their orbitals are
completely filled with electrons.
➢ A noble gas core with outer d10 configuration: ns2np6nd10
➢ But, this electronic configuration is not that of any noble gas pseudo-noble
gas configuration
➢ Example:
Cation with pseudo-noble gas configuration
Gallium, Ga is an element from Group 13:
Electronic configuration of Ga:
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p1
Electronic configuration of Ga3+:
1s2 2s2 2p6 3s2 3p6 3d10 = [Ar]3d10
iii. Half–filled orbitals configuration
➢ When a cation is formed from an atom of a transition metal, electrons are
always removed first from the ns orbital, then followed by the (n – 1)d orbitals.
➢ Some transition metal atoms form cations that have electron configuration
associated with half–filled d orbital (d5) .
102 | C h a p t e r 7 C h e m i s t r y 1 D K 0 1 4
EXAMPLE 3
Write the ionic symbol and ground–state electron configuration of the monoatomic ion most
likely to be formed by each:
a) Cl
b) K
c) Mg
d) O
ANSWER
a) CI : 1s2 2s2 2p6 3s2 3p5
CI– : 1s2 2s2 2p6 3s2 3p6 = [Ar]
b) K : 1s2 2s2 2p6 3s2 3p6 4s1
K+ : 1s2 2s2 2p6 3s2 3p6 = [Ar]
c) Mg : 1s2 2s2 2p6 3s2
Mg2+ : 1s2 2s2 2p6 = [Ne]
d) O : 1s2 2s2 2p4
O2– : 1s2 2s2 2p6 = [Ne]
EXAMPLE 4
Ferric oxide is the inorganic compound with the formula Fe2O3. Ferric refers to iron-containing
materials or compounds. In chemistry the term is reserved for iron with an oxidation
number of +3, also denoted iron(III) or Fe3+.
What type of stability of the electron configuration of ion Fe3+ ?
Note: Fe (Z = 26)
ANSWER
Fe (Z = 26)
Electron configuration:
1s2 2s2 2p6 3s2 3p6 4s2 3d6 = [Ar]3d6 4s2
Fe3+
Electron configuration:
1s2 2s2 2p6 3s2 3p6 3d5 = [Ar]3d5
Type of stability = half–filled d orbital
103 | C h a p t e r 7 C h e m i s t r y 1 D K 0 1 4
FORMATION OF IONIC BOND.
➢ Ionic bond is an attractive electrostatic forces between positive and
negative ions.
➢ Elements with large differences in electronegativity tend to form ionic
bonds
➢ Also known as electrovalent bond.
➢ Table below shows substance formed by ionic bond.
Anion Cl- OH- O2- SO42-
Cation NaCl NaOH Na2O Na2SO4
Na+ CaCl2 Ca(OH)2 CaO CaSO4
Ca2+ AlCl3 Al(OH)3 Al2O3 Al2 (SO4)3
Al3+
HOW DO IONIC BONDS FORMED??
➢ By transferring of electron from metals to non-metals
➢ Total number of electron lost by metal atoms equal to total number of
electron gained by the non-metal atoms
➢ Formation of ionic bond using Lewis symbol
EXAMPLE 5
Use Lewis symbols to describe the formation of Na2O
ANSWER
Using Lewis Symbol
Each Na atom loses an electron forming Na+ ion
O atom gains two electrons forming O2- ion
Electrostatic attraction between Na+ ion and O2- ion forming ionic bond
Formula of the compound formed = Na2O
104 | C h a p t e r 7 C h e m i s t r y 1 D K 0 1 4
FORMATION OF COVALENT BOND
➢ Atoms of elements with similar electronegativity tend to form covalent bonds.
➢ Pure covalent bonds result when two atoms of the same electronegativity bond.
➢ A covalent bond is a chemical bond that involves the sharing of electron pairs
between non-metal atoms.
➢ The stable balance of attractive and repulsive forces between atoms when they share
electrons is known as covalent bonding.
➢ A covalent bond is formed by two atoms sharing a pair of electrons. The atoms are
held together because the electron pair is attracted by both of the nuclei.
➢ In the formation of a simple covalent bond, each atom supplies one electron to the
bond - but that doesn't have to be the case. A co-ordinate bond (also called a dative
covalent bond) is a covalent bond (a shared pair of electrons) in which both electrons
come from the same atom.
➢
➢ Using Lewis Symbol
➢
➢
➢
➢
Covalent bond in anions:
105 | C h a p t e r 7 C h e m i s t r y 1 D K 0 1 4
LEWIS STRUCTURE
➢ Lewis Structure is the diagrams that show the bonding between atoms of a molecule
and the lone pairs of electrons that may exist in the molecule.
➢ In general, the following steps are taken when writing the Lewis Structure for
molecules and ions.
1) Count the total number of valence electron from all the atoms in the
molecules and ion.
(Add electrons if –ve charge and minus electrons if +ve charge)
2) Arrange all the atoms surrounding the central atom by using a pair of
electron per bond. The central atom is the most often the atom which is least
electronegative. Hydrogen is never a central atom.
3) Draw single covalent bond between central atom and surrounding atoms.
4) Complete the octets around all the surrounding atoms
5) Add remaining electrons (if any) to the central atom. Electrons not involved
in bonding shown as lone pairs
6) If the central atom is not octet yet, make a multiple bond (double or triple)
by using a lone pair from the surrounding atoms
EXAMPLE 6
How to draw the Lewis structure of ammonium ion, NH4+?
ANSWER
106 | C h a p t e r 7 C h e m i s t r y 1 D K 0 1 4
EXAMPLE 7
How to draw the Lewis structure of carbonate ion, CO32-?
ANSWER
107 | C h a p t e r 7 C h e m i s t r y 1 D K 0 1 4
EXERCISE
1. Draw the Lewis structure for each of the following:
a) H2O
b) ClF2-
c) CN-
2. Formic acid (also called methanoic acid) is the simplest carboxylic acid. Its chemical
formula is HCOOH. It is an important intermediate in chemical synthesis and occurs
naturally, most notably in ant venom.
Draw the Lewis structure of formic acid.
FORMAL CHARGE
➢ Certain molecules are possible to have more than one Lewis structure.
➢ However, the most stable structure is used to represent the molecule, which is
referred to as the most plausible Lewis structure.
➢ One way to determine the most plausible Lewis structure is based on concept of
formal charge.
➢ Formal charge is the difference between the valence electron in an isolated atom and
the number of electron assigned to that atom in a Lewis structure.
OR
➢ When several Lewis structures are possible, the most plausible Lewis structure is the
one with:
1. Zero formal charge on all atoms
2. Smaller formal charge (closest to zero)
3. Negative formal charges are placed on the more electronegative atoms.
➢ Taking the example of carbon dioxide, we can calculate the formal charge on each
atom and decide the most stable Lewis structure for carbon dioxide.
108 | C h a p t e r 7 C h e m i s t r y 1 D K 0 1 4
EXCEPTIONS TO THE OCTET RULE
➢ Some covalent molecules do not obey the octet rule.
➢ There are three exceptions to the octet rule.
1. Incomplete octet
2. Expanded octet
3. Odd number electrons
INCOMPLETE OCTET
➢ Occurs when the central atom has fewer than eight electrons in its valence shell.
➢ Elements in group 2 and 13 in period 2 with low metallic properties do not donate
but share the electrons.
➢ Metals: Beryllium (Be) and aluminium (Al), and the metalloid boron (B), exhibit
covalent character when they combine with non-metals.
➢ Be, B and Al cannot achieve octet configuration even after sharing electrons with
other atoms.
➢ Examples of incomplete octet compound are beryllium chloride,(BeCl2),aluminium
chloride (AlCl3) beryllium dihydride (BeH2) and boron trichloride (BCl3).
109 | C h a p t e r 7 C h e m i s t r y 1 D K 0 1 4
EXAMPLE 8:
Boron trifluoride is the inorganic compound with the formula BF3. This pungent
colourless toxic gas forms white fumes in moist air. It is a useful Lewis acid and a versatile
building block for other boron compounds.
Draw the Lewis structure of BF3 that obey the octet rule.
Calculate the formal charge of each atom.
Compare with the incomplete octet Lewisstructure of BF3 .
Answer:
Structure with incomplete octet is more stable because it has zero formal charges on all the atoms.
EXPANDED OCTET
➢ Occurs when the central atom has more than eight electrons in its valence shell.
➢ Usually involves non-metal atoms of 3rd period and beyond which have empty 3d
subshell.
➢ Example of expanded octet molecules are phosphorus pentachloride,(PCl5), and
sulphur hexafluoride, (SF6).
EXAMPLE 9:
Phosphorus pentachloride is the chemical compound with the formula PCl5. It is one of the
most important phosphorus chlorides, others being PCl3 and POCl3. PCl5 finds use as
a chlorinating reagent. It is a colourless, water- and moisture-sensitive solid, although
commercial samples can be yellowish and contaminated with hydrogen chloride.
Draw the Lewis structure of PCl5. State the type of exception to the octet rule.
Answer:
Expanded octet.
110 | C h a p t e r 7 C h e m i s t r y 1 D K 0 1 4
ODD NUMBER ELECTRONS
➢ Some molecules have an odd number of valence electron.
➢ Most odd–electron molecule have a central atom from an odd–numbered group, such
as N (Group 15) and Cl (Group 17)
➢ Example of odd number electrons molecules are nitric oxide and chlorine dioxide.
➢
EXERCISE:
3. By using Lewis dot symbol, show the formation of:
a) MgBr2
b) Al2O3
c) BaO
4. Draw the simplest Lewis structure for each of the following molecules. Comment on any
unusual features of the structure.
a) ClF3
b) BCl3
c) SeF4
7.2 MOLECULAR SHAPE
VALENCE SHELL ELECTRON PAIR REPULSION THEORY (VSEPR).
❑ The shapes of molecules and ions can be predicted by the Valence Shell Electron
Pair Repulsion(VSEPR) theory.
❑ Electron pairs around a central atom will repel one another and arrange themselves
as far as possible from each other in order to minimize repulsion.
❑ The followings are counted as one electron pair:
✓ Bonding pair no. of electron pairs = 4
- a single bond no. of electron pairs = 2
- a double bond no. of electron pairs = 4
- a triple bond no. of electron pairs = 4
✓ lone pair
❑ The order of the repulsion strength of lone pairs and bonding pairs of electrons is as
111 | C h a p t e r 7 C h e m i s t r y 1 D K 0 1 4
follows:
BASIC MOLECULAR SHAPES
There are five basic molecular shapes:
❑ Linear
❑ Trigonal planar
❑ Tetrahedral
❑ Trigonal bipyramidal
❑ Octahedral
❑ Electron pair arrangement is determined by the number of electron pairs around the
central atom.
Geometry No. of electron Arrangement General formula
pairs
2 Linear AX2
3 Trigonal planar AX3
4 Tetrahedral AX4
Trigonal
5 bipyramidal AX5
6 Octahedral AX6
112 | C h a p t e r 7 C h e m i s t r y 1 D K 0 1 4
SHAPES OF MOLECULE AND BOND ANGLES
How to predict the shapes of molecules using VSEPR theory?
❑ Draw the Lewis structure of molecule.
❑ Determine the electron pair arrangement.
❑ Use the electron pair arrangement to determine the molecular shape.
TWO ELECTRON PAIRS
Example : BeCl2 (gaseous beryllium chloride)
Lewis structure
❑ No. of electron pair 2 bonding pairs (AX2)
❑ Electron pair arrangement linear
❑ Repulsion of all bonding pair-bonding pair electrons are equal.
❑ Molecular shape linear; bond angle: 180o
THREE ELECTRON PAIRS
Example : BF3 (boron trifluoride)
❑ No. of electron pairs 3 bonding pairs (AX3)
❑ Electron pair arrangement trigonal planar
❑ Repulsion of all bonding pair - bonding pair electrons are equal.
❑ Molecular shape trigonal planar; bond angle: 120o
FOUR ELECTRON PAIRS
Example: CH4 (methane)
❑ No.of electron pairs 4 bonding pairs (AX4)
❑ Electron pairs arrangement tetrahedral
❑ Repulsion of all bonding pair - bonding pair electrons are equal.
❑ Molecular shape tetrahedral ; bond angle: 109.5o
FIVE ELECTRON PAIRS
Example: PCl5 (phosphorus pentachloride)
❑ No.of electron pairs 5 bonding pairs (AX5)
113 | C h a p t e r 7 C h e m i s t r y 1 D K 0 1 4
❑ electron pairs arrangement trigonal bipyramidal
❑ Repulsion of all bonding pair-bonding pair electrons are equal
❑ Molecular shape trigonal bipyramidal
❑ Bond angle: 120o, 90o
SIX ELECTRON PAIRS
Example: SF6 (sulfur hexafluoride)
❑ No.of electron pairs 6 bonding pairs, (AX6)
❑ Electron pairs arrangement octahedral
❑ Repulsion of all bonding pair-bonding pair electrons are equal
❑ Molecular shape octahedral
❑ Bond angle: 90o
EXERCISE:
5. Draw the molecular shape and predict the bond angles (relative to the ideal
angles) . What are the electron pairs arrangements and the molecular shapes of:
a) PF3
b) ClF2−
6. The nitronium ion, NO2+ is linear in shape but the nitrite ion NO2- is not linear.
Explain the observation.
7.3 INTERMOLECULAR FORCES
BOND POLARITY
➢ Bond polarity is a measure of how equally or unequally the electrons in a bond
are shared between the two atoms of a bond.
➢ A nonpolar covalent bond is one in which the electrons are shared equally, as in Cl2
and N2.
➢ A polar covalent bond, one of the atoms exerts a greater attraction for the bonding
electrons than the others.
➢ Atoms with different electronegativities form polar bonds.
➢ Depicted as a polar arrow:
➢ The arrow is denoted the shift in the electron density towards the more
electronegative atom.
114 | C h a p t e r 7 C h e m i s t r y 1 D K 0 1 4
➢
RESULTANT DIPOLE MOMENT
➢ Dipole moment is the quantitative measure of the magnitude of dipole and denoted as
➢ Its determined by molecular shape and bond polarity
EXAMPLE 10
ANSWER
Molecular shape of CO2 : linear
Symmetrical structure
2C=O bond are polar
The two bond dipoles cancel each other
Resultant dipole moment, µ= 0
CO2 is a nonpolar molecule
EXAMPLE 11
ANSWER
Molecular shape of OCS : linear
Unsymmetrical structure
The two bond dipole do not cancel each other
Resultant dipole moment, µ ≠ 0
OCS is a polar molecule
115 | C h a p t e r 7 C h e m i s t r y 1 D K 0 1 4
INTERMOLECULAR FORCES
➢ Attractive forces between neighbouring particles (atoms, molecules or ions).
➢ Exist when the molecules are sufficiently close to each other.
➢ Intermolecular forces of attraction are responsible for the physical properties such as
melting points, boiling points and solubility.
➢ Factors that influence the strength of van der Waals forces are molecular size,
molecular shape and molecular polarity.
➢
➢ There are two type of intermolecular forces which is van der Waals forces and
hydrogen bond.
➢ van der Waals forces are weaker than hydrogen bond
VAN DER WAALS FORCES
➢ van der Waals forces Is an attractive force between the neutral molecules
➢ Two type of van der Waals forces are dipole-dipole forces and London forces
or dispersion forces
Dipole-Dipole Forces
➢ The forces arise from the attraction of the permanently positive and negative end of a
neutral polar molecule.
➢ A more polar molecules (have large dipole moment), has a stronger dipole-dipole
forces.
➢ More energy is needed to overcome the strong dipole-dipole forces. Thus, its boiling
point increases.
➢ Thus, the more polar the molecules, the stronger the dipole-dipole forces, the higher
the boiling point.
EXAMPLE 12 Molecular Mr Dipole moment Boiling point (K)
Substance formula (D)
Propane CH3CH2CH3 44 0.1 231
Acetaldehyde CH3COH 44 2.7 294
Both propane and acetaldehyde have same relative molecular mass. However, acetaldehyde
has higher boiling point than propane. Explain.
116 | C h a p t e r 7 C h e m i s t r y 1 D K 0 1 4
Answer
• Acetaldehyde is polar molecule and has stronger dipole-dipole forces while propane
is non-polar molecule and has London dispersion forces.
• Dipole-dipole forces are stronger than London Forces. Therefore, the boiling point of
acetaldehyde is higher than propane.
LONDON FORCES @ DISPERSION FORCES
➢ Is a weak temporary force of attraction that occurs between non-polar molecules.
➢ Exist in all atoms and molecules.
➢ The forces arise because electrons in atoms or molecules are in constant motion.
➢ Are weak because the temporary dipole changes with the motion of the electrons.
➢ An atom has protons in its nucleus and electrons surrounding the nucleus. On
average, the negative charge of the atom is spread evenly around the nucleus.
➢ At any instant, one side of the molecules has higher electron density than the other
side.
➢ The atoms or molecules become a small instantaneous (temporary) dipole.
➢ The temporary dipole can induce a similar dipole on a neighbouring atom and can be
attracted to one another.
➢ This interaction is called London Forces @ Dispersion Forces and only significant
when atoms are close together.
HYDROGEN BOND
➢ Hydrogen bonding is a attraction between a hydrogen atom attached to a highly
electronegative atom F, O or N and partially negative lone pair electron on F, O, N of
another molecule.
➢ Hydrogen bonds are stronger than van der Waals forces but weaker than covalent
bonds.
➢ Hydrogen bond in HF
117 | C h a p t e r 7 C h e m i s t r y 1 D K 0 1 4
➢ Hydrogen bond in NH3
➢ Hydrogen bond in H2O
Hydrogen bond between
HF and H2O
Hydrogen bond between
NH3 and H2O
Hydrogen bond between
NH3 and HF
EXERCISE:
7. Name the type of Van der Waals forces that exist between molecules for each of the
following species:
a) I2
b) CH4
c) H2S
d) C6H6
8. Show hydrogen bonding in the following cases
a) Between ethanol molecule (C2H5OH)
b) between ethanol and water molecules.
118 | C h a p t e r 7 C h e m i s t r y 1 D K 0 1 4
TUTORIAL 7
SECTION A
1. The basic principle in the formation of electrovalent bond is
A the sharing of electrons which leads to the sharing of atomic orbitals.
B the transfer of electrons which leads to the formation of ions of opposite charges.
C the formation of positive ions which are bonded together in the negative charge of
electron cloud.
D the generation of instantaneous dipole moment arising from the differences of electron
affinities of two different atoms
2. The electronic configurations of element P and element Q are as follows:
P: 1s2 2s2 2p6 3s2
Q: 1s2 2s2 2p5
When P and Q combine to form a compound, determine the most possible formula of this
compound and the type of bonding between them.
Formula Types of bonding
A PQ2 covalent
B P2Q covalent
C PQ2 electrovalent
D P2Q3 electrovalent
3. The Lewis structure of SOCl2 is given as follows:
O
Cl S Cl
The formal charges on sulphur, chlorine and oxygen are
Sulphur Chlorine Oxygen
A0 0 0
B -1 +2 -1
C +2 -1 -1
D +2 -2 0
4. Identify the molecule that is T-shaped.
A IF3
B H2S
C SOCl2
D AlCl3
5. Which of the following pairs has the same molecular geometry?
A BF3 and PCl3
B PCl3 and NBr3
C NBr3 and AlCl3
D BeCl2 and O3
6. Choose a species with dipole moment NOT equal to zero ( μ ≠ 0 ).
A CCl4
B SF4
119 | C h a p t e r 7 C h e m i s t r y 1 D K 0 1 4
C XeF4
D PCl5
7. The structural formulae of three compounds are shown below:
H2O CH3CH2CH2-OH NH3
XY Z
Which of the following pairs forms the strongest hydrogen bond?
A X and X
B Y and Y
C X and Z
D X and Y
8. The boiling point of water is the highest among the hydrides of the Group 16 elements
because
A boiling point decreases down group 16.
B the size of water molecule is the smallest.
C water molecules can form hydrogen bond.
D the covalent bonds in the water molecules are strong.
SECTION B
1. (a) How is an ionic bond formed?
(b) By using Lewis dot symbol, show the formation of:
i. MgBr2
ii. Al2O3
iii. BaO
2. a) How is a covalent bond formed?
b) Use Lewis symbols to show the sharing of electrons for following molecules. Label the
electron pairs as bonding pair(s) and lone pair(s).
i. Cl2
ii. N2
iii. HBr
3. Draw the simplest Lewis structure for each of the following molecules. Comment on any
unusual features of the structure.
a) ClF3
b) BCl3
c) SeF4
d) NO
e) AlBr3
f) NCl3
120 | C h a p t e r 7 C h e m i s t r y 1 D K 0 1 4
4. Formaldehyde (CH2O), a liquid with an unpleasant odour, has been traditionally used to
preserve laboratory species. Draw the Lewis structure for formaldehyde. Determine formal
for each atoms.
5. Two resonance structures for NCO- are as follows:
Structure P Structure Q
N CO N CO
(a) Determine the formal charge of each atom in both structures P and Q.
(b) Determine the most plausible structure. Explain.
6. Using VSEPR model, predict the molecular geometries of:
(a) PBr5
(b) NO2+
7. a) How can a molecule be non-polar although it has polar bonds?
b) Explain whether each of the following molecules is polar or non-polar.
i) SF6
ii) BF3
iii) H2Se
iv) HBr
8. Name the type of van der Waals forces that exist between molecules for each
a) Br2
b) SO2
c) H2O
9. (a) What are dipole-dipole forces?
(b) How do polar molecules attract each other?
10. a) Describe hydrogen bond.
b) Ethanol, C2H5OH, and methyl ether, CH3OCH3, have the same molar mass.
Which has a higher boiling point? Explain.
Revised by:
Wan Rosilah binti Wan Llah
Zanarina Binti Sapiai
121 | C h a p t e r 7 C h e m i s t r y 1 D K 0 1 4
CHAPTER 8.0: CHEMICAL EQUILIBRIUM
GENERAL OVERVIEW
8.0 CHEMICAL 8.1 Dynamic a) Reversible i. Dynamic
EQUILIBRIUM equalibrium reaction equilibrium
8.2 Equilibrium b) ii. Law of mass
constant Characteristics action (law
chemical
c) Interpret the equalibrium)
curve
i. Homogeneous
a) Define(type)
ii. Heterogeneous
b) Deduce &
expressions i. Concentration,
c) Calculation of Kc
Kc, Kp at ii. Partial
equalibrium pressure, Kp
122 | Chapter 8 Chemistry 1 DK014
8.1 Dynamic Equilibrium
▪ Chemical equilibrium is achieved when the concentrations of reactant and
products cease to change and both the forward and reverse reactions are occuring
at the same rate.
▪ A reversible reaction is a chemical reaction that can proceed in both the forward
and reverse directions.
▪ For example
2SO2(g) + O2(g) ⇆ 2SO3(g)
▪ The symbol double-headed ⇌ is used to indicate a reversible reaction.
▪ The forward reaction is the one that goes to the right
2SO2(g) + O2(g) → 2SO3(g)
▪ The backward reaction is the one that goes to the left
2SO3(g) → 2SO2(g) + O2(g)
▪ SO2 and O2 combining to form SO3 at the same time as SO3 decomposed to form
SO2 and O2
▪ In a reversible reaction, initially the reaction proceeds toward the formation of
products
▪ As soon as some product molecules formed, the reverse process begins to take
place
▪ Reactant molecules are formed from product molecules.
Diagram 1
The graph of the concentration of A and B against time. Consider the reaction A ⇌ B
[A] decrease with time
[B] increase with time
After time, t1, [A] and [B] remains
unchanged.
Thus,the system is in the state of
equilibrium
EXAMPLE 1:
Nitrogen reacts with hydrogen to produce ammonia: N2(g) + 3H2(g) ⇌ 2NH3(g)
Write the balanced equation for the backward reaction.
Answer:
2NH3(g) → N2(g) + 3H2(g)
Dynamic equilibrium
Characteristics of Dynamic Equilibrium
i. Occurs in a closed system
ii. The rate of the forward reaction is equal to the rate of the reverse reaction
iii. The properties (such as concentration and partial pressure) for reactants
and products remain constant over time
▪ Does not mean that the reaction has stopped at equilibrium
123 | Chapter 8 Chemistry 1 DK014
Diagram 2
When the concentrations of product and
reactant are equal
EXAMPLE 2:
For a reaction,
P+Q ⇆ R
Sketch a graph to show the rate of reaction and product against time until reached
aquilibrium at condition
i. Concentration of reactant is larger than product
ii. Concentration of product is larger than reactant
Answer: ii.
i.
Law of Mass Action
▪ Also know as Law of Chemical Equlibrium
▪ The ratio of the product of equalibrium concentration of products to the product of
equalibrium concentrations of reactans, both raised to the power of their
stoichiometric in the balanced equation, is a constant at give temperature.
▪ A state of chemical equlibrium where both forward and reverse reactions proceed
at equal rates.
▪ It is applicable to both homogeneous and heterogeneous chemical equilibria.
2NO(g) + O2(g) ⇆ 2NO2(g)
▪ K = [NO2]2
[NO]2.[O2]
▪ The square brackets signify amount concentrations
▪ The constant K is the equlibrium constant.
EXAMPLE 3:
124 | Chapter 8 Chemistry 1 DK014
Write the the equilibrium constant expression for the reaction equation:
NH3(aq) + HCl(aq) ⇌ NH4+(aq) + Cl−(aq)
Answer:
[NH4+][Cl-] = K
[NH3][HCl]
Curve of Concentration Against Time for Reversible Reaction
N2O4 (g) ⇆ 2NO2 (g)
▪ For a particular system and temperature, the same equilibrium state is attained
whether forward or reverse reactions.
EXERCISE:
1. State the characteristic of a system at equilibrium
2. At equilibrium, there is no change in the concentrations of reactans and products,
yet it is considered dynamic. Explain
3. At some temperature, the equilibrium constant is 4.0 for the reaction
CO(g) + 3H2(g) ⇌ CH4(g) + H2O(g)
What is the equilibrium constant expression and value for the reaction,
CH4(g) + H2O(g) ⇌ CO(g) + 3H2(g) ? Answer: 0.25
8.2 EQUILIBRIUM CONSTANT
125 | Chapter 8 Chemistry 1 DK014
▪ Homogeneous Equlibrium is reactants and products are in the same phase
▪ Where all the species are in same physical state
▪ The concentration of all the species are included in the expression for equilibrium
constants
EXAMPLE 4:
Homogeneous Equilibrium
N2O4(g) ⇆ 2NO2(g)
2SO2(g) + O2(g) ⇆ 2SO3(g)
CH3COOH(aq) ⇆ CH3COO–(aq) + H+(aq)
▪ Heterogeneous Equlibrium is reactants and products are in different phases
▪ Where not all the species involved are in the same physical state
▪ The concentration of solids are not included in the expression for equilibrium
constants
EXAMPLE 5:
Heterogeneous Equilibrium
CaCO3(s) ⇆ CaO(s) + CO2(g)
2C(s) + 2H2O(g) ⇆ CH4(g) + CO2(g)
Equilibrium constant (K)
▪ Consider the following reaction
aA + bB ⇆ cC + dD
▪ At equilibrium : K = [ C ]c [ D ]d
[ A ]a [ B ]b
K = equilibrium constant
[ ] = concentration of reactants and products at equilibrium
a, b, c, and d = stoichiometric coefficients for the reacting species A, B, C and D.
▪ The value of an equilibrium constant will not change with the change in
concentration, pressure (volumes), or addition of catalyst.
▪ If the temperature change, the equilibrium constant will change.
▪ Increasing the temperature will increase the equilibrium constant, K as the product
concentrations increase and the reactant concentrations decrease.
Magnitude of Equilibrium constant (K)
▪ Indicate how far a reaction proceeds toward product at a given temperature.
▪ Large K value lies the equilibrium to the right (forward reaction) and favoring
formation of the products.
2CO (g) + O2 (g) ⇆ 2CO2 (g) K = 2.2 x 1022
▪ Small K value lies the equilibrium to the left (reverse reaction) and favoring the
formation of reactants
N2 (g) + O2 (g) ⇆ 2NO (g) K = 1 x 10-30
126 | Chapter 8 Chemistry 1 DK014
▪ Intermediate K value will make the equilibrium have significant amounts of both
reactant and product
2BrCl (g) ⇆ Br2 (g) + Cl2 (g) K = 5
Writing Kc Expressions
▪ Equilibrium involving concentrations, Kc
aA + bB ⇆ cC + dD
At equilibrium : Kc = [ C ]c [ D ]d
[ A ]a [ B ]b
Where [ ] Unit of concentration = M (mol/L)
The constant is known as the equilibria constant, Kc
▪ Equilibrium involving gases, Kp
At equlibrium ; Kp = (PC )c (PD )d
(PA )a (PB )b
Where P is the partial pressure of the gas at equlibrium
The constant is known as the equilibria constant, Kp
▪ Homogeneous Equilibrium
H2(g) + I2(g) ⇆ 2HI(g)
Kc = [HI ]2
[H2].[I2]
Kp = (PHI )2
(PH2).(PI2)
▪ Heterogeneous Equlibrium
▪ Where not all species involved are in the same physical state
▪ Since the density of any given solid and liquid is constant and changes very little
with temperature and it is constant
▪ The concentration of pure solids and pure liquids do not appear in the
equilibrium constant expression (can be neglected )
2C(s) + O2(g) ⇆ CO2(g)
Kc = [CO2 ]
[O2]
Kp = (PCO2 )
(PO2)
EXAMPLE 6:
127 | Chapter 8 Chemistry 1 DK014
Write the the expression Kc and Kp for the reaction.
i) 2SO2(g) + O2(g) ⇆ 2SO3(g)
ii) 4NH3(g) + 5O2(g) ⇆ 4NO(g) + 6H2O(g)
Answer: Kp = (PSO3)2
a) i) Kc = [SO3 ]2 (PSO2)2.(PO2)
[SO2]2.[O2] Kp = (PNO)4.(PH2O)6
(PNH3)4.(PO2)5
ii) Kc = [NO]4.[H2O]6
[NH3]4.[O2]5
EXAMPLE 7:
b) Write the the expression Kc and Kp for the reaction
CaCO3(s) ⇆ CaO(s) + CO2(g)
. Kp = PCO2
Answer:
a) Kc = [CO2]
Calculation Kc, Kp or quantities of species present at equilibrium
▪ Example
The equilibrium concentrations for the reaction between carbon monoxide, CO and
molecular chlorine, Cl2 to form COCl2 (g) at 74oC are [CO] = 0.012 M, [Cl2] = 0.054
M, and [COCl2] = 0.14 M. Calculate the equilibrium constants KC.
1. Write the expression Kc
Kc = [COCl2]
[CO][Cl2]
2. Fill the data (data must given [ ] at equlibrium )
= [0.14]
[0.012] x [0.054]
3. Calculate the final answer (no unit)
= 216
EXAMPLE 8:
The atmospheric oxidation of nitric oxide,
2NO(g) + O2(g) ⇆ NO2(g)
was studied at 184ºC with pressure at equilibrium, PO2 = 0.506 atm , PNO = 0.012 atm and
PNO2 = 0.988 atm. Calculate the equilibrium constants Kp.
Answer:
Kp = (P NO2 )
128 | Chapter 8 Chemistry 1 DK014
(PNO )2 (PO2 )
= 0.988
(0.012 )2 x 0.506
= 1.34 x 104
EXAMPLE 9:
In a study of the conversion of methane to other fuels, a chemical engineer mixes gaseous
CH4 and H2O in a 0.32 L flask at 1200 K. At equilibrium, the flask contains 0.8125 M of
CO, 0.2844 M of H2 and 0.1281 M of CH4 and Kc of the reaction is 0.26. What is [H2O] at
equilibrium?
CH4(g) + H2O(g) ⇆ CO(g) + 3H2(g)
Answer:
Kc = [CO] [H2]3
[CH4][H2O]
= (0.8125)(0.2844)3
(0.1281 ) . [H2O]
[H2O] = 0.56 M
EXAMPLE 10:
The equilibrium constant,Kp for the reaction,
2NO2(g) ⇆ 2NO(g) + O2(g)
Is 158 at 1000 K. What is the equilibrium pressure of O2 if the PNO = 0.270 atm and
PNO2 = 0.400 atm?
Answer:
Kp = (PNO )2 (PO2 )
= (PNO2 )2
(0.270)2. (PO2 )
(0.400)2
PO2 = 347 atm
EXAMPLE 11:
The equilibrium concentrations for the reaction between carbon monoxide, CO and
molecular chlorine, Cl2 to form COCl2 (g) at 74oC are [CO] = 0.012 M, [Cl2] = 0.054 M, and
[COCl2] = 0.14 M. Calculate the equilibrium constants KC.
CO(g) + Cl2(g) ⇆ COCl2(g)
Answer:
Kc = [COCl2]
[CO][Cl2]
= 0.14
0.012 x 0.054
= 216
▪ Guidelines of solving problems involving calculation of Kc and Kp quantities of
species at equilibrium
129 | Chapter 8 Chemistry 1 DK014
Step 1 – Analyse of problem
Step 2 – Construct ICE(Initial-Change-Equilibrium) table to determine the
equilibrium concentration/partial pressure(using quadratic if necessary).
Step 3 – Calcuate Kc, Kp or quantities (concentration/partial pressure) at
equilibrium
EXAMPLE 12:
The formation of ammonia, NH3 from nitrogen and hydrogen, H2 is according to following
equation:
N2(g) + H2(g) ⇆ 2NH3(g)
If 0.20 mol of N2 and H2 gas are placed into 1.0L vessel respectively at 400◦C, the
concentration of NH3 at equbrium is 0.12M. Calculate the equlibrium constant, Kc.
Answer:
N2(g) + 3H2(g) ⇆ 2NH3(g)
[ ]initial 0.20 = 0.20 0.20 = 0.20 0
[ ]change 1.0 1.0
[ ]equlibrium
-x -3x +2x
0.2-x 0.2-3x 2x
[NH3] = 2x = 0.12
x = 0.06
[N2] = 0.20-x
= 0.20-0.06
= 0.014
[H2] = 0.20-3x
= 0.20-3(0.06)
= 0.02
Kc = [NH3]2
[N2].[H2]3
= [0.12]2
[0.014].[0.02]3
= 1.29 x 104
EXAMPLE 13:
Initially, 1.0 mol of NO2 gas is allowed to reach equilibrium in 1.0L flask. Calculate the
concentration of NO2 and N2O4 gas at equilibrium if the KC for the following reaction is 8.3
2NO(g) ⇆ N2O4(g)
Answer:
[ ]initial 2NO(g) ⇆ N2O4(g)
[ ]change 1.0 0
[ ]equlibrium -2x
1.0-2x +x
x
Kc = [N2O4]2
[NO2]2
8.3 = x2
130 | Chapter 8 Chemistry 1 DK014
8.3(4x2 - 4x +1) [1.0 -2x]2
33.2x2 -33.2x + 8.3 =x
33.2x2 -34.2x + 8.3 =x
=0
x = 0.391(acceptable)
x @
[NO2] = 0.639 (not acceptable)
= 1.0 - 2(0.391)
[N2O4] = 0.218M
= 0.391M
EXERCISE:
1. The atmospheric oxidation of nitric oxide,
2NO(g) + O2(g) ⇌ NO2(g)
was studied at 184⁰C with pressure at equilibrium, PO2 = 0.506 atm , PNO = 0.012 atm
and P NO2 = 0.988 atm Calculate the equilibrium constants Kp.
2. Calculate KC for the following reaction given the following equilibrium concentrations of
H2, CO, and H2O with the concentration 0.030, 0.030 and 1.60 respectively.
C(s) + H2O(g) ⇌ CO(g) + H2(g)
3. N2O4 (l) is an important component of rocket fuel, At 25 ∘C N2O4 is a colorless gas that
partially dissociates into NO2. The color of an equilibrium mixture of these 2 gasses
depends on their relative proportions, which are dependent on temperature.
Equilibrium is established in the reaction N2O4(g) ⇌ 2NO2(g) at 25 °C, in 3.00 L of
container and 7.64 g of N2O4 and 1.56 g NO2. What is the Kc for this reaction?
4. For , Kp = 4.34 × 10−3 at 300 C . What is the value of Kp for the reverse reaction?
5. An aqueous solution of acetic acid is found to have the following equilibrium
concentrations at 25C: [CH3COOH] = 1.65 10−2 M, [H+] = 5.44 10−4 M; and
[CH3COO− ] = 5.44 × 10−4 M. Calculate the equilibrium constant Kc for the ionization of
acetic acid at 25 C. The reaction is CH3COOH(aq) ⇌ H+(aq) + CH3COO−(aq)
6. A closed system initially containing 1.0×10−3 M H2 and 2.0×10−3 M I2 at 448 C is
allowed to reach equilibrium, and at equilibrium the HI concentration is 1.87 10−3 M.
Calculate Kc at 448 C for the reaction taking place, which is H2(g) + I2(g) ⇌ 2HI(g)
Answer:
1. 1.34 x 104
2. 5.6 x 10-4
3. 4.61 x 10-3
4. 2.30 102
5. 1.79 10−5
6. 51
131 | Chapter 8 Chemistry 1 DK014
TUTORIAL 8
SECTION A:
1. Consider the following reversible reaction. In a 3.00 liter container, the following
amounts are found in equilibrium at 400 oC: 0.0420 mole N2, 0.516 mole H2 and
0.0357 mole NH3. Evaluate Kc.
N2(g) + 3H2(g) ⇆ 2NH3(g)
A. 0.202
B. 1.99
C. 16.0
D. 4.94
2. Which one of the following statements is correct?
A. The value of Kc gives us the rate of reaction.
B. Kc does not depend on temperature.
C. When Kc is very large, there are more products formed.
D. Kc never has units.
3. For the reaction
H2(g) + I2(g) ⇋ 2HI(g)
the equilibrium constant, Kc is 49 at a fixed temperature.
Two moles of hydrogen and two moles of iodine are allowed to reach equilibrium at
this temperature. What is the concentration of hydrogen iodide at equilibrium?
A. 9.2
B. 0.77
C. 1.55
D. 1.75
SECTION B:
1. (a) Define reversible reaction
(b) State two characteristics of a dynamic equilibrium
(c) Nitrogen (IV) dioxide dimerised as follows,
2NO2(g) ⇌ N2O4(g) ΔH = -58.84 kJ
Sketch a graph showing how the concentrations of the reactant and product of the
above reaction vary during the course of the reaction.
2 (a) Define homogeneous and heterogeneous equilibria.
(b) Determine whether the following reactions are homogeneous or heterogeneous.
i. 2PCl3(g) + O2(g) ⇌ 2POCl3(g)
ii. FeO(s) + CO(g) ⇌ Fe(s) + CO2(g)
132 | Chapter 8 Chemistry 1 DK014
iii. Ag+ (aq) + Fe2+(aq) ⇌ Ag(s) + Fe3+(aq)
iv. 3Fe(s) + 4H2O(l) ⇌ Fe3O4(s) + 4H2(g)
(c) Write the equilibrium law for the above reactions in terms of concentration, Kc, and
partial pressure, Kp.
3. At equilibrium, there are 2.50 moles of SO2 , 1.35×10-5 moles of O2 and 8.70 moles of
SO3 present in a 12.0 L flask. Calculate the equilibrium constant, Kc, for the reaction at
700oC.
2SO2(g) + O2(g) ⇌ 2SO3(g)
4. The partial pressures of PCl3, O2 and POCl3 at equilibrium are 5.05, 2.50 and 101
kPa respectively. Calculate the equilibrium constant, KP, for the reaction.
2PCl3(g) + O2(g) ⇌ 2POCl3(g)
5. Calculate the equilibrium constant, Kc for the following reaction if at 350 K the
equilibrium mixture contains 2.50 mol of N2, 3.50 mol of O2 and 0.06 mol of NO2 in
2.0 dm3 vessel.
N2(g) + 2O2(g) ⇌ 2NO2(g)
6. At 25oC, H2O(l) reached equilibrium according to the equation,
H2O(l) ⇌ H2O (g)
Calculate Kp. [vapour pressure of water at 25oC = 23.8 torr]
7. Nitrogen dioxide dimerised according to the equation,
2NO2(g) ⇌ N2O4(g) Kp = 7.13
At equilibrium, the partial pressure of NO2 in a container is 0.15 atm. Calculate the
partial pressure of N2O4 in the mixture.
8. 1.00 mol each of CO and Cl2 are introduced into an evacuated 1.75 L flask, at
395 0C. At equilibrium, the total pressure of the gaseous mixture is 32.4 atm.
Calculate Kp.
CO(g) + CI2(g) ⇌ COCl2(g)
9. Ammonium hydrogen sulfide decomposes according to the reaction.
NH4HS (s) ⇌ NH3 (g) + H2S (g) Kp = 0.11 at 250oC .
If 55.0 g of solid NH4HS is placed in a sealed 5.0 L container, what is the partial
pressure of NH3 and H2S at equilibrium?
10. Bromine gas is allowed to reach equilibrium according to the equation,
Br2(g) ⇌ 2Br (g) Kc = 1.1×10-3 at 1280oC
If the initial concentrations of Br2 and Br are 0.063 M and 0.012 M respectively,
calculate the concentrations of these species at equilibrium.
133 | Chapter 8 Chemistry 1 DK014
11. Consider the decomposition of dinitrogen tetroxide:
N2O4(g) ⇌ 2NO2(g)
In an experiment, the system reached equilibrium when 20% of dinitrogen
tetroxide in a closed vessel has dissociated. If the total pressure of the mixture is
800 kPa, what is the equilibrium constant, Kp at the temperature of the
experiment?
12. When the system A + B ⇌ C + D is at equilibrium,
A. Neither the forward nor the reverse reaction has stopped.
B. The forward reaction has stopped.
C. Both the forward and the reverse reactions have stopped.
D. The reverse reaction has stopped.
Revised by:
Khairul Aswari Bin Ab. Rahman
Zanarina Binti Sapiai
134 | Chapter 8 Chemistry 1 DK014
CHAPTER 9: REACTION KINETICS
GENERAL OVERVIEW a) Define reaction rate i. Experimental data
given
9.1 Rate of b) Determine
reaction the reaction ii. The graph of conc.
rate based against time
9.0 on:
REACTION i. Differential rate
KINETICS c) Write the rate equition.
expression in form
9.2 Factors of: ii. Rate Law
affecting reaction
d) Define order of i. Differential rate
rate reaction equation
e) Determine the rates ii. Rate Law
based on:
f) Determine the order of
reaction by initial rate
method.
a) Explain the i. concentration or pressure
effect of the ii. Temperature
following iii. Particle size
factors. iv. Catalyst
135 | Chapter 9 Chemistry 1 DK014
9.1 RATE OF REACTION
REACTION RATE
Reaction rate is the speed at which a chemical reaction occurs; a measure of the
decrease in a concentration of a reactant or the increase in concentation of product with
time. In other words, changes in concentrations of reactants or products as a function
of time.
Consider a reaction : A→B
A: reactant
B: product
As reaction proceed, the [A] is decreased while the [B] increased with time.
* [ ] = concentration
UNIT OF REACTION RATE
A→B
Rate = − [ ] = + [ ]
Explaination:
● The negative sign (-) indicates [A] decreases with time.
● The positive sign (+) indicates [B] increases with time.
● Unit [ ]: mol L–1 or mol dm−3 or M
● Unit of time: s (seconds), min, hr, day, year
● Unit of reaction rate: mol dm–3 min–1 or mol L–1s–1 or Ms–1
Diagram 1 : Reaction Rate: the centre focus of chemical kinetics.
136 | Chapter 9 Chemistry 1 DK014
GRAPH OF CONCENTRATION REACTANT AGAINST TIME
A(g) → B(g)
GRAPH OF CONCENTRATION PRODUCT AGAINST TIME
A(g) → B(g)
GRAPH OF CONCENTRATION REACTANT AND PRODUCT AGAINST TIME
A(g) → B(g)
INSTANTANEOUS RATE
Instantaneous rate is the rate at a particular instant during reaction. This instantaneous
rate is determined by slope of a line tangent to the curve at a particular point.
INITIAL RATE
Initial rate is the instantaneous rate at the moment reactants are mixed. This initial rate
is determined by slope of a line tangent to the curve at t = 0.
137 | Chapter 9 Chemistry 1 DK014
AVERAGE RATE
Average rate is the change in concentration of reactant or product over time.
Rate = ∆[ ]
∆[ ]
EXAMPLE 1:
C2H4(g) + O3(g) → C2H4O(g) + O2(g)
Table shows the concentration of O3 at various times.
Times (s) Concentration of O3 (mol/L)
0.0 3.2x10-5
10.0 2.42x10-5
20.0 1.95x10-5
30.0 1.63x10-5
40.0 1.40x10-5
50.0 1.23x10-5
60.0 1.10x10-5
138 | Chapter 9 Chemistry 1 DK014
(a) Determine the reaction rate during the first minute after introduce O3 gas.
Solution:
Rate = − [ 3]
= (1.10 X 10-5 mol/L ) – (3.20 X 10-5 mol/L)
(60.0s – 0.0s)
= 3.50 X 10-7 molL-1s-1
*ignore – sign
(b) Determine the reaction rate for first 10 second after introduce O3 gas.
Solution:
Rate = − [ 3]
= (2.42 X 10-5 mol/L ) – (3.20 X 10-5 mol/L)
(10.0s – 0.0s)
= 7.80 X 10-7 molL-1s-1
(c) Determine the reaction rate during the last 10 second after introduce O3 gas.
Solution:
Rate = − [ 3]
= (1.10 X 10-5 mol/L ) – (1.23 X 10-5 mol/L)
(60.0s – 50.0s)
= 1.50 X 10-7 molL-1s-1
EXAMPLE 2:
Calculate the reaction rate if the concentration of a product is 0.5 M at 10 s reaction time
and 1.0 M at 20 s reaction time?
Solution:
Rate = Concentrations change
Time change
Rate = ( 1.0 – 0.5 ) M = 0.05 Ms-1
( 20 – 10 ) s
EXAMPLE 3:
The reaction between C and D is represented as follows :
C+D→E
The experiment results are shown in the table below. Calculate the rate of reaction for each
experiment:
139 | Chapter 9 Chemistry 1 DK014
Experiment Initial Concentration (M) Time Interval The change in
[C] [D] (minutes) concentration
1
2 0.10 1.0 30 of C (M)
3 0.10 2.0 30
0.05 1.0 120 2.5 X10-5
1.0 X 10-2
5.0 X 10-3
Solution:
Rate = Concentrations change
Time change
Experiment 1 : Rate = 2.5 X 10-5 M = 8.33 X 10-5 M min-1
30 min
Experiment 2 : Rate = 1.0 X 10-2 M = 3.33 X 10-4 M min-1
30 min
Experiment 3 : Rate = 5.0 X 10-3 M = 4.17 X 10-5 M min-1
120 min
EXERCISE 1:
The decomposition of nitrosyl bromide is:
NOBr(g) → NO(g) + ½Br2(g)
Use the data in the table below to answer the following:
Time (s) [NOBr] (mol/L)
0.00 0.0100
2.00 0.0071
4.00 0.0055
6.00 0.0045
8.00 0.0038
10.00 0.0033
(a) Determine the average rate over the entire experiment.
(b) Determine the average rate between 2.00s and 4.00 s.
(c) Use graphical methods to estimate the initial reaction rate. (hint: use graph paper)
(d) Use graphical methods to estimate the rate at 7.00 s. (hint: use graph paper)
Answer:
a) 6.7 x 10–4 mol/Ls-1
b) 8.0 x 10–4 mol/Ls-1
c) +/-1.5 x 10–3 mol/Ls-1 *depend on graph
d) +/-2.9 x 10–4 mol/Ls-1 *depend on graph
140 | Chapter 9 Chemistry 1 DK014
DIFFERENTIAL RATE EQUATION
The rate expression in the form of differential rate equation shows the relationship between
the rate of disappearance of reactants and formation of products.
Example:
Let consider; aA + bB → cC + dD
● The negative sign indicates concentration of A and B decreases with time.
● The positive sign indicates concentration of C and D increases with time.
● Unit : mol L–1s–1 or mol dm−3 s−1 or Ms–1
EXAMPLE 4:
The formation of NH3,
N2(g) + 3H2(g) → 2NH3(g)
The differential rate equation is;
Solution:
d [N2] =- 1 d [H2] =+ 1 d [NH3]
Rate = - 3 dt 2 dt
dt
EXERCISE 2:
Write the rate expression (differential rate equation) for the following reaction:
(a) H2(g) + I2(g) → 2HI(g)
(b) CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
Answer: d [H2] d [I2] 1 d [HI]
(a) =- =-
Rate = - dt 2 dt
(b) dt
Rate = - d [CH4] = - 1 d [O2] = + d [CO2] = + 1 d [H2O]
dt 2 dt
dt 2 dt
RATE LAW
The rate expression in the form of rate law shows an equation that relates the rate of a
reaction to the rate constant and the concentrations of the reactants raised to some
powers. It show how the rate depends on the reactant concentration. For general equation:
aA + bB → cC + dD
The rate law generally has the form
Rate = k [A]m [B]n
141 | Chapter 9 Chemistry 1 DK014
Where :
[A] = concentration of A
[B] = concentration of B
k = rate constant
m = order of reactions with respect to A
n = order of reaction with respect to B
m + n = overall order of reaction
EXAMPLE 5:
2O3(g) → 3O2(g)
Write the general rate law for the above equation.
Solution:
Rate = k [ O3 ]m
EXERCISE 3:
N2(g) + 3H2(g) → 2NH3(g)
Write the rate equation for the above equation if the reaction is first order reaction respect
to N2 and second order reaction respect to the H2.
Answer:
Rate = k [ N2 ] [ H2 ]2
RATE CONSTANT (k)
For general reaction:
aA + bB → cC + dD
the rate law generally has the form
Rate = k [A]m [B]n
Notice that only the concentration of the reactant generally appear in the rate law. The
constant k is called rate constant. The magnitude of k changes with temperature and
therefore determines how temperature affect rate. So, rate constant is proportionality
constant for a given reaction and temperature. Does not change as reaction proceeds. You
can always determine the unit of k mathematically.
EXAMPLE 6:
What is the unit of k for rate = k[A]3 ?
Solution:
Rate = k[A]3
M/s = k (M)3
k = M-2s-1
142 | Chapter 9 Chemistry 1 DK014
EXERCISE 4:
Without consulting textbooks, give the units of the rate constants for reactions with the
following orders. Show the calculation. (Concentration: in M & Time: in s)
(a) Rate = k[A]
(b) Rate = k[A]2
(c) Rate = k[A]5/2
Answer:
(a) k = s–1
(b) k = M–1 s–1
(c) k = M-3/2s–1
REACTION ORDER: THE EXPONENTS IN THE RATE LAW
The rate law for most reaction has the form
Rate = k [reactant 1]m [reactant 2]n......
The exponent of m and n are called reaction orders. For example, consider again the rate
law for the reaction of F2 with ClO2:
rate = k [F2][ClO2]
Because the exponent of [F2] is 1, the rate is first order in F2. The rate is also first order in
ClO2. (The exponent 1 is not shown in rate laws). The overall order is the sum of the orders
with respect to each reactant represented in the rate law. Thus for the reaction between F2
and ClO2 the rate law has an overall order reaction order of 1 + 1 = 2, and the reaction is
second order overall.
The exponents in a rate law indicate how the rate affected by each reactant
concentration. The reaction orders cannot be predicted from the stoichiometry of the
balanced equation; it must be determined experimentally. In the most rate laws, reaction
order is 0, 1, or 2. However, we also occasionally encounter rate laws in which the reaction
order is fractional or even negative.
i. ZERO-ORDER REACTION
For reaction;
aA → Products
Rate = k [A]0 or Rate = k
This reaction is zero-order with respect to A. A zero-order reaction is one in which the rate
of disappearance of A is independent of [A]. The most common type of zero-order reaction
occurs when a gas undergoes decomposition on the surface of a a solid.
143 | Chapter 9 Chemistry 1 DK014
ii. FIRST-ORDER REACTION
For reaction;
aA → Products
Rate = k [A]1
This reaction is first-order with respect to A. A first-order reaction is one whose rate
depends on the concentration of a single reactant raised to the first power. It means The
rate depends linearly on [A]. If [A] is double, the rate doubles. Example:
[ A ] = [ 2A ]
Rate = k [2A]
Rate = 2 k[A]
Example of first-order reaction is radioactive reaction (Uranium, Plutonium and etc)
iii. SECOND-ORDER REACTION
For reaction;
aA → Products
Rate = k [A]2
This reaction is second-order with respect to A. A second-order reaction is one for which
the rate depend either on a reactant concentration raised to the second power. It
means When [A] is doubled, the rate increases by factor of 4. Example:
[ A ] = [ 2A ]
Rate = k[2A]2
Rate = 4 k[A]2
Example of second-order reaction is a dimerization reaction
EXAMPLE 7:
For the following reactions, determine the reaction order with respect to each reactant and
the overall order from the given rate law:
2NO(g) + 2H2(g) → N2(g) + 2H2O(g)
Rate = k[NO]2[H2]
Solution:
● Second order with respect to NO
● First order with respect to H2
● Overall reaction is third order reaction
144 | Chapter 9 Chemistry 1 DK014
EXERCISE 5
For each of the following reactions, determine the reaction order with respect to each
reactant and the overall order from the given rate law:
(a) 2NO(g) + O2(g) → 2NO2(g) Rate = k[NO]2[O2]
(b) CH3CHO(g) → CH4(g) + CO(g) Rate = k[CH3CHO]3/2
(c) H2O2(aq) + 3I–(aq) + H+(aq) → I3–(aq) + 2H2O(l) Rate = k[H2O2][I–]
Answer:
(a) Second order with respect to NO
First order with respect to O2
Overall reaction is third order reaction
(b) 3/2 order with respect to CH3CHO
Overall reaction is 3/2 order reaction
(c) First order with respect to H2O2
First order with respect to I-
Zero order with respect to H+
Overall reaction is second order reaction
DETERMINATION OF REACTION RATES BASED ON:
I. DIFFERENTIAL RATE EQUATION
Differential rate equation can be relating rates at which product appear and reactant
dissappear. For example:
Hydrogen sulfide burns in oxygen to form sulfur dioxide and water will produce the equation
below;
2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(g)
If sulfur dioxide (SO2) is being formed at a rate of 0.30 mol L–1 s–1, what are the rates of:
i. Disappearance of hydrogen sulfide (H2S)?
ii. Formation of water (H2O)?
Solution:
2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(g)
1 d [H2S] = - 1 d [O2] 1 d [SO2] 1 d [[H2O]
=+ =+
Rate = -
2 dt 3 dt 2 dt 2 dt
Given; d [SO2]
rate = + dt = 0.30 mol L-1 s-1
145 | Chapter 9 Chemistry 1 DK014
i. Disappearance of hydrogen sulfide (H2S).
Rate = - 1 d [H2S] =+ 1 d [SO2]
2 dt 2 dt
Rate = - d [H2S] 1 d [SO2] X2
=+
dt 2 dt
- d [H2S] = + 1 (0.30 mol L -1 s-1) X 2
dt 2
= 0.30 molL-1s-1
Rate of disappearance of H2S = 0.30 mol L–1s–1
ii. Formation of water (H2O).
Rate = 1 d [H2O] 1 d [SO2]
+ 2 dt = + 2 dt
Rate = + d [H2O] = + 1 d [SO2] X 2
dt 2 dt
+ d [H2O] = + 1 (0.30 mol L -1 s-1) X 2
dt 2
= 0.30 molL-1s-1
Rate of formation of H2O = 0.30 mol L–1s–1
II. RATE LAW
If both of the rate law and the k value for a reaction are known, we can calculate the reaction
rate for any set of concentration. For example:
The reaction A + B → C + D is second order in A and zero order in B. The value of k is
0.0103 M–1 min–1. What is the rate of this reaction when [A] = 0.116 M and [B] = 3.83 M.
Solution:
Rate = k[A]2[B]o
= k[A]2
= 0.0103 M–1 min–1[0.116 M]2
= 1.39 x 10–4 M min–1
146 | Chapter 9 Chemistry 1 DK014