WAEC SSCE - Chemistry - 1993 (Essay)

Bayanin Amsa

(a)(i) Strong acid vs concentrated acid: a strong acid is one that ionises (dissociates) completely into ions in aqueous solution, whereas a concentrated acid is one containing a large amount of acid (solute) in a small amount of water (little water). Strength refers to degree of ionisation; concentration refers to amount of solute per unit volume.

(a)(ii) Amphoteric oxide: an oxide that can react with both acids and bases to form a salt and water, e.g. aluminium oxide, Al₂O₃ (or ZnO, PbO).

(b)(i) Contact process for H₂SO₄:

1. Sulphur (or a sulphide ore) is burnt in air to give sulphur(IV) oxide: \(\text{S} + \text{O}_2 \to \text{SO}_2\).
2. The purified, dry SO₂ and excess air are passed over a catalyst of vanadium(V) oxide (V₂O₅) at about 450 °C and 1-2 atm, oxidising it to sulphur(VI) oxide: \(2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3\).
3. The SO₃ is absorbed in concentrated H₂SO₄ to form oleum (fuming sulphuric acid): \(\text{SO}_3 + \text{H}_2\text{SO}_4 \to \text{H}_2\text{S}_2\text{O}_7\).
4. The oleum is diluted carefully with water to give concentrated H₂SO₄: \(\text{H}_2\text{S}_2\text{O}_7 + \text{H}_2\text{O} \to 2\text{H}_2\text{SO}_4\).

(b)(ii) Equations:

• As a dehydrating agent (removing the elements of water from sugar): \(\text{C}_{12}\text{H}_{22}\text{O}_{11} \xrightarrow{\text{conc. H}_2\text{SO}_4} 12\text{C} + 11\text{H}_2\text{O}\).
• As an oxidizing agent (hot concentrated acid with copper): \(\text{Cu} + 2\text{H}_2\text{SO}_4 \to \text{CuSO}_4 + \text{SO}_2 + 2\text{H}_2\text{O}\).

(b)(iii) Tetraoxosulphate(VI) acid is classified as a heavy chemical because it is manufactured and used in very large (industrial) quantities and is relatively cheap and impure.

Bayanin Amsa

(a) Air pollutant that causes:

• (i) Blood poisoning: carbon(II) oxide (carbon monoxide, CO), which combines with haemoglobin and prevents oxygen transport. (Lead compounds also poison the blood.)
• (ii) Acid poisoning / acid rain: sulphur(IV) oxide (SO₂) (also oxides of nitrogen, NO_x).
• (iii) Blackening of the walls of buildings: carbon particles/soot (unburnt carbon from incomplete combustion).

(b) Major chemical industry using as raw material:

• (i) Petrochemicals: the plastics/polymer (petrochemical) industry (also the synthetic fibre or synthetic rubber industry).
• (ii) Cellulose: the paper industry (also the rayon/artificial silk or nitrocellulose industry).

Bayanin Amsa

The four compounds shown are:

• (a) \(CH_3-CH{=}CH-CH_3\) (but-2-ene);
• (b) \(CH_3-CH_2-CHBr-CH_2Br\) (1,2-dibromobutane);
• (c) \(HOOC-CH(CH_3)-CH_2-COOH\) (2-methylbutanedioic acid);
• (d) \(HOOC-CH{=}CH-COOH\) (butenedioic acid).

Compounds that exhibit geometric (cis-trans) isomerism: (a) and (d).

Reason. Geometric isomerism requires two conditions to be met together:

1. a carbon-to-carbon double bond (\(C{=}C\)), because the double bond prevents (restricts) free rotation about that bond; and
2. each of the two doubly-bonded carbon atoms must carry two different groups.

In (a) each doubly-bonded carbon carries an \(H\) and a \(CH_3\) group (two different groups), so cis- and trans-but-2-ene exist. In (d) each doubly-bonded carbon carries an \(H\) and a \(-COOH\) group, so the cis form (maleic acid) and the trans form (fumaric acid) exist. Both conditions are satisfied, so (a) and (d) show geometric isomerism.

Compounds (b) and (c) contain only single (C-C) bonds. Single bonds allow free rotation, so the atoms cannot be locked into fixed cis/trans positions; therefore (b) and (c) do not exhibit geometric isomerism.

Bayanin Amsa

(a)(i) Four characteristic properties of transition metals: they form coloured ions/compounds; they show variable oxidation states; they and their compounds act as catalysts; they form complex ions. (They are also hard, dense metals with high melting points and are often paramagnetic.)

(a)(ii) Two metals extracted by electrolysis: sodium and aluminium (also potassium, calcium, magnesium).

(b)(i) Oxidation number of Cr in Cr₂O₇^2-:

\[ 2x + 7(-2) = -2 \Rightarrow 2x = +12 \Rightarrow x = +6 \]

(b)(ii) Adding dilute H₂SO₄ (increasing H⁺) to the equilibrium \(\text{Cr}_2\text{O}_7^{2-} + \text{H}_2\text{O} \rightleftharpoons 2\text{CrO}_4^{2-} + 2\text{H}^+\) shifts the position of equilibrium to the left (Le Chatelier's principle). The colour therefore changes from yellow (chromate) to orange (dichromate).

(c)(i) Bubbling H₂S into acidified K₂Cr₂O₇: the orange solution turns green. H₂S is a reducing agent and reduces Cr⁶⁺ (orange dichromate) to green Cr³⁺, while sulphur is deposited (a pale yellow precipitate):

\[ \text{Cr}_2\text{O}_7^{2-} + 8\text{H}^+ + 3\text{H}_2\text{S} \to 2\text{Cr}^{3+} + 7\text{H}_2\text{O} + 3\text{S} \]

(c)(ii) Precaution: generate and use the hydrogen sulphide in a fume cupboard (fume chamber) so that the poisonous gas is drawn away and not inhaled.

Bayanin Amsa

(a)(i) The rate of a chemical reaction is the change in concentration (or amount) of a reactant used up, or product formed, per unit time.

(a)(ii) Interpreting “vision theory” as collision theory: particles must collide for a reaction to occur. Increasing temperature increases the particles’ average kinetic energy, so they move faster and collide more frequently. It also means that a greater proportion of collisions have energy at least equal to the activation energy. Therefore, there are more successful collisions each second and the reaction rate increases.

(b)(i)

\[2\text{H}_2\text{O}_{2(aq)} \rightarrow 2\text{H}_2\text{O}_{(l)}+\text{O}_{2(g)}\]

(b)(ii) Use manganese(IV) oxide, \(\text{MnO}_2\), as the catalyst.

1. Place equal volumes and concentrations of hydrogen peroxide solution into two identical conical flasks at the same temperature.
2. Fit each flask with a bung and delivery tube connected to a gas syringe.
3. Add a measured mass of \(\text{MnO}_2\) to one flask, immediately fit the bung, and start the timer. Leave the other flask without catalyst as a control.
4. Record the volume of oxygen collected in each gas syringe at regular time intervals.

The flask containing \(\text{MnO}_2\) produces oxygen more rapidly: it will give a steeper volume-of-oxygen-against-time graph and reach a given oxygen volume in a shorter time. This shows that manganese(IV) oxide increases the rate of decomposition. Keeping the volume and concentration of hydrogen peroxide, temperature, and apparatus the same makes this a fair comparison.

(b)(iii) The reaction is exothermic, so the products have lower energy than the reactants. A catalyst provides an alternative pathway with a lower activation energy, but it does not change the enthalpy change, \(\Delta H\).

(c)(i) Enthalpy change alone cannot predict spontaneity because spontaneity depends on both enthalpy change and entropy change, as well as temperature:

\[\Delta G=\Delta H-T\Delta S\]

A process is thermodynamically spontaneous when \(\Delta G<0\). An endothermic reaction, with positive \(\Delta H\), can still be spontaneous if the entropy increase is sufficiently large at a suitable temperature. Conversely, a favourable enthalpy change alone does not guarantee spontaneity under all conditions. Also, spontaneity does not mean that a reaction is fast: hydrogen peroxide decomposition may be thermodynamically feasible but slow without a catalyst because of its activation energy.

Examination reminder: Use \(\Delta H\) to describe heat energy change, but use \(\Delta G\), including \(\Delta S\) and \(T\), to decide thermodynamic spontaneity.

Bayanin Amsa

• (a) Fractional distillation of liquefied air: physical change (the components are only separated by boiling point; no new substance is formed).
• (b) Cracking of petroleum fractions: chemical change (large molecules are broken into new, smaller molecules).
• (c) Conversion of rhombic sulphur to monoclinic sulphur: physical change (only a change in the crystalline form/allotrope; the substance is still sulphur).
• (d) Chromatographic separation of chlorophyll: physical change (the pigments are merely separated; no new substance is formed).

Bayanin Amsa

X is a Group I element (a metal, e.g. sodium) and Y is a Group VII element (a non-metal, e.g. chlorine); both are in the same period.

(a)

• (i) A good oxidizing agent: Y (the Group VII element).
• (ii) Smaller atomic volume / smaller atomic radius: Y (across a period the nuclear charge rises and the atom contracts).
• (iii) Higher ionization potential (ionization energy): Y.

(b) Explanation for (a)(i): Y (Group VII) has 7 electrons in its outermost shell and needs to gain only one electron to attain a stable octet. It therefore has a strong tendency to accept (gain) electrons from other species; a species that gains electrons is an oxidizing agent. X (Group I) instead loses its single valence electron easily and acts as a reducing agent. Hence Y is the good oxidizing agent.

Bayanin Amsa

• (a) Diamond for drilling rocks: diamond is extremely hard (the hardest natural substance) because each carbon atom is covalently bonded to four others in a rigid three-dimensional structure, so it can cut and drill hard rocks.
• (b) Diamond as jewels: diamond has a very high refractive index and disperses light strongly, so it sparkles brilliantly when cut, making it valuable as a gem.
• (c) Graphite as electrodes: graphite is a good conductor of electricity (it has delocalised/mobile electrons between its layers) and has a high melting point, so it is used as an inert electrode.
• (d) Graphite for slowing down neutrons in nuclear reactors: graphite is a good moderator; its carbon atoms slow down fast neutrons without absorbing them, and it withstands the high temperatures involved.
• (e) Wood charcoal in gas masks: wood charcoal is highly porous with a very large surface area, so it adsorbs (takes onto its surface) poisonous gases, purifying the air breathed.

Bayanin Amsa

(a) Phenomena:

• (i) The spreading of the smell of hydrogen sulphide gas through the laboratory illustrates diffusion (the movement of gas particles from a region of high concentration to a region of low concentration).
• (ii) The existence of atoms of the same element having different mass numbers illustrates isotopy (the atoms are isotopes).

(b) Relative atomic mass

Atomic number = 17, so each atom has 17 protons. The two isotopes have mass numbers:

• 17 protons + 18 neutrons = 35
• 17 protons + 20 neutrons = 37

\[ \text{RAM} = \frac{(35\times75)+(37\times25)}{100} = \frac{2625 + 925}{100} = \frac{3550}{100} = 35.5 \]

The relative atomic mass of the element is 35.5 (the element is chlorine).

Bayanin Amsa

(a)

• (i) \(X = \dfrac{\text{mass of L molecules of gas or vapour}}{\text{mass of L molecules of hydrogen}}\) is the relative density (vapour density) of the gas. Because L molecules is one mole, this ratio equals the molar mass of the gas divided by the molar mass of hydrogen, i.e. relative density \(= \tfrac{1}{2}\times\) molar mass.
• (ii) \(Y = -nFE^\circ\) is the standard change in Gibbs (free) energy of the cell reaction, \(\Delta G^\circ\), where n is the number of moles of electrons transferred and F is the Faraday constant.

(b) Two differences between a primary cell and a secondary cell:

Primary cell	Secondary cell
Cannot be recharged; used once and discarded.	Can be recharged and used repeatedly.
The cell reaction is irreversible.	The cell reaction is reversible.
Example: Leclanche (dry) cell.	Example: lead-acid accumulator.

Bayanin Amsa

The compound shown is a glyceryl ester (triglyceride): a glycerol backbone CH2-CH-CH2 esterified with three fatty-acid chains.

• The top and bottom chains, -COO(CH2)12CH3, are saturated (no C=C).
• The middle chain, -COOCH=CHCH=CHCH=CHCH=CHCH3, is unsaturated and contains four carbon-carbon double bonds.

(a) Change in physical state on hydrogenation (H₂, finely divided Ni, ~180 °C, long time)

The four C=C double bonds in the unsaturated chain add hydrogen and become saturated (this is hardening, the industrial process used to make margarine). The soft (semi-liquid) fatty substance therefore changes into a hard solid fat. In short: soft/oily → hard solid.

(b) Amount of hydrogen consumed by one mole of the component

Each carbon-carbon double bond adds exactly one mole of hydrogen; the ester (C=O) bonds are not reduced under these conditions.

\[\text{Number of C=C bonds}=4 \;\Rightarrow\; \text{moles of }H_2 = 4\]

\[\text{One mole of the triglyceride} + 4\,\text{mol }H_2 \longrightarrow \text{saturated triglyceride}\]

Amount of hydrogen consumed = 4 moles.

(c) Reaction with hot concentrated sodium hydroxide solution

Hot concentrated NaOH hydrolyses the ester linkages (saponification), giving:

• Soap - the sodium salts of the fatty acids (sodium tetradecanoate from the saturated chains and the sodium salt of the unsaturated acid), and
• Glycerol (propane-1,2,3-triol).

Bayanin Amsa

(a)(i) Polymerization: the chemical process in which many small molecules (monomers) join together to form a very large molecule (a polymer).

(a)(ii) Conditions for polymerization of ethene: high pressure, high temperature (moderately raised temperature), and a catalyst (e.g. a trace of oxygen or a Ziegler-Natta catalyst).

(a)(iii) The common property is that such compounds are unsaturated (they contain a carbon-carbon double bond, C=C).

(b) From propan-1-ol (CH₃CH₂CH₂OH):

• (i) Propene, by dehydration with concentrated H₂SO₄ (or hot Al₂O₃): \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4} \text{CH}_3\text{CH}=\text{CH}_2 + \text{H}_2\text{O}\). Type: elimination (dehydration).
• (ii) Propyl methanoate, by esterification with methanoic acid: \(\text{HCOOH} + \text{C}_3\text{H}_7\text{OH} \rightleftharpoons \text{HCOOC}_3\text{H}_7 + \text{H}_2\text{O}\). Type: esterification (condensation).

(c)(i) Molecular formula from 40.0% C, 6.7% H, 53.3% O:

Element	%/Ar	ratio
C	40.0/12 = 3.33	1
H	6.7/1 = 6.7	2
O	53.3/16 = 3.33	1

Empirical formula = CH₂O (mass = 30). \(n = \dfrac{180}{30} = 6\), so the molecular formula is C₆H₁₂O₆.

(c)(ii) Ethanoic acid molecules contain O-H groups and form strong intermolecular hydrogen bonds (indeed dimers) between molecules, whereas butane, being non-polar, has only weak van der Waals forces. Much more energy is needed to overcome the hydrogen bonds, so ethanoic acid boils at a much higher temperature despite the similar molar mass.

Bayanin Amsa

(a) One disadvantage of:

• (i) Hard water: it wastes soap because it forms an insoluble scum with soap before it can lather (it can also fur/scale kettles and boilers).
• (ii) Soft water: it dissolves lead from lead pipes, giving water that can cause lead (plumbo) poisoning (soft water is more corrosive to pipes).

(b) Clear lime water contains dissolved calcium hydroxide, Ca(OH)₂, which supplies Ca²⁺ ions responsible for hardness. When carbon(IV) oxide is bubbled through and the lime water turns milky, the Ca²⁺ ions are precipitated as insoluble calcium trioxocarbonate(IV):

\[ \text{Ca(OH)}_2 + \text{CO}_2 \to \text{CaCO}_3\downarrow + \text{H}_2\text{O} \]

Removing Ca²⁺ from solution as the insoluble carbonate lowers the concentration of hardness-causing ions. Hence the milky sample has a lower degree of hardness than the clear lime water.

Bayanin Amsa

(a) Residue on strong heating:

• (i) ZnCO₃ in an open crucible decomposes to give zinc oxide (ZnO) as the residue: \(\text{ZnCO}_3 \to \text{ZnO} + \text{CO}_2\).
• (ii) CuSO₄·5H₂O loses its water of crystallisation to give anhydrous copper(II) tetraoxosulphate(VI), CuSO₄: \(\text{CuSO}_4\!\cdot\!5\text{H}_2\text{O} \to \text{CuSO}_4 + 5\text{H}_2\text{O}\).

(b) Colour changes:

• (i) ZnO: white when cold, turns yellow on heating, and returns to white on cooling.
• (ii) CuSO₄·5H₂O: the blue hydrated crystals turn white (anhydrous CuSO₄) on heating; on cooling in a desiccator it stays white (it would turn blue again if water is added).