WAEC SSCE - Chemistry - 1988 (Essay)

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(a) Major raw materials

(b) One problem associated with oil producing areas

Environmental pollution caused by oil spillage, which destroys aquatic life, farmland and drinking water. (Air pollution from gas flaring is also acceptable.)

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(a) The collision theory and the rate of reaction

The collision theory states that for a chemical reaction to occur, the reacting particles must collide with one another. However, not every collision leads to a reaction. Only collisions in which the particles possess energy equal to or greater than the activation energy, and in which the particles are correctly orientated, are effective and lead to product formation. The rate of a reaction therefore depends on the frequency of effective collisions: the greater the number of effective collisions per unit time, the faster the reaction.

(b) Effect of the following on rate of reaction

• (i) Surface area: increasing the surface area of a solid reactant (for example by powdering it) exposes more particles to collision. This increases the frequency of effective collisions and so increases the rate of reaction.
• (ii) Catalyst: a catalyst provides an alternative reaction pathway of lower activation energy. More colliding particles then possess the required energy, so the number of effective collisions and hence the rate increases.

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(a) What is a carbohydrate?

A carbohydrate is an organic compound containing carbon, hydrogen and oxygen, in which the hydrogen and oxygen are usually present in the ratio 2 : 1 (the same ratio as in water). Carbohydrates have the general formula C_x(H₂O)_y and are polyhydroxy aldehydes or ketones (or compounds that yield them on hydrolysis).

(b) Two types of carbohydrates with one example of each

Any two types with their correct examples are acceptable.

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(a) Properties of transition metals

• (i) High melting points: transition metals have strong metallic bonds because both the outer s electrons and some inner d electrons take part in bonding, giving a large number of delocalised electrons and a strong lattice that needs much energy to break.
• (ii) Variable oxidation states: the energies of the 3d and 4s electrons are close, so different numbers of these electrons can be lost during bonding, giving several stable oxidation states.
• (iii) Paramagnetism: they have unpaired electrons in their d orbitals, and these unpaired electrons cause the atoms to be attracted into a magnetic field (paramagnetic).

(b) Bauxite and aluminium oxide

• (i) Impurities in bauxite: iron (III) oxide (Fe₂O₃) and silicon (IV) oxide (SiO₂).
• (ii) Removal of impurities: the powdered bauxite is treated with hot concentrated sodium hydroxide solution. The amphoteric aluminium oxide dissolves to form sodium aluminate, while the insoluble impurities (iron (III) oxide) are filtered off. Pure aluminium oxide is then precipitated and heated.
• (iii) Amphoteric nature of aluminium oxide: aluminium oxide is amphoteric because it reacts with both acids and bases to form a salt and water. With acid: Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O. With base: Al₂O₃ + 2NaOH + 3H₂O → 2NaAl(OH)₄.

(c) Electrolysis of copper (II) tetraoxosulphate (VI) using copper electrodes

(i) At the cathode, copper ions are discharged and copper is deposited: Cu²⁺_(aq) + 2e^- → Cu_(s). At the anode, the copper electrode itself dissolves (goes into solution) rather than oxygen being evolved: Cu_(s) → Cu²⁺_(aq) + 2e^-. Copper thus dissolves from the anode and is deposited on the cathode. This is the principle used in the electrolytic purification of copper.

(ii) No, the blue colour does not change. For every Cu²⁺ ion removed at the cathode, one Cu²⁺ ion enters the solution from the anode. The concentration of copper (II) ions therefore remains constant, so the blue colour of the solution is unchanged.

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(a) Functions of the substances in the Solvay Process

• (i) Limestone: it is heated (decomposed) to supply carbon (IV) oxide for the process and to produce calcium oxide (quicklime), which is later used to recover ammonia.
  CaCO_3(s) → CaO_(s) + CO_2(g)
• (ii) Ammonia: it makes the brine alkaline (forms ammoniacal brine) so that it can absorb carbon (IV) oxide, and it is recovered and recycled at the end of the process.
  NH_3(g) + H₂O_(l) + CO_2(g) → NH₄HCO_3(aq)
• (iii) Brine: it supplies the sodium ions (as concentrated sodium chloride solution) needed to form sodium hydrogen trioxocarbonate (IV).
  NH₄HCO_3(aq) + NaCl_(aq) → NaHCO_3(s) + NH₄Cl_(aq)

(b) Why the reaction is a neutralization

Aqueous sodium trioxocarbonate (IV) is a base (it is the salt of a strong base and a weak acid and reacts as a base towards acids), while dilute hydrochloric acid is an acid. Their reaction produces a salt and water only, which is the definition of neutralization:

Na₂CO_3(aq) + 2HCl_(aq) → 2NaCl_(aq) + H₂O_(l) + CO_2(g)

The essential ionic change is H⁺ + OH^- (from the carbonate acting as a base) forming water; hence it is a neutralization.

(c) Mass of Na₂CO₃ from 16.8 g of NaHCO₃

Decomposition equation:

2NaHCO_3(s) → Na₂CO_3(s) + H₂O_(l) + CO_2(g)

Molar mass of NaHCO₃ = 23 + 1 + 12 + 48 = 84 g mol^-1

Moles of NaHCO₃ = 16.8 ÷ 84 = 0.2 mol

From the equation, 2 mol NaHCO₃ give 1 mol Na₂CO₃, so:

Moles of Na₂CO₃ = 0.2 ÷ 2 = 0.1 mol

Molar mass of Na₂CO₃ = 46 + 12 + 48 = 106 g mol^-1

Mass of Na₂CO₃ = 0.1 × 106 = 10.6 g

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(a) Two allotropes of carbon

Allotropes are different forms of the same element existing in the same physical state but with different crystalline structures.

• Diamond
• Graphite

(Buckminsterfullerene, C₆₀, is also acceptable.)

(b) Products of the destructive distillation of coal and one use of each

Destructive distillation of coal is the heating of coal to a high temperature in the absence of air. It yields the following:

Any two products with their matching uses are acceptable.

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(a) The element X, represented as ⁴⁰₂₀X (atomic number 20, mass number 40)

• (i) Electrons and neutrons: number of electrons = atomic number = 20. Number of neutrons = mass number - atomic number = 40 - 20 = 20.
• (ii) Electronic configuration: 2, 8, 8, 2 (that is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²). X is calcium.

(b) Reaction with chlorine (atomic number 17)

• (i) The bond formed is an electrovalent (ionic) bond.
• (ii) X has 2 electrons in its outermost shell and each chlorine atom needs 1 electron to complete its octet. The X atom loses its 2 outer electrons to become X²⁺, and two chlorine atoms each gain one electron to become Cl^-. The oppositely charged ions are held together by strong electrostatic attraction.
• (iii) Formula of the compound: XCl₂ (calcium chloride, CaCl₂). Three properties: it is a crystalline solid with a high melting point; it is soluble in water; and it conducts electricity when molten or in aqueous solution.

(c) Possible relative molar masses of a chlorine molecule

Chlorine exists as Cl₂. Using isotopes of mass 35 and 37 the possible molecular masses are:

• 35 + 35 = 70
• 35 + 37 = 72
• 37 + 37 = 74

(d) Mass of one atom of carbon

1 mole (6.02 × 10²³ atoms) weighs 12.0 g, so:

Mass of one atom = 12.0 ÷ (6.02 × 10²³) = 1.99 × 10^-23 g

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(a) Ionic equation for the reaction between zinc powder and silver trioxonitrate (V) solution

Zinc displaces silver from the solution because zinc is higher than silver in the electrochemical series:

Zn_(s) + 2Ag⁺_(aq) → Zn²⁺_(aq) + 2Ag_(s)

(b) Oxidation and reduction

• (i) Oxidized: zinc (Zn). Its oxidation number rises from 0 to +2; it loses electrons.
• (ii) Reduced: the silver ion (Ag⁺). Its oxidation number falls from +1 to 0; it gains electrons.

(c) Two applications of oxidation numbers

• They are used in writing and balancing redox equations.
• They are used in the systematic naming (IUPAC nomenclature) of compounds, for example iron (II) and iron (III).
• They are used to identify the species oxidized or reduced and the oxidizing and reducing agents in a reaction.

Any two are acceptable.

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(a) Equation for the reaction

S_(s) + O_2(g) → SO_2(g)

(b) Volume of oxygen used up

From the equation, the mole ratio of O₂ to SO₂ is 1 : 1. By Gay-Lussac's law of combining volumes, gases combine in volume ratios equal to their mole ratios (at the same temperature and pressure). Therefore:

1 volume of O₂ produces 1 volume of SO₂

Since 50 cm³ of SO₂ is produced, the volume of oxygen used up = 50 cm³.

(c) Gas law applicable

Gay-Lussac's law of combining volumes. It states that when gases react, they do so in volumes which bear a simple whole number ratio to one another and to the volumes of the gaseous products, provided the temperature and pressure remain constant.

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(a) Order of increasing energy

The energy carried by the three radiations increases in the order:

alpha particles < beta particles < gamma rays

Alpha particles are the least energetic (heavy, slow moving), beta particles are of intermediate energy, while gamma rays are the most energetic (high frequency electromagnetic radiation).

(b) One difference between nuclear fission and nuclear fusion

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(a) Examples

• Heavy chemical: tetraoxosulphate (VI) acid, H₂SO₄ (sodium hydroxide, ammonia or sodium trioxocarbonate (IV) are also acceptable). Heavy chemicals are produced on a very large scale and sold cheaply.
• Fine chemical: a drug such as aspirin (dyes, perfumes and photographic chemicals are also acceptable). Fine chemicals are produced in small quantities, are of high purity and are relatively expensive.

(b) Compounds with the formula C₄H₉Cl

The molecular formula C₄H₉Cl has four structural isomers:

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Gay-Lussac's law of combining volumes

The law states that when gases react, they do so in volumes which bear a simple whole number ratio to one another and to the volumes of the gaseous products, provided the temperature and pressure remain constant.

Illustration with a chemical reaction

Consider the formation of ammonia:

N_2(g) + 3H_2(g) → 2NH_3(g)

Here 1 volume of nitrogen combines with 3 volumes of hydrogen to give 2 volumes of ammonia. The ratio 1 : 3 : 2 is a simple whole number ratio, which illustrates the law.

(The reaction H_2(g) + Cl_2(g) → 2HCl_(g), with ratio 1 : 1 : 2, is also acceptable.)

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(a) Structural formulae

• (i) 2,2,4-trimethylpentane: (CH₃)₃C-CH₂-CH(CH₃)-CH₃
• (ii) Ethyl methanoate: HCOOCH₂CH₃
• (iii) 2,3-dimethylbut-2-ene: (CH₃)₂C=C(CH₃)₂

(b) Straight-chain isomer of 2,2,4-trimethylpentane

The molecular formula is C₈H₁₈, so the straight-chain isomer is octane: CH₃CH₂CH₂CH₂CH₂CH₂CH₂CH₃

(c) Oxidation equations

• (i) Secondary alkanol (gives a ketone), using propan-2-ol:
  CH₃CH(OH)CH₃ + [O] → CH₃COCH₃ + H₂O
• (ii) Dihydric alkanol, using ethane-1,2-diol:
  HOCH₂CH₂OH + 4[O] → HOOC-COOH + 2H₂O

(d) Cassava sequence

• (i) Compound D belongs to the carbohydrates (a sugar/monosaccharide, glucose).
• (ii) E is ethanol; F is ethyl ethanoate.
• (iii) The functional group in F is the ester (alkanoate) group, -COO-.
• (iv) C₂H₅OH + CH₃COOH → CH₃COOC₂H₅ + H₂O
• (v) The reaction is esterification (a condensation reaction).

(e) Order of increasing boiling point

The compounds are butane (CH₃CH₂CH₂CH₃), hexane (CH₃CH₂CH₂CH₂CH₂CH₃) and propan-1-ol (CH₃CH₂CH₂OH).

Increasing order: butane < hexane < propan-1-ol

Explanation: butane and hexane are non-polar alkanes held together only by weak van der Waals forces; hexane has a larger molecule with more electrons than butane, so its van der Waals forces are stronger and its boiling point higher. Propan-1-ol contains the -OH group and its molecules are held together by strong hydrogen bonds, which require the most energy to break, so it has the highest boiling point despite its smaller size.

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Distinction between cracking and reforming

Importance in the petroleum industry

• Cracking converts the less useful heavy fractions (such as heavy naphtha and diesel) into more valuable products, especially petrol, and provides alkenes (ethene, propene) as raw materials for the petrochemical and plastics industries.
• Reforming converts straight-chain hydrocarbons into branched-chain and aromatic hydrocarbons, thereby raising the octane number of petrol and improving its anti-knock quality, and supplies aromatics for the chemical industry.