Chemistry Lesson Note For SS2 (First Term) 2024

Chemistry lesson note for SS2 First Term is now available for free. The State and Federal Ministry of Education has recommended unified lesson notes for all secondary schools in Nigeria, in other words, all private secondary schools in Nigeria must operate with the same lesson notes based on the scheme of work for Chemistry.

Chemistry lesson note for SS2  First Term has been provided in detail here on schoolgist.ng

For prospective school owners, teachers, and assistant teachers, Chemistry lesson note is defined as a guideline that defines the contents and structure of Chemistry as a subject offered at SS level. The lesson note for Chemistry for SS stage maps out in clear terms, how the topics and subtopics for a particular subject, group works and practical, discussions and assessment strategies, tests, and homework ought to be structured in order to fit in perfectly, the approved academic activities for the session.

To further emphasize the importance of this document, the curriculum for Chemistry spells out the complete guide on all academic subjects in theory and practical. It is used to ensure that the learning purposes, aims, and objectives of the subject meant for that class are successfully achieved.

Chemistry Lesson note for SS2 carries the same aims and objectives but might be portrayed differently based on how it is written or based on how you structure your lesson note. Check how to write lesson notes as this would help make yours unique.

The SS2 Chemistry lesson note provided here is in line with the current scheme of work hence, would go a long way in not just helping the teachers in carefully breaking down the subject, topics, and subtopics but also, devising more practical ways of achieving the aim and objective of the subject.

The sudden increase in the search for SS2 Chemistry lesson note for First Term is expected because every term, tutors are in need of a robust lesson note that carries all topics in the curriculum as this would go a long way in preparing students for the West African Secondary Examination.

This post is quite a lengthy one as it provides in full detail, the Chemistry-approved lesson note for all topics and sub-topics in Chemistry as a subject offered in SS2.

Please note that Chemistry lesson note for SS2 provided here for First Term is approved by the Ministry of Education based on the scheme of work.

I made it free for tutors, parents, guardians, and students who want to read ahead of what is being taught in class.

SS2 Chemistry Lesson Note (First Term) 2024

FIRST TERM E-LEARNING NOTES

SUBJECT: CHEMISTRY                                                                       CLASS: SS 2

SCHEME OF WORK

WEEK     TOPIC

THEME: THE CHEMICAL WORLD

1. Periodic Table:Periodic law, Electronic configuration of first thirty elements, Blocks

of elements – metals, non-metals, metalloids and transition metals, families of

elements

2. Periodic Table (cont’d):Atomic properties and the periodic table, Diagonal Relationship between elements in the periodic table.
3.      Chemical Reactions:Basic concept (Activation energy) and Introduction to collision

theory, Factors affecting rate of chemical reactions, Types of chemical reactions.

4.         Chemical Reactions (cont’d):Chemical equilibrium: characteristics of a chemical

system at equilibrium, Introduction of chemical equilibrium using simple equations,

Factors affecting equilibrium of chemical reaction and Equilibrium constant, Kc

5. 5.        Mass – Volume Relationships:I unit of quantities and Basic concepts (mole, molar

quantities, molality, standard temperature and pressure (S.T.P), Calculations involving

mass and volume.

6.       Acid-Base Reactions:Common indicators and their pH ranges, standard solution and

Simple acid-base

titration, , Heat of Neutralization.

7. 7.       Mid –Term Break and Revision

THEME: CHEMISTRY AND ENVIRONMENT

8.     Water:Structure of water and Solubility (basic concepts: solute, solvent, solution),

Factors affecting solubility and uses of solubility curve, Hardness of water and removal

sof hardness, Municipal water supply.

9.            Air:Constituents of air, Percentage composition, Properties of air, Flame, noble

gases and their uses

10. Revision.
11. Examination.

WEEK 1:

SUBJECT: CHEMISTRY

CLASS: SS2

TOPIC: PERIODIC TABLE

CONTENT:

1. Periodic law
2. Electronic configuration of the first thirty elements
3. Blocks of elements – metals, non-metals, metalloids and transition metals
4. Families; S-P-D-F ( according to group I-VIII to group IA- Alkalis metals, group IIA- Alkali earth and other family names)

PERIOD 1 AND 2 : PERIODIC LAW

Periodic Table: This shows the arrangement or grouping of elements in order of increasing atomic number.

Periodic Law: Is the basic assumption behind the modern periodic table; it states that the properties of the elements are the periodic function of their atomic number.

ELECTRONIC CONFIGURATION OF THE FIRST THIRTY ELEMENTS

The lectronic configuration  of an atom is the representation of the arrangement of the electrons distributed among the orbital shells and subshells. Commonly, the electronic configuration is used to describe the orbitals of an atom in its ground state

MEANING OF ATOMIC ORBITAL

Orbital is the region of space around the nucleus where there is a high probability of finding electron. The four different types of orbitals are s, p, d, and f. These orbitals have different shapes and one orbital can hold a maximum of two electrons. The p-orbital has three degenerate orbitals, with a maximum of six electrons, d has five sub orbitals with a maximum of ten electrons and the f-orbital has seven sub-orbitals with maximum of fourteen electrons.

RULES   AND PRINCIPLES FOR FILLING IN ELECTRONS

1. Aufbau Principle:states that electrons are filled in their orbital in order of increasing energy level. The order is as follows:

1s < 2s < 2p< 3s < 3p <4s < 3d <4p etc.

2. Pauli’s Exclusion Principle: states that two electrons in the same orbital of an atom cannot have same values for all quantum numbers. Usually the first electron in an orbital is assumed to enter with up-spin (clockwise ),↑ while the second electron enters the same orbital but with a down-spin (anti-clockwise ), ↓.
3. Hund’s Rule of maximum Multiplicity:This rule states that electrons occupy each orbital singly first before pairing takes place in a degenerate orbital. Hence, no pairing in degenerate orbital until each orbital is singly occupied with parallel spin.

EVALUATION:

1. Draw a periodic table showing the 1^st20 elements
2. Define periodic law
3. Define periodic table.
4. Write the electronic configuration of the following elements: Cu, Na, Al, and Cl

PERIOD 3: BLOCKS OF ELEMENTS

From the position of the various elements on the periodic table and the electrons arrangements or configuration seem to stand out. Elements that have one and two electron(s) in   their last s- orbital i.e. group 1 and 2 elements are called s-block elements.  P-block elements have their last electron (s) in the p-orbital are called P block elements.

Elements that have their last electrons in the d orbital are called d blocks elements while f block elements have their last electrons in the f- orbital.

Moving across a particular period on the periodic table, two elements are present in the S block, six elements in the P block, ten elements in the d- block and fourteen elements in the f- block. This corresponds with the maximum number of electrons in the s, p, d, and f orbitals respectively. Also, S block elements are metals, P block elements are mostly non-metals d- block elements  are transition metals while f-block elements are lanthanides (rare earth metals) and actinides (heavy rare earth metals).

There is therefore a diagonal division of the elements into metals and non-metals as shown below. Metals are found on the left side of the thick boundary line and the non-metals on the right with metalloids occurring along the boundary line.

CHARACTERISTICS PROPERTIES OF TRANSITION ELEMENTS

1. Paramagnetism: Transition elements exhibit paramagnetism  because of the presence of unpair elecrons in the d-orbital. Paramagnetism is the ability of an element to align to the poles of a magent.
2. Variable oxidation states: Transition elements exhibit variable oxidation states because they can lose electrons from both 4s and 3d orbitals for bond formation e.g we have Fe²⁺ and Fe³⁺
3. Complex ion formation:Transition elements form complex ions due to the presence of vacant or empty d- orbitals in their ions.
4. Catalytic ability:The catalytic ability of transition metals is due to the fact that they exist in different oxidations states, hence they are used as catalyst. The partially filled d-orbital allows the exchange of electrons to and from molecules which enable them to act as catalyst.

EVALUATION:

1. State the blocks of the following elements:  H, F, Na, Ca, Fe and Mn
2. State three properties of transition elements and explain any two of the properties.

PERIOD 4:  FAMILIES OF ELEMENTS.

Elements are arranged into groups or families and periods. Each group has been given a name to allow proper understanding during discussions on the periodic table.

Names of the different group (1-8) on the periodic table are shown below.

Group (i) i.e. Li, Na, K etc. ——– alkaline metals

Group (ii) i.e. Be, Mg, Ca, etc. —— Alkaline earth metals

Group (iii) i.e. B, Al, Ga, etc. ——- Boron family

Group (iv) i.e. C, Si, Ge, etc. ——— Carbon family

Group (v) i.e. N, P, As, etc. ———- Nitrogen family

Group (vi) i.e. O, S, Se, etc. ——— Oxygen family or Chalcogens

Group (vii) i.e. Fl, Cl, Br, etc. —— Halogens

Group (viii) or group O i.e. He, Ne, Ar, etc. —— Rare, inert, noble gases.

Transition Elements: These are between groups 2 and 3 on periods 4 and 5.

Lanthanides (Rare- earth elements): these are fifteen elements La- Lu.

Actinides: These are found on the seventh period AC –Lr.

Artificial Elements: These are elements with atomic numbers 93 to 103. They are products formed during chemical reactions. They are unstable and disintegrate in a short time e.g. Plutonium (Pu) and curium (Cm)

EVALUATION:

1. Four elements P, Q, R and S, have 1, 2, 3 and 7 electrons in their outermost shell respectively. What is the element that is unlikely to be a metal?
2. The elements listed below belong to the same group in the periodic table; ₉F, ₁₇Cl, ₃₅Br, ₅₃I

• What is the electronic structure of the first-member?
• What is the family name of the elements?
• Which of the elements has the strongest oxidizing ability?

GENERAL EVALUATION

OBJECTIVE TEST;

1. The period table is an arrangement of elements in order of their. (a) Relative molecular mass (b) mass numbers (c) molecular masses (d) atomic numbers.
2. Four elements T, X, Y, Z have 1, 2, 3 and 6 electrons in their outmost shells respectively. The element that is likely to be a non-metal is.      (a) X (b)T (c) Z (d) Y
3. An atom with the electronic configuration: Is², 2s², 2p⁶, 3S², 3d¹, 4s²is that of:

(a) an alkali metal (b) an alkali earth metal (c) a transition metal (d) a halogen

4. An element whose atomic number is 19 is

(a) a noble gas (b) an alkali earth metal (c) An alkali metal (d) a transition metal

5. Group 0 or 8 elements are called.  (a) Active elements (b) lanthanides (c) noble gas (d) alkali earth metals.

ESSAY QUESTIONS

1. A shortened form of the periodic table is shown below. Use it to answer questions a and b.

I    II III IV V VI VII O

1. A

2. B

C

3.

4.           D                E

• Which of the elements represented as A to E in the table above is
• A transition metal; (ii) an alkaline earth metal

(iii) The least reactive (iv) the most electronegative

2. State the four properties of transition elements (d-block)
3. Name one element in each case which

(a) forms an amphoteric oxide;

(b) Liberates hydrogen from cold water;

(c) Exists as a liquid at room

4. Give the names of the members of the alkaline earth metal family.

WEEK 2

TOPIC: PERIODIC TABLE

CONTENT:

1. Periodic properties
2. Periodic gradation of the element in the third period
3. Diagonal relationships

PERIOD 1 AND 2 : PERIODIC PROPERTIES. Some properties of the atom change along a group or across a period on the periodic table. Atomic radius which is measured of the size is one of such properties. The orbiting electrons in an atom are best represented by an electron cloud which has no distinct limit as the size of an action cannot be defined easily.

1. Atomic radius:This has been defined as the distance of closest approach to another identical atom in a given bonding situation. There are two types of atomic radii. Covalent radius and Van der Waals radius. Covalent radius is half the distance between two identical atoms which are not chemically bonded. For the two types of atomic radius two variations are noticeable:

• The atom radius increases down a group
• The atomic radius decreases along a period.

This is because going down any group on the periodic table the number of valence electrons remains constant while the shells increase in size (radius) despite increase in nuclear charge. The atomic radius of potassium is greater than that of Sodium. The atomic radius of caesium is greater than that of rubidium.

Across a period, electrons are added to orbitals in the same shell, all the valence electrons are therefore at the same energy level. As atomic number increase the positive charge of the nucleus increases giving rise to greater attraction between the positive nucleus and negative electrons. This is turn result in contraction of the electrons cloud resulting in a smaller atom. Atomic radii therefore decrease across a given period on the periodic table.

2. Ionic Radius: Ions are formed by a loss or gain of electrons by an atom. A positive ion (cation) is smaller than the original metal atom because electrons are pulled in due to increase in effective nuclear charge.

A negative ion (anion) is bigger than the corresponding non- metal atom because the effective nuclear charge is reduced.

As we move across the second short period, the cationic radii decrease from sodium to aluminium while the anionic radii increase from phosphorous to chlorine.

3. Ionization energy: Ionization occurs when gaseous atom loses electrons from its outer most shell to become positively charged

K

The energy required to do this is called ionization energy or ionization potential.

First ionization energy of an element is the energy needed to remove one mole of electron(s) from one mole of atoms in the gaseous state. It is expressed in kilo- joules per mole of atoms ionized.

First ionization energy increase  across the period with noble gases having the highest. As we go down he the group, the value of first ionization energy decreases.

FIRST IONIZATION ENERGIES OF ALKALI METALS

FIRST IONIZATION ENERGIES OF THE ELEMENTS IN THE THIRD PERIOD OF THE PERIODIC TABLE

Three factors that affect the ionization potential of an atom

Ionization potential of an atom is affected by.

• Distance of the outer most electrons from the nucleus.
• Size of the positive or effective nuclear change.
• Screening effect of the inner electrons.

Moving from left to right across a period, there is a general rise in the first ionization energy. This is due to the fact that the nuclear charge is increasing across the period. This in turn causes a decrease in atomic radius that is a decrease in the distance of the outermost electrons from the nucleus. The screening effect is almost the same across the period.  Down a group of the periodic table, ionization energy decreases because the nuclear charge on the outermost electron is reduced. The outermost electron are properly shielded from the effect of nuclear charge

4. Electron Affinity:is the energy released when an electron is added to gaseous atom in its lowest energy state. Its unit is kJmol^-1 or electron volts (ev). Electron affinity increase across a period from left to right and decrease down the group on the periodic table.

Group 1 elements, alkali metals have the least tendency to add electrons to their neutral atoms.

Elements in groups VI and VII have greatest tendency to accept electron. Noble gases (group 8 or 0) have stable electronic configuration

5. Electronegativity:Electronegativity is the ability or power of that atom in a molecule to attract shared pair of electrons. It is more pronounced in heteronuclear molecules where two dissimilar atoms share one or more pairs of electrons.

Electronegativity increases across the period, i.e. going from left to right of the Periodic Table but decreases down the group i.e. going down the Periodic Table. The steady increase as one goes across the period is due to a steady increase in nuclear charge and decrease in atomic size. Consequently, the halogen atom, Fluorine, has the highest electronegativity in the period, due to the strong affinity for electrons. But down the group, the increase in atomic size due to screening effect of the inner shells of electrons decreases the nuclear attraction for shared electrons. The noble gases of group O are not assigned electronegativity values since they have completed shells of electrons.

EVALUATION

1. List three periodic properties of elements that generally increase the across the period of the Periodic Table.
2. Explain the term electron affinity and discuss how it varies across the period and down the group of the Periodic Table.

PERIOD 3: PERIODIC GRADATION OF THE ELEMENTS IN THE THIRD PERIOD (Na-Ar)

Gradation in properties is not confined only to the elements, but it is also found in their compounds with increasing atomic number.

The extent of hydrolysis of the chlorides changes across the third period. Sodium chloride is not hydrolyzed at all in aqueous solution. The same applies to magnesium chloride although hydrated crystals undergo hydrolysis when heated given off HCl and leaving a basic salt. An aqueous solution of aluminium chloride shows appreciable hydrolysis and turns blue litmus red. The chlorides of silicon, phosphorus and Sulphur hydrolyze completely in water.

The general conclusion from the above is therefore as follows: From left hand side to right hand side across any period of representative elements, the metallic character, i.e. tendency to lose electron(s) decreases, and the non-metallic character, i.e. tendency to gain electron(s) increases. Also, as one goes across the period, ionic property decreases while covalent property increases.

PERIOD 4: DIAGONAL RELATIONSHIP

Because metallic character increases down a group and decreases from left to right along a period, there exists a diagonal relationship between the chemical properties of the first member of a group and that of the second member of the next group as in the cases of lithium and magnesium on one hand, and beryllium and aluminium on the other (see the periodic table)

EVALUATION          

1. The atoms of four elements are represented as ₂₀Q, ₁₆R, ₁₀S and ₈ Which of the elements would be unreactive?
2. Explain the meaning of the diagonal relationship the periods 2 and 3 of elements in the Periodic Table.

GENERAL EVALUATION

OBJECTIVE TEST:

1. An element which has atomic number 11, is likely to have similar chemical properties with another element having atomic number.     (a) 7 (b) 9 (c) 18 (d) 19
2. If the electronic configuration of an element X is 1s², 2S², 2P⁵, which of the following statements about X is false? It   (a) Have five electrons in its valence shell. (b) Has atomic number 9   (c) Is a halogen  (d) Is a non-metal
3. The electronic configuration of an element X is 1s², 2s², 2p⁶ , 3s²,3P⁴. It can be deduced that X (a) Belongs to group 6 of the periodic table.(b) Belongs to period IV of the periodic table.(c)  Has 3 unpaid electrons in its atom. (d) belongs to group 4
4. Which of the properties of the atom increases across the period and down the group in the periodic table? (a) Atomic radius (b) electronegativity (c) electropositivity (d) Ionic radius
5. Which of the following group 1 elements has the highest Ionization energy? (a) LI (b) Na (c) K (d) Rb

ESSAY QUESTIONS

1. (a) What are isoelectronic ions? (b) Give two examples, one each of a cation and an anion, which are isoelectronic with neon.
2. (a) Which of the following elements: calcium fluorine, iodine, neon, magnesium and helium are:  (a) Halogens (b) noble gas (c) alkaline earth metal
3. List three periodic properties of elements that generally increase across the periodic table.
4. Arrange the following elements in order of decreasing electro negativity.

₁₆S, ₁₃Al, ₁₁Na, ₁₅P

5. State the periodic law.

WEEK 3

TOPIC: CHEMICAL REACTIONS

CONTENTS:

1.     Basic concept (Activation energy) and
2.     Introduction to collision theory.
3.      Factors affecting rate of chemical reactions.
4.     Types of chemical reactions – Endothermic and Exothermic.

PERIOD 1: BASIC CONCEPTS: REACTANTS AND PRODUCTS

Chemical reactions can be represented simply on paper by chemical equations involving only formula and symbols e.g.

4NH_3(g)       +        O_2(g)                            4NO_(g) + 6H₂O_(g)

Reactants                                                Products

Reaction time: The total time taken for a particular reaction to take place.

Reaction rate: is the number of moles of reactions converted or product formed per unit time.  Unit is moldm^-3s^-1

In all chemical reactions, existing bonds in the reactants particles must be broken first before new bonds can be formed to form products. The breakings of bonds require energy. An initial energy input is required to activate the reactant particles. This energy is the activation energy of the reaction. The activation energy of a reaction is equivalent to that energy barrier that must be overcome before bonds are broken to enable the reaction to occur. Hence, activation energy can be defined as the minimum amount of energy required for a reaction can occur.  When this activation is acquired by the reactant particles, they form complex particles of high energy content. This complex particle is known as the activated complex. Therefore activated complex is an unstable molecule with high energy contents which gives the product  It is unstable because of its high energy content and so will readily decompose to give the products or the reactants, depending on the nature if the reaction.

EVALUATION

Define activation energy.

PERIOD 2:   INTRODUCTION TO COLLISION THEORY

The collision theory was developed from the kinetic theory of gases to account for the kinetic theory of gases to account for the influence of concentration and temperature on reaction rates. The theory is based on the following postulates:

1. Reactions occur as a result of the collision of reactant particles.
2. A reaction results only if collision attains certain minimum energy. This minimum energy is called activation energy.

iii. Collision will not give rise to a reaction unless the colliding particles are correctly oriented to one another.

1. The rate of reaction is proportional to the number of effective collisions. A collision is said to be effective if the energy of the colliding molecules is greater than or equal to the activation and the molecules are correctly oriented to one another.

The collision theory assumes that for a chemical reaction to occur there must be collisions between reactant particles.

EVALUATION

Explain the collision theory.

PERIOD 3: FACTORS AFFECTING THE RATE OF CHEMICAL REACTION

The rate at which reaction takes place will be affected by the following factors:

1. Nature of reactant: The chemical nature of the reactants taking part in the reaction influences the rate of reaction. For instance if iron, zinc and gold metals are placed in different beaker continually Hydro chloric acid, there will be rapid evolution in the beaker containing the acid and zinc metal wherein the beaker containing Iron metal and the acid there will be slower evolution of hydrogen gas and in the beaker containing gold metal and the acid solution there will be no reaction.
2. Effect of concentration / pressure (gases) of reactants: If reactant particles are crowded in a particular place, their frequency of collision will be faster than if the particles are far from one another. Thus, the more the concentration of reactant particles the higher the rate of reaction.

Pressure affects the concentration of gaseous reactant, the higher the pressure of gaseous reactants, the higher the frequency of collision of the particles the higher the rate of reaction.

3. Effect of surface area of the reactants: the more exposed the area of contact of reacting particles to each other the faster the rate of reaction. For solid reactants the exposed surface area must be increased by subdividing or breaking the solid into smaller pieces.

Where the reactants are gases, liquids or solids dissolved in solution, the thoroughness of mixing is of vital importance so as to ensure maximum contact between the reactants.

4. Effect of temperature of reaction mixture: The higher the temperature the higher the rate of reaction of most reactions. When the temperature of a chemical reaction is increased, the number of particles with energies equal to or greater than the activation energy increases. Also the kinetic energy of the reactant particles increase which implies increase in frequency of effective collision and increase in the rate of reaction.
5. Effect of the presence of light: Reactions whose rates are affected by light are called photochemical reactions. They are thus able to overcome the activation energy barrier and react rapidly by a chain reaction.

Examples: i. The decomposition of hydrogen peroxide

1. The reactions between methane and chlorine Photosynthesis

iii. The conversion of silver halides to grey metallic silver

1. The halogenations of alkanes
2. Effect of catalyst:catalyst will alter the rate of a chemical reaction but itself does not    undergo any permanent change at the end of the reaction.

Catalysts that speed up the rate of a chemical reaction are called positive catalyst. They lower the activation energies of the reactant particles  by providing alternative pathway so more reactant particles are able to collide effectively to produce more products. E.g. Manganese(IV) oxide, MnO₂ catalysts is used in the production of oxygen thermal decomposition of KClO₃

EVALUATION

List and explain 5 factors that affect the rate of chemical reaction.

PERIOD 3: TYPES OF CHEMICAL REACTIONS: ENDOTHERMIC AND EXOTHERMIC REACTIONS

An endothermic reaction is one during which heat is absorbed from the surrounding. Most decomposition processes are endothermic reaction.  Dissolution of ammonium chloride, ammonium tetraoxosulphate(vi) etc are endothermic process. In endothermic reaction, the enthalpy of product(s)  is greater than that of reactant(s). change in enthalpy is equal to the sum of heat of product minus the sum of heat of reactant i.e △H =      –   . Where    is sum of heat of products and  is sum of heat of reactants. Enthalpy change in this regard is positive. i.e △H = +ve

E

Examples of Endothermic reaction

• N_2(g)+ O_2(g)                       2NO_(g)                                        ∆H= +180.6KJmol ^-1

Exothermic reaction is one during which heat is released to the surrounding. Examples of exothermic reaction is combustion  reaction. Dissolution of H₂SO₄,  NaOH,  KOH, neutralization  and AlCl₃ are exothermic process. In exothermic reaction, the enthalpy of product(s)  is less than that of reactant(s). change in enthalpy is equal to the sum of heat of product minus the sum of heat of reactant i.e △H =      –   .  Where     is  sum of heat of products and  is sum of heat of reactants. Enthalpy change in this regard is negative. i.e △H = -ve

• Example of exothermic reaction.

C_(s) + O_2(g)                                            CO_2(g)

∆H = – 408KJmol ^–1

Note: The negative and the positive signs indicated in the above figures indicate exothermic and endothermic reactions respectively

Heat of formation: Heat of formatiom of a compound is the heat change that occurs when one mole of a compound is formed from its elements in their standard states. The standard condition are 25^Oc and 1 atm. The heat of formation of water is -286kJmol^-1. The equation is given below:  H_2(g)  +  ½O_2(g0   →  H₂O_(l)      ΔH_f = -286kJmol^-1

Heat of combustion: This is the heat evolved when one mole of a substance is completely burnt in oxygen under state condition. The heat of combustion of carbon is -393kJmol^-1

Example:

Calculate the heat of reaction in the equation below given that the heat of combustion of carbon and hydrogen are -393kJmol^-1 and -286kJmol^-1 respectively and heat of formation of butane is -125kJmol^-1.

C₄H_10(l)  +   O_2(g)      4CO₂  +  5H₂O_(l)

Solution

△H =      –

=    [-393+(-286)]  –   (-125+0)

=    -2877kJmol^-1

NOTE: The enthalpy of formation of an element is zero

EVALUATION

1. Differentiate between exothermic and endothermic reaction.

GENERAL EVALUATION

OBJECTIVE TEST:

1. Endothermic reaction may be defined as except

• Reaction in which heat is absorbed
• Reaction in which heat is released’
• Reaction in which     H is positive
• Reactant than energy of the reactant.

2. Rate of chemical reaction depends on the following factors except.

• Rate at which gas is evolved
• Rate at which products are formed
• Rate at which colour of reactions change.
• Rate at which the reactants diminish.

3. The unit of rate of chemical reaction is (a) Moldm^-3S^-1(b) Mol^-1S^-1(c) Mol^-1 (d) Smol^-1
4. The rate of a reaction is proportional to the number of effective collisions occurring per second between the reactants. This statement is associated with the.

• Kinetic theory (b) rate law (c) atomic theory (d) collision theory

5. Which of the following statement is not true about the reaction between dilute hydrochloric acid and marble chips calcium trioxocarbonate(IV)?

• It is faster after three seconds than it is after ten seconds.
• It slows down with time.
• It eventually stops.
• It proceeds at a constant rate

ESSAY QUESTIONS

1. Use the arguments of the collision theory to explain how the following factors affect the rate of a chemical reaction. (i) concentration (ii) surface area
2. (a) State in three short sentences the main ideas of the collision theory.
3. State how the rate of a chemical reaction is affected by the following factors.

• Concentration (b) surface area

4. (i) What is a catalyst?

• Write a relevant equation to show the industrial process in which V₂O₅was used as a catalysts.

5. Explain briefly how temperature affects the rate of a chemical reaction.

WEEK 4

TOPIC: CHEMICAL REACTIONS

CONTENTS:

1. CHEMICAL EQUILIBRIUM:
2.  CHARACTERISTICS OF A CHEMICAL SYSTEM AT EQUILIBRIUM,
3. LE CHATELIER’S PRINCIPLE AND FACTORS AFFECTING EQUILIBRIUM OF CHEMICAL REACTION
4.  EQUILIBRIUM CONSTANT, KC

PERIOD 1:   CHEMICAL EQUILIBRIUM

Studies have shown that some reactions are such that all the reactants are never completely converted into products. In such reactions, the products are converted into the reactant as they are been formed.

A reaction is said to be reversible, if it can be made to proceed forward and backward, under a given set of conditions.

For a chemical reaction involving a gas to be reversible, it must be carried out in a closed vessel, i.e. an isolated or closed system. Two opposing arrows in a chemical equation represent a reversible reaction:

1. a) H₂O_(l) H₂O_(g)
2. b) N_2(g) + 3H_2(g)2NH_3(g)
3. c) CaCO_3(s)  CaO_(s)+ CO_2(g)

Such reactions usually have relatively low activation energies for the reverse reactions.

A reaction in which the products cannot readily be combined to give the reactants is said to be irreversible. In such a reaction, a single arrow is used pointing to the product(s), e.g. combustion reactions

1. a) C_(s) +  O_2(g)CO_2(g)
2. b) 2Mg_(s) + O_2(g) 2MgO_(s)

Such reactions usually have relatively high activation energies for the reverse reaction.

REACTIONS AT EQUILIBRIUM

Dynamic Equilibrium

When the rates of forward and backward reactions in a reversible reaction are equal, the reaction is said to be at equilibrium.  At equilibrium, both the forward and backward reactions are still going on, but because there rates are the same, the concentrations of the reactants and products do no longer change with time.  Hence, chemical equilibrium is dynamic and not static.

How to Identify a Reaction at Equilibrium

At equilibrium, certain observable properties become constant depending on the type of reaction. Such properties include;

• Concentrations of the reactants and products;
• Pressure of gases;
• Density or intensity of colour of a solution;
• Temperature of the system.

EVALUATION

1. When is a reaction said to be reversible?
2. Describe a natural process that is reversible.

PERIOD 2: CHARACTERISTICS PROPERTIES OF A SYSTEM AT EQUILIBRIUM

1. Equilibrium is dynamic; that is,

• The reactants and products are present at equilibrium, i.e. equilibrium can be achieved from either direction:

H_2(g) + I_2(g)   2HI_(g)

2HI_(g)                H_2(g) + I_2(g)  

2. The concentrations of the reactants and the products, pressure, density or intensity of colour, and the temperature of the system are constant- depending on the type of the reaction.
3. The rate of forward and backward reactions are equal.
4. The equilibrium position is not affected by the presence of the catalyst, it only quickens the rate at which equilibrium is achieved.
5.  At equilibrium, the free energy change, ∆G, of the reaction is zero.
6.  A system at equilibrium will resist a change.

Information Obtained from an Equilibrium Equation

Consider a typical reversible reaction:

N_2(g) + 3H_2(g)      2NH_3(g)        ∆H  = – 92kJ

The following information can be obtained:

1.  The two opposing reactions are occurring at the same time – as indicated by the opposing double arrow.
2. The two opposing reactions are at equilibrium.
3. The amounts, in moles, of the reactants and products at equilibrium are proportional to the stoichiometry in the balanced equation.
4. The forward reaction is exothermic (∆H is negative; hence, the backward (or reverse) reaction will be endothermic (∆H is positive), i.e.

2NH_3(g)          N_2(g) + 3H_2(g)     ∆H  = + 92kJ

EVALUATION

1. What is meant by an equilibrium system?
2. Give three properties of a chemical system at equilibrium.

LE CHATELIER’S PRINCIPLE

Le Chatelier’s principle stares that if an external constraint such as a change to temperature, pressure or concentration is imposed on a chemical system in equilibrium, the equilibrium will shift so as to annul or neutralize the constraint.

FACTORS AFFECTING EQUILIBIUM OF A CHEMICAL REACTION

1. EFFECT OF A CHANGE IN TEMPERATURE e.g. For the following systems at equilibrium

• 2SO_2(g)+ O_2(g)              ⇌        2SO_3(g)                      H = – 395.7 KJmol ^–1

The forward equation is exothermic i.e. it involves increase in temperature whereas the backward equation is endothermic or it involves decrease in temperature.

If the temperature of the system is increased:

Increase in the temperature of the system will shift the equilibrium position o backward reaction, that is, reactant formation is favoured.

Increasing the temperature of the system will shift the equilibrium position to the left, favouring the backward reaction i.e. reactants formation (i.e. SO₂ and O₂).

If the temperature of the system is decreased:

Decrease in the temperature of the system will cause the equilibrium position to shift to the right favouring the forward reaction, that is, product formation.

• N2(g) + O₂⇌       2NO(g)        ΔH   =   +90.4KJmol ^–1

When the system is at equilibrium, the forward reaction is endothermic and the backward reaction is exothermic.

If the temperature of the system is increased:

Increase in the temperature of the system will cause the equilibrium position to shift to the right favouring forward reaction. i.e. product formation.

If the temperature of the system is decreased:

Decrease in the temperature of the system will cause the equilibrium position to shift to the left favouring the backward reaction.

2. EFFECT OF A CHANGE IN PRESSURE:

For a change in pressure to affect a chemical system in equilibrium.

• One of the reactants in the or products in the reversible reaction must be gaseous.
• The total number of gaseous molecules on the left side of the equation must be different from the total number of moles of gaseous molecules on the right side.

For the following systems at equilibrium

• 3H₂(g) + N₂(g)        ⇌2NH₃(g)

The forward reaction is from high pressure to low pressure and the backward reaction is from low pressure to high pressure.

Increase in pressure of the system:

When the pressure of the system is increased the equilibrium position will shift to the right favouring the forward reaction that is, the product formation.

Decrease in the pressure of the system:

When the pressure of the system is decreased the equilibrium position will shift to the left favouring the backward reaction that is reactant formation.

• N₂O₄(g)       ⇌2NO₂(g)

When the system is at equilibrium, the forward reaction involves an increase in the pressure of the system and the backward reaction involved a decrease in the pressure of the system.

If the pressure of the system is decreased:

A decrease in the pressure of the system will cause the equilibrium position to shift to the right favouring the forward reaction that is, product formation.

If the pressure of the system is increased:

An increase in the pressure of the system will cause the equilibrium position to shift to the left favouring the backward reaction that is, reactant formation

• H₂(g) + I₂(g)       ⇌2HI(g)

A change in pressure will not affect this system at equilibrium because the number of moles of the reactants is the same as the number of moles of the product.

• 3Fe(s) + 4H₂O(g)         ⇌Fe₃O₄(s) +4H₂(g)

A change in pressure will not affect this system at equilibrium because the number of moles of gaseous reactant and product is the same.

3. EFFECT OF A CHANGE IN CONCENTRAION

E.g. 3Fe(s) + 4H₂O (g)         ⇌      Fe₃O4(s) + 4H₂(g)

If the concentration of any of the reactants is increased the equilibrium position will shift to the right but if it is decreased it will shift to the left.

Also removal of any of the reactant or product will also cause the equilibrium position to shift. E.g. removal of hydrogen gas from the system will cause the equilibrium position to shift to the right.

4. EFFECT OF A CATALYST

Addition of a catalyst to a system in equilibrium will not affect the equilibrium position. Instead, addition of catalysts will only make equilibrium state to be reached attained faster.

EQUILIBRIUM CONSTANT, K_C

In a reversible reaction, there is a fixed relationship, at constant temperature, between the concentrations in moldm^-3 of the products and the reactants. This relationship is accounted for in the study carried out by Guldberg and Waage. This law, known as the Law of Mass Action states that at constant temperature, the rate of reaction is proportional to the active masses (concentration raise to the power of its coefficient) of each of the reactants.

EXPRESSION OF EQUILIBRIUM CONSTANT, K_C

In a reaction: aA + bB       ⇌      cC +dD

The equilibrium constant K_C is expressed as follows:

K_C   =     . In the cause of a chemical reaction, it the concentration of gases and aqueous species the normally change. The concentrations of solid and pure liquid are always constant, hence, thay cannot appear in equilibrium constant expression.

Note: The species must be written in a square bracket as shown above e.g. in the reaction:

• 2SO_2(g)+ O_2(g)            ⇌      2SO_3(g)

The equilibrium expression

K_C =

• N_2(g)+3H_2(g)      ⇌     2NH_3(g)

K_C =

(c).   CaCO_3(s)   ⇌  CaO_(s)   +   CO_2(g)

K_C   =  ]

IMPORTANCE OF K_C

KC determines the yield of chemical reaction at equilibrium. If Kc is greater than one, it more product is formed 1.e forward reaction is favoured. But if Kc is less than one, it less product is formed (poor yield) i.e backward reaction is favoured. Kc is temperature dependent and change with change in temperature.

RELATIONSHIP BETWEEN KC AND △G

Free energy (G) is the energy for doing useful work. Gibb’s free energy cannot be measured directly but the change in free energy is usually measured. Change in free energy determines the spontaneity of a given reaction. For a spontaneous reaction, △G is negative.

△G  =  -RTInKc. Where n = no of mole of specie,  R = gas constant (8.314Jmol^-1K^-1 and T = temperature in Kelvin

EVALUATION:

1. Write the expression for the equilibrium constant K_C for the following

• 2H_2(g)+  O_2(g)         ⇌       2H₂O_(g)

• 2CO_(g)+ O_2(g)        ⇌      2CO_2(g)

GENERAL EVALUATION:

OBJECTIVE TEST:

1. Two boys balanced in a sea-saw game is an example of

(a) Static equilibrium (b) dynamic equilibrium (c) homogenous equilibrium (d) mutual equilibrium

2. In the reaction, 2SO_2(g)+ O_2(g)⇌      2SO_3(g)

(a) The reaction is in physical equilibrium

(b) Contact process

(c) The reaction is the rate determining step

3. In the decomposition of  CaCO_3(s)⇌CaO(_s) + CO_2(g)

The reaction will attain dynamic equilibrium

• When the reaction is in open system
• When the reaction is heated strongly
• When the reaction is in a closed system
• When the reaction is catalysed

4. Factors affecting equilibrium reaction include the following except.

(a) Pressure (for solid system)

(b) Concentration

(c) Temperature

(d) Pressure (for gaseous system)

5. All except one is not a condition for considering pressure in an equilibrium system.

(a) There must be concentration gradient between the reactants and the products.

(b) The reactants could be gas while the products may be solids

(c) Both products and reactants must be gaseous.

(d) The reaction takes place in a closed system

ESSAY QUESTIONS

1. What is the effect of each of the following on the equilibrium position of the system indicated below?

A_(s) + B_(g)    ⇌    C_(g)    Δ H = – xKLmol^-1

• Cooling the system
• Removing
• As soon as it is formed

2. Write the expression of equilibrium constant for

2AB_2(g) + B_2(g)      ⇌   2AB_3(g)

3. Consider the reaction represented by the equation

N₂O_4(g)    ⇌    2NO_2(g)            △H = +57.2KJmol^-1

• When is the reaction said to be at equilibrium
• Mention two conditions that can favour the forward reaction.
• Name the principle involved in (b) above
• Addition of catalyst.

WEEK 5

TOPIC: STOICHIOMETRY

1.  S.I. UNIT OF QUANTITIES: LENGTH, MASS VOLUME ETC
2. BASIC CONCEPTS: MOLE, MOLAR QUANTITIES, MOLALITY, STANDARD TEMPERATURE AND PRESSURE (S.T.P.)
3. CALCULATIONS INVOLVING MASS AND VOLUME
4. MORE CALCULATIONS ON MASS AND VOLUME RELATIONSHIP.

PERIOD 1:  S.I. UNIT OF QUANTITIES: LENGTH, MASS VOLUME ETC

S.I. UNITS OF QUANTITIES.

S.I.  unit means international system [used to describe units of measurements from French ‘system international’] S.I. units

For example

Relationship between quantities

(a) M =   i.e molar mass =

Molar volume  =

(b) Molar volume =  =

(c) Avogadro’s constant = L =  ==

TYPES OF MASS AND VOLUME PROBLEMS

There are two types of mass and volume relationship problems namely:

1. Mass- mass problem:

In this type of problem, the reacting mass of substance and its molar mass is usually given and we may be required to determine the mass of another substance produced and its molar mass also given. The reversed form of this is when the mass of another substance product produced is given and one is asked to find the mass of the substance which produced it.

2. Mass- gas volume problem:

In this case, you may be given a certain reacting mass of a reactant and required to find the volume of another substance produced at S.T.P. or under any other conditions. The reverse of this type is when you are given the volume of a certain gas liberated at S.T.P or other conditions and the gas produced.

Note: This method is related to simple proportion as in mathematics.

3. Gas volume – gas volume problem

In this class of reactions, only the volumes of the reactants and products are involved in the calculations.

4.  Mass – Liquid volume problem

In this type, the mass I given and the volume of liquid has to be determined.

5. Liquid volume – volume problem

In this group of calculations, one volume is given while another is to be found.

EVALUATION

1. What is a mole of a chemical substance?
2. Mention two elementary units of a chemical substance.
3. What mass, in grams, of aluminium contains the same number of atoms as in 12.0 g of carbon? (C = 12.0, Al = 27.0).

PERIOD 2: BASIC MOLE CONCEPT

Definition: A mole is the amount of a chemical substance that contains as many elementary particles as there atoms in 12.0 g of C-12 isotope. It means that a mole of an element are collections of its atoms such that the total mass in grams, is numerically equal to its relative atomic mass. That is: if 1 mole of carbon atoms weighs 12.0 g, then, 1 mole of oxygen atoms will weigh 16.0 g; 1 mole of sodium atoms will weigh 23.0 g, e.t.c.  The number of elementary units or particles in 1 mole of a chemical substance is constant, and it is called Avogadro’s constant, L. It is numerically equal to 6.02 x 10²³=. Hence:

• 1 mole of carbon atoms weighs 12.0 g and contains 6.02 x 10²³
• 1 mole of sodium atoms weighs 23.0 g and contains 6.02 x 10²³
• 1 mole of sulphur atoms weighs 32.0 g and contains 6.02 x 10²³
• The molar mass of every element contains the number of atoms.
• The molar mass of every compound contains the number of molecule
• One faraday of electricity (9630 coulombs) is said to have flowed past a point in a circuit when this number of electrons have flowed through.
• The molar volume of every gas, 22.4 dm³at s.t.p contains this umber of molecules. One    mole of very particle, atom, molecule, ion or electron contains the Avogadro’s number of particles.

The formula of a compound tells us the elements present in it. It also tells us the elements present in it. It also tells us the amounts of the different elements present in it. These amounts are usually expressed as a mole ratio of the different elements in the compounds.

For example:

Formula

Element present           Sodium              sulphur     oxygen   

Number of Moles             2                          1              4

Mole ratio                          2                           1             4

A balanced equation of a chemical reaction tells us the relationship of the amounts of the reactants to one another and for the products is known as of stoichiometry reactions.

MOLE RATIO AND MASS RELATIONSHIP

Again, a mole is the amount of substance which contains and many elementary particles as there are carbon atoms in 0.012kg of carbon-12.

The mole is the amount of substance which contains as many formula units as their atoms in 12gram of carbon-12.

The numerical coefficients of a balanced equation represent the numbers of moles of reactants and products. From coefficient, we get the mole ratio of the reactants and product in a reaction. For example

Equation 1

Number of moles                    1                 2                   1

Mole ratio                                1                 2                    1

Therefore, the mole ratio in which reactants combine and products are formed gives the stoichiometry of the reaction

Equation 2.

Mg_(s) + 2HCl_(aq)     MgCl_2(s) + H_2(g)

Number of moles      1              2              1               1

Mole ratio                 1    :         2             1    :            1

Molar mass              24g          36.5g        95g            2g

Reacting mass          24g         (36.5 ×2)     95g           2g

From the equation 2, we see that

Amount × molar mass = reacting mass

Amount =

Molality– molality is defined as the amount of substance (in mole) of solute divided by the mass in kilogram of the solvent

Molality =

Molality should not be confused with molarity as they are both measures of concentration of solutions but Molarity is the ratio of moles to the volume while molality is the ratio of the moles to the mass of the solution. Molality is the number of solute per kilogram of solvent. It is important that the mass of solvent issued and not the mass of the solutions.

WORKED EXAMPLE

1. A 4 g of sugar cube (sucrose) – C₁₂H₂₂O₁₁is dissolved in a 350molteacup of 8⁰OC water. What is the molality of the sugar solution?

(Given- density of water at 80⁰ =0.975gmol^-1)

Solution

Molality is the number of moles of solute per kilogram per solvent.

Step 1: Determine the number of moles of sucrose in 4g

Molar mass =  12(12)   +    1(22)  +  16(11)

=    144  +   22   +  176

=  342 gmol^-1

Divide this amount into the size of the sample.

=    0.0117mol

Step 2 Determine mass of solvent in kg

Density =  ;                       mass    =    density    ×   volume

Mass   =     0.975gmol ×350mol

Mass  =   341.25g

Change gramme to kilogramme = 0.341kg

Step 3: Determine the molarity of the sugar solution

Molality ==

Molality = 0.034mol

EVALUATION

1. How many molecules are there in 14g of nitrogen at s.t.p.? [N=14. Avogadro’s number =  6.02 ×10²³]
2.  What amount, in mole, of copper is deposited when 13.0g of zinc reacts with excess copper(II) tetraoxosulphate(VI) solution according to the following reaction?

Zn_(s)  +  CuSO_4(aq)                 ZnSO_4(aq)  +     Cu_(s)    [Cu=63.5, Zn =65 ]

3. Calculate the volume of chloride at s.t.p. that would be required to react completely with 3.70g of dry slaked lime according to the following equation: Ca(OH)_2(g)+ Cl_2(g)→       CaCl₂  + H₂O  [H=1; O=16; Ca=40; 1mole of gas occupies 22.4dm3 at s.t.p.]

PERIOD 3: CALCULATION INVOLVING MASS AND VOLUME

Example 1: mass- mass problem

1. What mass of copper is deposited when 6.5g of granulated zinc reacts with excess copper(II) tetraoxosulphate(vi) solution according to the following equation

Zn(s) + CuSO₄(aq)            ZnSO₄ + Cu(s)

Cu = 64, Zn =65

Given: molar mass of Zn =65gmol-, molar

Mass of Cu = 64gmol-1

Reacting mass of Zn=6.5g

Zn + CuSO₄                    ZnSO₄  + Cu

From the above, 65g of Zn deposited 64g of Cu.

1g of Zn will deposit bof cu

65g of Zn will deposit g of cu

= 6.4g

2. What volume of carbon(IV) oxide is produced at s.t.p. when 2.5g of calcium trioxocarbonate (iv) reacts with excess acid according to the following equation

CaCO₃ +2HCl                            CaCl₂ + H₂O + CO₂

[CaCO_{3 =} 100, molar volume of gas at s.t.p= 22.4dm₃]

Solution

CaCO₃ +2HCl                               CaCl₂ + H₂O + CO₂

100g   22.4dm₃

From the equation, 100g of CaCO₃ produces 22.4 dm₃ of CO₂

1g of CaCo₃ will produce  of CO₂

= 0.56dm³

EVALUATION

1. CaO_(s)+ H₂O_(l)        Ca(OH)_2(s).

From the equation above, calculate the mass in grams of calcium hydroxide produced by 5.6g of calcium oxide.   [Ca = 40, O = 16, H = 1].

2. What volume in dm³will 0.5g of H₂occupy at s.t.p? [H = 1, 1 mole of gas at s.t.p = 22.4dm³].
3. Consider the reaction represented by the following equation:  2NaCl_(s)+ H₂SO_4(aq)→ Na₂SO_4(aq) + 2HCl_(g). Calculate the volume of HCl gas that can be obtained at s.t.p. from 5.85 g of sodium chloride. [ Na = 23.0, Cl = 35.5, Molar volume of gas s.t.p. = 22.4 dm³]

PERIOD 4: MORE CALCULATIONS ON MASS- VOLUME

Phosphorous burns in the limited supply of air to form phosphorous(iii) oxide, P₄O₆ and in unlimited supply of air to form phosphorous(v) oxide P₄O₁₀.If 3.3g of phosphorus(III) oxide is formed

(a) what mass oxygen from air is used?

(b) How many molecules of oxygen are contained in this mass?

(c) How many molecules of phosphorous are converted to the oxide?

(d) How many molecules of oxygen are needed to convert this oxide into phosphorous v oxide?

(e) What volume would the oxygen in (d) above occupy at s.t.p.?

Solution:                                                                                                  

Molar mass of P_{4   =} 31 × 4 = 124gmol^-1

O₂ =16 × 2 = 32 g.mol^{– 1}

P₄ O_{6 =}  gmol^–

• 220g of P₄O₆ used (32 ×3) 96g of oxygen

3.3 of P₄O₆ requires g of oxygen

= 1.44g

(b) 1.44g of oxygen =mole

1 mole contain 6.02 ×  molecules

Number of moles in 1.44g of oxygen =

= 2.71 molecules

(c) Mass of phosphorous in 3.3g of P₄O₆  =  (3.3 – 1.44)g

= 1.86g

1.86 g of phosphorous is mole

The number of molecules in (c) above i.e. 1.86g of phosphorus

= = 9.03 molecules

• P₄O₆ +20₂             P₄O₁₀

220 of P₄O₆ require 2 moles of oxygen

3.3g of P₄O₆ will require

Number of molecules of oxygen contained in this

olecule

= 1.81 molecules

• 02molecules occupy 22.4dm³ at s.t.p

1.81  molecule occupy  22.4dm³

= 0.067 dm³ or 6.73.49 cm³ 

EVALUATION

1. Calculate the mass of sodium trioxocarbonate(IV) produced by the complete decomposition of 16.8g of sodium hydrogen trioxocarbonate(IV) (H= 1, O= 16, Na= 23, S= 33).
2. On carrying out a chemical analysis discovered that the component of egg shells is CaCO₃^.  If 25.0 cm³ of 0.700 moldm^-3 HCl reacted with all the CaCO₃in a sample of egg shells, calculate the:

(i) mass of CaCO₃ in the egg shells; (ii) volume of CO₂ evolved at s.t.p.

The equation for the reaction is:

CaCO_3(aq) + 2HCl_(aq)        CaCl_2(aq)    + H₂O_(l)  + CO_2(g)

( H = 1.00; O = 16.0; Ca = 40.0;  1 mole of gas occupies 22.4dm³ at s.t.p.)

GENERAL EVALUATION

OBJECTIVE TEST:

1. What volume of steam is produced when 10g of propyne is burnt in excess oxygen at s.t.p?

(a) 44.8dm (b) 11.2 dm³ (c) 22.4 dm³  (d) 11.2cm³

2. What mass of anhydrous sodium trioxocarbonate(iv), Na₂CO₃, present in 500cm³ of 0.1 moldm^-3?[ Na = 23, C= 12, O = 16]    (a) 10.6g (b) 212g (c) 53 (d) 106g.
3. The mass of potassium hydroxide required to make 300.00cm³of 0.4moldm^-3 solution is [KOH = 56.0] A.26.88 g   B. 13.44 g C.  6.72 g    D. 3.36 g
4. A solution of sodium hydroxide containing 6.0g in 250cm³of solution has a concentration of A. 0.04 mol. dm³B. 0.60 mol. dm³ C. 0.96 mol. dm³ D. 0.15 mol. dm³ [molar mass of NaOH = 40g mol.^-1
5. The volume occupied by 17g of H₂S at s.t.p. is [H=1.00,  S=32.0, Molar volume =22.4 dm³] A. 11.2 dm³  B.17.0 dm³C. 34.0 dm³D.  44.8 dm³

ESSAY QUESTIONS

1. If 11 g of a gas occupies 5.6 dm³at s.t.p. calculate the vapour density [1 mole of a gas occupies 22.4 dm³at s.t.p]
2. 100 cm³of a gas at 27exert a pressure of 750mnHg. Calculate its pressure. If its volume is  increased to 250 cm³ at 127.
3. What is an S.I unit?

4(a) Write down five examples of quantities with their S.I units. (b) Define mole.

5. What volume will 22.4dm₃ of oxygen measured at 25^oC and and 1.12 × 10⁵Nm    occupy at standard pressure of 1.01 ×10⁵Nm.
6. What volume of steam is produced when 10g of propyne is burnt in excess?

WEEK 6

Topic: Acid- base reactions

CONTENT:

1. COMMON ACID –BASE INDICATORS AND THEIR pH  RANGE
2. ACID-BASE TITATION
3. HEAT OF NEUTRALIZATION (INTRODUCTORY) CONSTRUCTION OF WOODEN RETORT STAND

PERIOD 1:  1: COMMON ACID –BASE INDICATORS AND THEIR pH RANGE

Acid-base indicator is a weak organic acid or weak organic base (organic dyes) which shows colour change depending on the pH of the solution or organic dye which show one colour in acidic medium and another colour in basic medium. It can defined as an organic dye which as colour in ionized state and another colour in unionized state.  Examples of indicators are methyl orange, phenolphthalein, methyl red etc.

CHOICE OF INDICATOR FOR ACID-BASE TITRATION

The choice of an indicator for acid-base titration depends on the pH of the mixture at equivalent point and the pK of the indicator. When a strong acid is titrated against a strong base, at the equivalent point, the salt formed does not under hydrolysis and so any suitable indicator can be used. The salt formed when strong acid is titrated against a weak base undergoes hydrolysis to produce excess hydrogen ions. This makes the solution acidic and any indicator that shows colour change in acidic medium is used in the case. Methyl orange changes colour within a pH of 2.9—4.6  and so suitable for this titration. The salt formed when weak acid is titrated against strong base undergoes hydrolysis to produce excess hydroxide ions making the solution alkaline. Phenolphthalein changes colour in basic medium and so suitable to the kind of titration.

Indicators are used in acid-base titration to show the end point when the acid would have materialised  the base. Colour changes of common indicator are shown below

S.A = Strong acid, S.B = Strong base, W.A = weak base

This is volumetric analysis which aim at estimation of the qualities of certain substances in solution

PERIOD 2 AND 3: ACID- BASE TITRATION

This involves titrating a solution (usually the acid) from the burette into a fixed volume (20 or 25cm³) of base until the two solutions have completely reacted. The concentration of one of the solution is normally known. An indicator is used to determine the point of complete reaction.

For titration, more dilute solutions containing 0.1moldm^-3 or 0.05mold^-3 are used.

STANDARD SOLUTION

In any titration, standard solution must be used to react with a solution of unknown concentration.

A standard solution is a solution of known concentration.

PRIMARY STANDARD

A primary standard is a reagent that is extremely pure, stable and has no water of hydrate. A primary solution is used for the preparation of standard solution. There a primary standard solution is a solution which is prepared from a non-hygroscopic, non-deliquescent and pure substance. Examples are sodium trioxocarbonate(iv), sodium oxalate and benzoic acid

Features of a primary standard

1. High purity
2. Stability
3. High solubility
4. High relative molecular mass

SECONDARY STANDARD

Secondary standard solution refers to a solution that has that has its concentration measured titration with a primary standard solution.

The concentration of a solution is the amount of solute in 1dm³ or 1000cm³ the solution. It can be expressed as moldm^-3 or gdm^-3

The above diagram is an example of the apparatus used in acid-base titration. Some materials used in acid- base titration are:

1. Weighing bottle
2. chemical balance
3. pipette
4. Burette
5. retort stand
6. filter paper
7. funnel
8. white tile
9. standard volumetric flask
10. conical flask

PRECAUTIONS IN USING THE PIPETTE, BURETTE AND CONICAL FLASK

(a)PIPETTE: (i) Rinse the pipette with the solution it should be used to measure i.e. base

(ii) Avoid air bubbles in the pipette

(iii) Make sure that the mark to be read is at level with your eye.

Do not blow the last drop on the pipette.

(b) BURETTE:

(i) Rinse the burette with acid or allow it to drain after rinsing with distil water

(ii) Make sure the burette jet is filled

(iii) Make sure that the burette is not leaking

(iv)Take your burette readings with your eyes at the same level as the measures to avoid error due to parallax

(v) Remove the funnel before taking your reading

(vi)Avoid incorrect reading.

CONICAL FLASK

1. Do not rinse it with any of the solutions used in the titration but with distilled water.
2. Wash down with distilled water any drop of the solution that stick by the sides of the conical flask during titration.

Example of titration problem solved.

A is a solution of hydrochloric acid containing 4.00g in 1dm³ of solution B is a solution containing sodium hydrochlorid

• Put A into a burette and titrate with 25cm³portion of B using methyl orange indicator. Tabulate you burette reaching. Determine the average volume of solution A used
• From your result and information provided
• Find concentration of A in moldm^-3
• Find concentration of B in moldm^-3
• Find the amount in mass of NaOH present in 1dm³of solution B

Equation of reaction:

HCl +  NaOH           NaCl + H₂O

[H = 1, Cl= 355, K=39, Na= 23, O = 16]

STUDY ANSWER

Indicator used: methyl orange

Size of pipette: 25cm3

Average litre volume (cm³) =

= == 25.45cm³

B (i) Concentration of HCl of A in moldm^-3

Conc. in gdm^-3 = 4 gdm^-3

Molar mass of HCl=  1+35.5 =36.5

Con. In moldm^-3 of HCl = = 0.110 moldm^-3

b(ii) V_a= 25.45 cm³

Ca= 0.110 moldm^-3

V_b= 25cm³

C_b =? Of resultant concentration of effect of the two alkalis in solution B.

2HCl +            HCl + NaCl + 2H₂O

=

C_b V_b: na = Ca Va: nb

Make C_b the subject formula.

Cb=  =

0.0559 moldm^-3 (38.8 fig)

b(v) Amount in mass of NaOH on 1dm³  of solution B

Molar mass of NaOH = 23 +1 + 16=40g/mol

Amount in mass of NaOH in 1dm³ of solution B=

40 × 0.0559 moldm^-3

= 2.23gdm^-3

EVALUATION

State five precautions you need to take while using. (a) Pipette   (b) burette and 2 precautions while using (c) conical flask

PERIOD 4:   HEAT OF NEUTRALIZATION

When an aqueous solution of a acid is added to alkalis, there is an increase in temperature. This increase in temperature is called heat of neutralization.

Heat of neutralization of strong acid and strong alkalis are of the same value. The value of weak acids by strong base and strong acids by weak bases are different and varies due to the fact that energy for the complete ionisation of such weak bases and weak acids before neutralization is drawn from the solution.

PROCEDURE

Measure the temperature of the acid solution and that of the alkali solution before mixing. Find the average temperature of both acid and base.

Determine volumes of acid and alkali to mix in order to get a neutral, solution [by titration]. Mix these volumes (say 50cm³ of acid + 50cm³ of alkali; find the temperature of mixture; note the increase in temperature.

If t₁ = temperature of acid, and t₂= temperature of alkalis

Average of the two = = t₃

If t₄ is the temperature of mixture, then the increase in temperature      = t₄ –t₃ =t₅

Assume that the solution, have the same density and specific heat as water.

Therefore, mass of mixture=100g

Specific heat of water = 4.2joulesg^-k

Heat of neutralization= specific heat x mass x temperature rise=4.2 x 100 × t₅ joules

Heat evolved when 1 mole of each reacts=100 x t₅ x 4.2 x 10joulesmol^-1

EVALUATION

1. What is heat of neutralization?
2. State three indicators with their pH ranges.
3. What indicators are as applied in volumetric analysis.

GENERAL EVALUATION

OBJECTIVE TEST :

1. A student prepares 0.5M solution each of hydrochloric and ethanoic acids and then measured their pH the result show that

• pH values are equal
• The solution has a higher pH
• Sum of the pH value is 14
• Ethanoic acid solution has a higher pH

2. What volume of 0.5M H₂SO₄will exactly neutralize 20cm³of 0.lM NaOH solution?

• 0cm³ (b) 5.0 cm³  (c) 6.8 cm³  (d) 8.3 cm³  (e) 10.4 cm³

3. Filling the burette for titration involves these except.

• Wash with water (b) rinse with acid (c) eject air bubbles (d) read at eye level

4. In the titration of acid against base solutions, averaging must involve.

• Rough reading (b) concordant reading (c) higher reading (d) all the titres obtained

5. What mass of anhydrous sodium trioxocarbonate(IV), Na₂CO₃; present in 500cm³ of 0.1moldm^-3? (Na=23, C=12, O=16)

• 6g (b) 212g (c) 5.3g (d) 106g.

ESSAY QUESTIONS

1. (a) In an acid –base titration (i) which solution is normally put in the burette and why. (ii) To which of the solutions is the indicator added?  (iii) Why is the conical flask rinsed with distilled water and not with the solution to be place in it?

• What is the colour of methyl orange in (i) distilled water (ii) lime water (iii) vinegar?

2. A is dilute tetraoxosulphate (vi) acid. B contains 1.50g of sodium hydroxide per 250cm³ of solution. 25 cm³ of B required 15.5cm³ of A for complete neutralization

The equation for reaction:

H₂S0_4(aq) + 2NaOH _{(a q})            Na₂SO_4(aq) + 2H_2(l)

• Calculate:
• The concentration of solution B in moldm^-3[H=1, O= 16, Na=23]
• The concentration of A in moldm^-3
• The number of hydrogen ions in 1.0dm^-3of solution A [Avogadro’s constant 6.0
• The volume of distilled water that should be added to 25cm³ of solution A in order to dilute it ten times

1. What would be the colour of methyl orange indicator in:

(i) Solution A  (ii) solution B (iii) titration mixtures of A and B at the end point.

3. Calculate the mass concentration in gdm^-3of the ions in the following solutions:

(a) Cl^–ions in 0.2 moldm^-3 sodium chloride solution

(b) NO₃^– ions in 0.75 mol dm^-3 trioxonitrate (v) acid.

(c) SO₄^2- ions in 2moldm^-3 potassium tetraoxosulphate (vi) solution. Cl= 35.5, N=14, O=16, S=32]

WEEK 8

TOPIC: WATER

CONTENT

1. Structure of water, Hardness of water and removal of hardness of water
2. Solubility- Basic concept
3. Types of Solutions
4. Factors affecting solubility and uses of solubility curve.

PERIOD 1: STRUCTURE OF WATER, HARDNESS OF WATER AND REMOVAL OF HARDNESS OF WATER

A molecule of water consists of two atoms of hydrogen and one atom of oxygen chemically combined to form a covalent molecule. A molecule of water has a V-shape. The repulsion between the two lone pairs of electrons on the oxygen atom makes the structure of water molecular to be V- shaped

Structure of water

O

(a) Bonding            (b) Shape of water molecule

The shared pair of electrons between each hydrogen atom and oxygen is drawn more electronegative, making the oxygen atom to be partially negatively charged and hydrogen atoms to be partially positively charged. This leads to hydrogen bonding between the molecules of water the attendant relatively high melting point, high boiling point and low vitality.

Hydrogen bonding in water molecules

HARDNESS OF WATER

A hard water is water which will not readily form lather with soap. For example, river water, lake water, stream water and sea water. Hardness of water is due to the presence of calcium, magnesium and iron in water. There are two types of hard water: temporary hardness and permanent hardness.

TEMPORARY HARDNESS: This is caused by the presence of dissolved calcium or magnesium hydrogen trioxocarbonate(IV), HCO₃^– of Ca, Mg or Fe. Temporary hardness is so called because it is easily removed by boiling.

Removal of temporary hardness: Temporary hardness can be removed by:

1. Boiling: The calcium hydrogen trioxocarbonate(IV) which causes the temporary hardness is decomposed by heating.
2. Addition of calculated amount of slaked lime, CaCOH)₂

Ca(HCO₃)₂ + Ca(OH)₂            2CaCO₃ + 2H₂O

Soluble    slightly soluble                       insoluble

Effects of temporary hardness:

1. Furring of kettle: When a kettle or boiler has been used to boil hard water for some time, the inner surface becomes coated with a white fur-like layer. The layer is due to the gradual deposit of CaCO₃from the decomposition of Ca(HCO₃)₂
2.  Stalagmites and stalactites: These are pillars of limestone, CaCO_3.

Found in hot caves. They are formed when hard water flow temporary over the top of the cave and drop off water then becomes decomposed by the heat inside the cave, leaving deposits of CaCO₃ on the roof and floor of the cave. The deposits that grow downward are termed stalactites and the ones that grow upward are termed stalagmites. Pillars of CaCO₃ are formed when both stalactites and stalagmites meet.

PERMANENT HARDNESS: This is caused by the presence of dissolved calcium and magnesium ions in form of soluble tetraoxosulphate(VI) and chlorides and certainly cannot be removed by boiling. When water containing any of these substances is evaporated, a white solid deposit of calcium or magnesium tetraoxosulphate(VI) and/ or calcium trioxocarbonate(IV) is left behind.  Calcium trioxocarbonate(IV) causes the ‘furring’ in kettles that occur in hard water.  This furring may be removed by the addition of a dilute acid:

2H⁺   +  CO₃^2-_(aq)                        CO_2(g)     +  H₂O_(l)

Blockages in hot water pipes are caused by similar process to the furring of kettles. A thick deposit of limescale builds up.

Removal of permanent hardness: permanent hardness as well as temporary hardness can be removed by:

1. Addition of washing soda, (Na₂CO₃.10H₂O) crystals

In each type, the calcium or magnesium ion, which actually causes the hardness, is removed as a precipitate and can, therefore, no longer cause hardness.

Na₂CO_3. 10H₂O + CaSO₄                            CaCO₃ + Na₂SO₄

Soluble                            insoluble

Na₂CO₃ + MgSO₄                                  MgSO₄ + Na₂SO₄

2. Addition of caustic soda, NaOH

2NaOH + CaSO₄                                 Ca(OH)₂ + Na₂SO₄

Soluble                                          insoluble

2NaOH + MgSO₄                          Mg (OH)₂ + Na₂SO₄

3. Ion- exchange method: Zeolite or permutitis an ion exchange resin used industrially and in the home for softening water. When the hard water is passed through the resin, the sodium ions in it goes into the solution and the calcium and magnesium ions take their place in the complex salt.

CaSO₄      +        Na₂Y                 CaY      +     NaS₂O₄

Soluble            resin                                insoluble

Sodium aluminium trioxosilicate (iv)

EVALUATION

1. Why is water molecule V-shaped?
2. Name the constituent elements in water with their respective number of atoms.
3. What type of chemical combination exists between water molecules?

PERIOD 2: SOLUBILITY

Definition: The Solubility of a solute (solid) in a solvent (liquid) is the concentration of the saturated solution. Solubility can be defined thus:

1. It is the maximum amount of the solid that dissolves in 1 dm³of the solution at a given temperature. It is expressed in mol.dm^-3.
2. It is the maximum mass, in grams, of the solid that dissolve in 100 g of the solvent at a given temperature.

When some common salts is added to a beaker of water and the mixture stirred, the salt gradually disappear, and the clear colourless mixture is obtained. The salt is said to have dissolved in the water. Thus, the salt that dissolved in the water is called solute and the water that does the dissolving is known as the solvent and the product obtained by dissolving the salt in the water is called a solution.

Therefore, a solution is a homogenous mixture which is formed when a solute is completely dissolved in a solvent. A solution can be saturated or supersaturated.

Determination of solubility of KNO₃ at 30^oC

Stage 1: Preparing of Saturated Solution

1. Put 100 cm³of distilled water in a beaker, add the salt little at a time and warm on a Bunsen burner with stirring. Continue the addition of the salt with stirring and keeping the beaker warm at about 50^oC, until the salt can no longer dissolve.
2. Allow the saturated solution to cool to 30^oC; this is the saturated solution of the salt at 30^oC. It will be noticed that as the hot saturated solution cools, the excess solid in the solution separates out, to give a saturated solution at a lower temperature.
3. Stage 2: Evaporation of Saturated Solution to Dryness.
4. Weigh accurately a clean dry evaporating dish: xg.
5. Rinse a 25 cm3 pipette with the saturated solution, in order to warm it.
6. Pipette 25 cm³of the saturated solution at 30Oc, as quickly as possible (without pipetting any undissolved solid) into the evaporating dish and re-weigh: yg
7. Place the dish on a steam bath and evaporate to dryness. Allow to cool in a charged desiccators, and then re-weigh.
8. Repeat the process of heating and cooling, until a constant mass is obtained: wg.

Calculation of the solubility at 30^oC

The following hypothetical values will be used to illustrate how to calculate the solubility of a salt.

Data required:

• Mass of the evaporating dish:   x= 15.20 g
• Mass of evaporating dish + 25 cm³of saturated solution:    y = 40.70 g
• Mass of dish  + anhydrous salt: w = 20.70 g

Solubility in mol.dm^-3

Mass of anhydrous salt  = (w – x)   = (20.70  – 15.20) g = 5.50 g

i.e. 25 cm³ of saturated solution contain 5.50 g of salt,

Hence, 100 cm³ of saturated solution contain 5.50  x 1000/25  = 220 g of salt.

To convert 220 g to moles

Amount (mol)  = (Mass of salt)/ (Molar mass).

Molar mass of KNO₃ = 39 + 14 + 48 = 101 g.mol^-1

i.e. Amount, in moles = 220/ 101  = 2.18 moles

Therefore, Solubility of KNO₃ at 30^oC = 2.18 mol.dm^-3

Solubility in grams per 100 g of water

Mass of water used = (y – w)  = (40.70  – 20.70 ) g = 20.0 g

Mass of salt used =      (w – x)  = (20.70  – 15.20) g = 5.50 g

i.e. 20.0 g of water at 30^oC saturated 5.50 g of salt

Therefore, 100 g of water will saturate 5.50 x 100/20 g of salt

i.e.                            5.50 x 5 g = 27.5 g of salt.

Hence, the solubility of the salt at 30^oC is 27.5 g per 100 g of water.

WORKED EXAMPLES

1. If 11.87g of potassium trioxonitrate(V) were dissolved in 43.4 g of distilled water at 60, calculate the solubility of the solute in moldm^-3

Solution:

Molar mass of KNO₃ = 101g

1g of water = 1cm3 of water

Mass of dissolved KNO₃ = 11.87g

Concentration in moldm^-3. Of dissolved KNO₃

= 0.118moldm_-3

43.4g or cm³ of water at 60 dissolve 0.118 moldm_-3 of KNO₃

1000cm³ of water at 60 will dissolve:

= 2.7 moldm^-3 of KNO₃

Therefore, the solubility of KNO₃ at 60 in water is 2.7moldm^-3

EVALUATION

1. Explain the following terms: (a) solute (b) solvent (c) solution
2. What is solubility? List the different types of solution.
3. Calculate the solubility of Pb(NO₃)₂in moldm^-3 if 12.2g of the solute were dissolved in 21g of distilled water 20.

PERIOD 3: TYPES OF SOLUTION

A solution is a homogenous mixture which is formed when a solute is completely dissolved in a solvent. A solution can be saturated, unsaturated or supersaturated.

1. A saturated solution is one which contains, at a given temperature, as much solute as it can hold in the presence of the dissolving substance. That is, a given solute dissolves  in a given solvent, such that no more of the solute will dissolve in the solvent at the temperature of dissolution.
2. An unsaturated solution is one which contains less solute than it can hold at the given temperature.
3. A supersaturated solution is one which contains more of the dissolves substance than it can hold at a given temperature, when the saturated solution is in contact with the solid solution. Super saturated solution are unstable. The excess solute can be separated out by slightly shaking the solution or creating centres of crystallization for the excess solute.

Relationship Between solubility and crystallization

The different solubilities of substances are utilized in their purification by the process of crystallization. In manufacturing KNO₃, solutions of KCl and NaNO₃ are mixed. The resulting solution is then concentrated at boiling point and the NaCl is then deposited. The equilibrium in the equation is displaced to the right.

K⁺ + Cl^– +  Na⁺ +  NO₃^–                    K⁺ +  NO₃^– + NaCl

The solution is filtered hot and later cooled. KNO₃, with the lowest solubility, crystallizes out first and is purified by recrystallization :

KCl  +  NaNO₃    ⇌   KNO₃   +   Na⁺ + Cl^–

A saturated solution is in dynamic equilibrium with the excess solid present in the solution. The dissolution equilibrium can be expressed in terms of an equilibrium constant, e.g. in the case of silver chloride

AgCl_(s)     ⇌     Ag⁺_(aq)  + Cl^–_(aq)

EVALUATION

1. Define the term saturated solution.
2. Distinguish between saturated and unsaturated solutions.

PERIOD 4.  FACTORS AFFECTING SOLUBILITY OF SOLIDS IN LIQUIDS

 The four factors that influence or affect the solubility of substances are:

1. Nature of solvent: Generally, ionic compounds are more soluble in polar solvent (e.g. water), than in non- polar increases the solubility of the solvent e.g. NaCl is more soluble in water than in ethanol.
2. Nature of the solute: Ionic compounds are more soluble than covalent compounds and vice-versa. For instance, NaCl is more soluble in water than sugar at room temperature.
3. Temperature: In most cases solubility increases with increase in temperature. In endothermic reactions solubility increases with increasing temperature and vice versa e.g. calcium oxide.
4. Common ion effect: The solubility of an ionic compound in water is affected by the presence of another compound if both compounds have the same cation or anions. For example, the solubility of NaCl in water is much higher than its solubility in dilute HCl because there is a common chloride in NaCl and HCl.

Uses of Solubility curves

1. Solubility curves enable pharmacists to determine the amount of solid drugs in a solution of drug mixture of solids
2. It helps separate and purify mixtures of solids.
3. Used to determine the most suitable solvents for the extraction of solutes from natural sources.

EVALUATION:

1. Mention three factors that affect the solubility of a solid in a liquid.
2. Explain the term common ion effect.

GENERAL EVALUTION

OBJECTIVE TEST:

1. A substance that dissolves in a solvent to form a solution is called (a) acid (b) suspension (c) glue (d) solute
2. The shape of a water molecule is: (a) K-shaped (b) V-shaped (c) N-shaped (d) S-shaped (e) Y-shaped
3. Which of the following birds water molecules together?  (a) ionic bond (b) covalent bond (c) hydrogen bond (d) van der Waals forces
4. When a crystal was added to its solution, it did not dissolve and the solution remained unchanged, showing that the solution was. (a) concentrated (b) colloidal (c) saturated (d) unsaturated (e) supersaturated
5. Calculate the solubility of KCl in moldm^-3at 30 if 20cm³ sample of a saturated solution of KCl contains 4.50g of the salt. (a) 2.5 (b) 3.0 (c) 3.02 (d) 2.03

ESSAY QUESTIONS:

1. If the solubility of KHCO₃is 0.4 moldm^-3 at room temperature, at 20 (KHCO₃ = 100g.mol^-1)
2. (a) What is solubility curve? (b) Sketch a curve to show how the solubility of a gas varies with an increase in temperature. (c) List two uses of solubility curve.
3. Name and explain any three factors that affect the solubility of a solute in a solvent.
4. Explain why a solid dissolves faster in hot water than cold water
5. (a) How is the solubility of a solute determined? (b) If the solubility of KHCO₃ at 20is 3.10moldm^-3 determine whether a solution of KHCO₃ containing 303gdm^-3 at 20 is saturated or unsaturated.

WEEK 9

TOPIC: AIR

CONTENTS:

1. CONSTITUENTS OF AIR
2. PROPERTIES OF AIR
3. AIR POLLUTION
4. FLAME

PERIOD 1:  CONSTITUENTS OF AIR

Air is a mixture of gases, composed mainly of nitrogen and oxygen with small amounts of Carbon (iv) oxide, noble gases and water vapour. The constituents of air can be separated easily by physical methods. Nitrogen accounts for about 4/5 of the atmosphere by volume, while oxygen occupies the remaining 1/5

Percentage composition

The chief natural constituents of air and their percentage composition by volume are as follow:

PERIOD 3: PROPERTIES OF AIR

1. Water vapour in air: The presence of water vapour or moisture in the atmosphere is due to the evaporation of water from the oceans, rivers, lakes, sea and water reservoirs. The amount present in air varies with temperature, the prevailing weather and the position of the place.
2. Noble gases in air: Argon is the most abundant of the bobble gases in the air, then neon. Noble gases are generally in reactive. They are used to produce colourful light for advertisement.
3. Carbon (iv) oxide air: The presence of carbon (iv) oxide in air can be proved by the passage of air through aqueous solution of calcium hydroxide (lime water) or KOH. The percentage of carbon (iv) oxide in air is balanced and maintained by the process of respiration and photosynthesis
4. Oxygen in air: oxygen is the most active component of air. It supports life. The process of burning a substance in air is called combustion. Its major reactions include corrosion (rusting), respiration and combustion.

Corrosion: This is the change on metallic surface that is observed after being exposed to air for few days. The corrosion of iron is commonly known as rusting. This can be prevented by (a) Applying grease (b) painting exposed surface (c) coating with metal e.g. chromium

EVALUATION

1. Mention two substances that can be used to remove carbon (iv) oxide from air.
2. List 5 properties of air.
3. What is corrosion? How can it be prevented?
4. State the percentage by volume of CO₂ in the atmosphere.

PERIOD 3: AIR POLLUTION

Air pollution is the released of substances into the atmosphere in quantities that are harmful to lives.

Causes of air pollution

1. COMBUSTION OF FOSSIL FUEL SUCH AS COAL, GASOLINE AND PETROLEUM GASES.
2. RELEASE OF FREONS FROM AEROSOL CANS.
3. RELEASE OF POISONOUS GASES FROM CHEMICAL WARFARE

Air pollutants:

1. Particulate matter: these are materials which exist as micro-sized solids in the air. They include dust, smoke, soot, asbestos, heavy metals particles etc. some of these heavy metals are lead, cadmium, arsenic and mercury. Lead particles are released to the atmosphere from the combustion of petrol in the form of lead (ii) bromide. Exposure to lead pollution over a long period of time is dangerous. This can cause brain damage among children. It also cause irritability, aggressive tendencies and gastric disorder. Mercury causes poison while cadmium causes severe gastric disorder.
2. Oxides of carbon: carbon (ii) oxide is released into the atmosphere through incomplete combustion of fossil fuel. It causes brain damage when inhaled even in low concentration. In high concentration, it results to death because it combines with the haemoglobin thereby depriving the body of oxygen. Increased amount of carbon (iv) oxide in the atmosphere leads to increased retention of infra-red rays and hence further warming of the earth-a phenomenon known asgreenhouse effect. Therefore carbon (IV) oxide is called greenhouse gas.
3. Oxides of Sulphur and nitrogen: Sulphur (IV) oxide is derived from burning of coal, the combustion and refining of petroleum and the smelting of mineral ores for extraction. Once Sulphur (IV) oxide is present in air, it is oxidized by atmospheric oxygen to Sulphur (VI) oxide. The Sulphur (VI) is washed out by rain which converts it to tetraoxosulphate (VI) acid. This is called acid rain. The effect of inhaling the Sulphur (IV) oxide and acid rain are as follow:

(a) Sulphur (IV) oxide causes respiratory disorder.

(b). Acid rain causes death of plants.

(c). Acid rain leads to corrosion of metals, aging of fabrics, stiffening and crack in of leather, and destruction of buildings made of limestone

Nitrogen (ii) oxide causes depletion of ozone layer, and nitrogen (iv) oxide results to acid rain

4. Gaseous hydrocarbons and chlorofluorocarbons: some hydrocarbons in the air are the volatile ones such methane. They have carcinogenic effect in man and animal. Methane is also a greenhouse gas. Chlorofluorocarbons deplete the ozone layer. This exposes the earth to ultra-violet rays which causes skin cancer, sunburns and their ailments.

EVALUATION:

1. What is air pollution?
2. State four air pollutants
3. What is global warming? Name two greenhouse gases.

PERIOD 4: FLAMES

Flames are produced when substances burn. A flame is a region where gases combine chemically, with the production of heat and light. The type of flame produced depends on the nature of the substance that is burning. A flame may be luminous (brightly seen and yellow) or non – luminous (barely seen). Types of flame can be explained further using hydrogen, candle and Bunsen flames as examples.

• Hydrogen flame: Hydrogen burns with a very faint, non- luminous flame. The structure is simple, consisting of only regions: the unburnt gas zone and the zone of complete combustion.
• Candle flame: A candle burns with luminous flame four zones can be identified in the flame.
• The zone of unburnt gas around the wick
• The bright yellow luminous zone where there is incomplete burning of the hydrocarbon due to insufficient air supply
• The barely visible, non- luminous zone on the outside where complete of carbon particles take place due to sufficient air supply
• The blue zone at the base of the flame which also a region of complete combustion.

• Bunsen flame: A Bunsen burner is built with an air inlet of the base of the burner tube so that a stream of air can be supplied to the flame together with the fuel gas.

It is important equipment in a school laboratory used for heating. The fuel for the burner is a mixture of hydrocarbon gas (methane and butane) and some hydrogen and carbon (II) oxide

Luminous Bunsen flame: to produce luminous Bunsen flame, the air hole at the base of the burner tube should be closed. The flame produced is high, bright, wavy and yellow but not hot.

Non- luminous Bunsen flame: To produce non- luminous Bunsen flame, the air hole should be kept open. The flame is much hotter, cleaner and more compact than the luminous Bunsen flame.

EVALUATION:

1. Define flame
2. Compare the luminous and non- luminous flame Bunsen flames.
3. Describe the structure of a candle flame and explain the formation the products obtained during the burning.

GENERAL EVALUATION:

OBJECTIVE TEST:

1. These gases burn with blue flames except.

(a) C₂H₂ (b) H₂ (c) CH₄ (d) CO (e) SO₂

2. All these are the causes of luminosity in flames except.

(a) Solid particles (b) size of material (c) solid particles and increased pressure (d) pressure (e) solid particles and increased temperature

3. Which of these is not necessary for metallic corrosion?

(a) SO₂ (b) water (c) heat (d) oxygen (e) CO₂

4.  The most abundant noble gas in nature is

(a) Neon (b) Argon (c) Helium (d) Krypton (e) Radon

5. When air is passed through a tube containing finely divided copper, the component of air absorbed is (a) Nitrogen (b) Water vapour (c) noble gas (d) carbon (IV) oxide (e) oxygen

ESSAY TEST:

1. (a) List four examples of noble gases. (b) Compare the structure of luminous and the non- luminous.
2. Give the constituents of atmospheric air, and indicate its approximate percentage composition.
3. Define combustion. Combustion in air would be impossible in the absence of which gas?
4. Draw and label the structures of (a) hydrogen flame (b) candle flame
5. What is rusting and how can it be prevented. (b) Explain any two properties of air.
6. Revision.
7. Examination.

Hope you got what you visited this page for? The above is the lesson note for Chemistry for SS2 class. However, you can download the free PDF file for record purposes.

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