WAEC SSCE - Chemistry - 1998

Bayanin Amsa

Bayanin Amsa

The products of the reaction between CH₃CH ₂COOH (propanoic acid) and CH₃OH (methanol) are CH ₃CH₂COOCH₃ (methyl propanoate) and H₂O (water). This is an esterification reaction where the carboxylic acid (propanoic acid) reacts with an alcohol (methanol) in the presence of an acid catalyst (such as sulfuric acid) to form an ester (methyl propanoate) and water. The general equation for this reaction is: R-COOH + R'-OH ⇌ R-COOR' + H₂O In this specific case, the reaction between propanoic acid and methanol produces methyl propanoate and water as the products.

Bayanin Amsa

When Cr^2- is oxidized to CrO₄^2-, it gains oxygen atoms and loses electrons. To determine how many electrons are lost, we need to look at the change in oxidation state of chromium in the reaction. In Cr^2-, the oxidation state of chromium is +2. In CrO₄^2-, the oxidation state of chromium is +6. Therefore, chromium has lost 4 electrons in the reaction (going from +2 to +6). So, the correct answer is 4 electrons.

Bayanin Amsa

Substances which absorb water from the atmosphere without dissolving in it are said to be hygroscopic. This means that they have the ability to attract and hold water molecules from the surrounding environment due to their physical and chemical properties. Examples of hygroscopic substances include salt, sugar, and silica gel. Hygroscopic substances are commonly used as desiccants, which are materials that are used to absorb moisture from the air and keep products dry. They are also used in certain industrial processes to remove water from various substances.

Bayanin Amsa

To solve this problem, we can use the formula: mass of substance deposited = (current × time × molar mass) ÷ (Faraday’s constant × number of electrons transferred) First, we need to find the number of moles of copper deposited by dividing the charge passed by the Faraday’s constant: Q = I × t = 1.0 A × 965 s = 965 C n = Q ÷ (Faraday’s constant × number of electrons transferred) n = 965 C ÷ (96500 C/mol × 2) n = 0.005 mol Next, we can use the number of moles to calculate the mass of copper using its molar mass: mass = n × molar mass mass = 0.005 mol × 63.5 g/mol mass = 0.318 g Therefore, the mass of copper deposited by the current is 0.318 g. The correct option is (a) 0.318g.

Bayanin Amsa

An aqueous solution is acidic if it contains more H₃O- ions than OH- ions. This is because the concentration of H₃O- ions determines the acidity or basicity of a solution. In an acidic solution, there is an excess of H₃O- ions, which can donate a proton (H+) to other substances, making it corrosive and able to change the color of litmus to red. A high pH value indicates that the solution is basic, not acidic. A bitter taste is not a reliable indicator of acidity, as many substances can have a bitter taste without being acidic.

Bayanin Amsa

Bayanin Amsa

Out of the given options, dilute sodium chloride solution is the one that will produce oxygen and hydrogen during its electrolysis using platinum electrodes. During electrolysis, the ions present in the solution are attracted to the electrodes and undergo reduction or oxidation, depending on their charge. In the case of dilute sodium chloride solution, it contains Na+ and Cl- ions. At the cathode (negative electrode), the Na+ ions are attracted and undergo reduction to form sodium atoms and hydrogen gas: 2Na+ + 2e- → 2Na 2H+ + 2e- → H2(g) At the anode (positive electrode), the Cl- ions are attracted and undergo oxidation to form chlorine gas and release oxygen gas: 2Cl- → Cl2(g) + 2e- 2H2O → O2(g) + 4H+ + 4e- Since platinum electrodes are inert, they do not participate in the chemical reactions and do not affect the electrolysis process.

Bayanin Amsa

Metals are said to be malleable because they can be beaten into thin sheets without breaking. This is due to the nature of metallic bonding, where positively charged metal ions are surrounded by a sea of delocalized electrons that are free to move around. When a force is applied to a metal, these delocalized electrons act as a lubricant between the metal ions, allowing them to slide past each other and change shape without breaking. This property is important for the shaping of metals into various forms and structures for use in construction, manufacturing, and other applications.

Bayanin Amsa

Hexane has the highest boiling point among the given options. Boiling point increases with increasing molecular size and branching decreases boiling point. Hexane has the highest molecular size and the least branching compared to the other options, making it have the highest boiling point.

Bayanin Amsa

Rusting of iron is an example of a redox reaction. It involves the oxidation of iron, where it loses electrons and reacts with oxygen to form iron oxide. The process is also accompanied by reduction, where oxygen gains electrons to form hydroxide ions. The overall reaction is: Fe + 2H2O + O2 → Fe(OH)2 2Fe(OH)2 + 1/2O2 → Fe2O3.H2O (rust) Therefore, rusting of iron involves both oxidation and reduction, which is the basis of redox reactions.

Bayanin Amsa

Benzene is the only compound that can undergo both addition reactions and substitution reactions. This is because of its unique structure, which consists of a ring of six carbon atoms with alternating double bonds. The double bonds in benzene are considered to be "delocalized," meaning that the electrons involved in the bonds are spread out over the entire ring rather than being localized between specific carbon atoms. This delocalization makes benzene relatively stable and resistant to addition reactions, but it also makes it susceptible to substitution reactions. In a substitution reaction, one or more of the hydrogen atoms in the benzene ring can be replaced with another atom or group of atoms. In an addition reaction, on the other hand, two molecules combine to form a single molecule, usually with the addition of a double or triple bond. Because benzene already has a stable ring structure with delocalized double bonds, it is not easily able to undergo addition reactions, but it can undergo substitution reactions.

Bayanin Amsa

The device used in the laboratory for intermittent production of gases without heating is Kipp's apparatus. The Kipp's apparatus is a laboratory equipment used for generating gases such as hydrogen, carbon dioxide, and sulfur dioxide, etc. without heating. It is made up of a glass bottle with three parts: a bulb-shaped bottom, a cylindrical middle section, and a funnel-shaped top. The middle section contains a tube with a tap or stopcock, while the bottom is filled with the reactants needed to generate the gas. When the tap is opened, water or another liquid enters the bottom bulb and reacts with the solid to generate the gas, which then bubbles up and is collected in the top funnel. Kipp's apparatus is often used for qualitative analysis in the laboratory.

Bayanin Amsa

When alkanols react with sodium, the gas evolved is hydrogen. This is because sodium is more reactive than hydrogen and reacts with alkanols to liberate hydrogen gas. The reaction is as follows: R-OH + Na → R-ONa + 1/2H2 where R represents the alkyl group. The reaction is a redox reaction, in which sodium is oxidized and alkanols are reduced. The hydrogen gas evolved can be tested using the 'squeaky pop' test, which involves holding a lighted splint near the gas. The hydrogen gas ignites with a squeaky pop sound due to its high reactivity with oxygen.

Bayanin Amsa

Pipe-borne water is usually chlorinated in order to kill harmful bacteria in the water. Chlorine is a strong oxidizing agent that can destroy the cell walls and enzymes of bacteria, viruses and other microorganisms that might be present in the water. By killing these microorganisms, chlorination helps to prevent the spread of waterborne diseases and make the water safe for consumption.

Bayanin Amsa

The two metallic ions commonly present in hard water are Mg²⁺ and Ca²⁺. Hard water is water that contains high levels of dissolved minerals, such as calcium and magnesium. These minerals come from the ground and rock formations that the water flows through. When hard water is heated or evaporates, it leaves mineral deposits behind that can build up in pipes, appliances, and other surfaces. Mg²⁺ and Ca²⁺ ions are the most common minerals that contribute to the hardness of water. They form insoluble precipitates with soaps, reducing their lathering ability, and can also cause scaling in pipes and boilers. Therefore, it is important to remove or reduce these ions from water for many domestic and industrial applications.

Bayanin Amsa

The liquid product of the destructive distillation of coal is ammoniacal liquor. This is because during the destructive distillation of coal, coal is heated in the absence of air to produce several by-products such as coke, coal tar, and coal gas. Ammoniacal liquor is a by-product of coal gas production. It is a brown liquid that contains ammonia, carbon dioxide, and other impurities. The ammonia in ammoniacal liquor is usually recovered through a process called scrubbing, which involves passing the gas through water to dissolve the ammonia.

Bayanin Amsa

Bayanin Amsa

The compound that is used as a gaseous fuel is CH _3-CH _2-CH _2-CH ₃. This is the chemical formula for butane, a hydrocarbon that is commonly used as a fuel in lighters and camping stoves. Butane is a gas at room temperature and pressure, which makes it ideal for use as a gaseous fuel. It burns cleanly and efficiently, producing mainly carbon dioxide and water vapor as byproducts.

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Dehydration of ethanol produces ethene. During the reaction, a molecule of water is eliminated from ethanol, and the remaining atoms combine to form ethene. This is an example of an elimination reaction, where a small molecule is removed from a larger molecule to form a new product. The reaction is typically carried out at high temperatures in the presence of a catalyst such as concentrated sulfuric acid or alumina. Ethene is an important industrial chemical that is used as a feedstock for the production of many other organic chemicals.

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A positive reaction to Fehling's test indicates the presence of reducing sugars. Fehling's test is a chemical test used to detect the presence of reducing sugars such as glucose, fructose, and maltose in a solution. The test involves adding Fehling's reagent, which is a mixture of copper sulfate and potassium tartrate, to the solution being tested and heating it. If reducing sugars are present, they will react with the copper ions in the Fehling's reagent, causing a red precipitate of copper(I) oxide to form. Therefore, the presence of a red precipitate after the Fehling's test indicates the presence of reducing sugars in the solution.

Bayanin Amsa

Petrol consists mainly of hydrocarbons. Hydrocarbons are organic compounds made up of carbon and hydrogen atoms only. In the case of petrol, these hydrocarbons are typically made up of between 5 and 12 carbon atoms per molecule, with some additional small amounts of other compounds. When petrol is burned in an engine, the hydrocarbons react with oxygen in the air to produce energy, carbon dioxide, and water vapor.

Bayanin Amsa

The method suitable for the preparation of an insoluble salt is the Double decomposition method. In this method, two soluble salts are mixed together to form an insoluble salt, which is then separated from the mixture through filtration. This method is particularly useful for preparing salts that are insoluble in water, as it does not require the use of a precipitating agent.

Bayanin Amsa

Aluminium is extracted from bauxite. Bauxite is an ore that is rich in aluminium oxide (Al2O3). The process of extracting aluminium involves the following steps: first, the bauxite ore is purified to remove any impurities, then it is converted into alumina (Al2O3) by the Bayer process. Next, the alumina is electrolyzed in molten cryolite (Na3AlF6) to produce pure aluminium metal. The process of extracting aluminium is energy-intensive and requires large amounts of electricity.

Bayanin Amsa

In the given reaction, sulphur (IV) oxide (SO₂) acts as an oxidizing agent. It oxidizes hydrogen sulphide (H₂S) gas to form solid sulfur (S) and water (H₂O). The oxidation state of sulphur in SO₂ is +4, while in H₂S, it is -2. During the reaction, sulphur (IV) oxide is reduced to sulphur (S) with a decrease in its oxidation state, while hydrogen sulphide is oxidized to sulphur (S) with an increase in its oxidation state. Therefore, sulphur (IV) oxide acts as an oxidizing agent in the given reaction.

Bayanin Amsa

The spontaneity of a chemical reaction is determined by the change in the free energy of the system. Free energy is the energy available to do work, and a chemical reaction is spontaneous when the change in free energy is negative. In other words, the products of the reaction have lower free energy than the reactants, and the reaction can occur without the input of additional energy. Concentration, temperature, pressure, and volume can affect the spontaneity of a reaction, but the driving force behind the spontaneity is the change in free energy.

Bayanin Amsa

(a) Two postulates of the kinetic theory of gases:

• A gas consists of a very large number of tiny particles (molecules) which are in continuous, rapid, random motion.
• The actual volume of the gas particles is negligible compared with the total volume of the container (there is a lot of empty space between them). (Also acceptable: there are negligible forces of attraction between the particles, and collisions between them are perfectly elastic; the average kinetic energy is proportional to the absolute temperature.)

(b) Two chemical properties common to both carbon(IV) oxide, \(CO_2\), and sulphur(IV) oxide, \(SO_2\):

• Both are acidic oxides: they dissolve in water to form acids (\(CO_2\) forms trioxocarbonate(IV) acid, \(SO_2\) forms trioxosulphate(IV) acid) and turn moist blue litmus red.
• Both react with alkalis (bases) to form salt and water, e.g. \(CO_2 + 2NaOH \rightarrow Na_2CO_3 + H_2O\) and \(SO_2 + 2NaOH \rightarrow Na_2SO_3 + H_2O\). (Both are also non-supporters of combustion.)

Bayanin Amsa

(a) IUPAC name of \(Fe_2(SO_4)_3\)

The compound contains the iron(III) cation, \(Fe^{3+}\), and the sulphate anion, \(SO_4^{2-}\). Because iron shows more than one oxidation state, the oxidation number of the metal must be shown in Roman numerals, and the anion is named in full as the tetraoxosulphate(VI) ion. The IUPAC name is therefore iron(III) tetraoxosulphate(VI) (commonly, iron(III) sulphate).

(b)(i) Reaction of hydrogen sulphide with iron(III) chloride solution

Hydrogen sulphide is a reducing agent. It reduces iron(III) to iron(II), and is itself oxidised, so that its sulphur is deposited as free (elemental) sulphur:

In ionic terms the essential change is \(2Fe^{3+} + H_2S \rightarrow 2Fe^{2+} + 2H^+ + S\), which shows clearly that iron is reduced from the \(+3\) to the \(+2\) state while sulphide sulphur (\(-2\)) is oxidised to sulphur (\(0\)).

(b)(ii) One observed change

The yellow-brown iron(III) chloride solution turns pale green, showing that iron(III) has been reduced to iron(II). At the same time a pale yellow deposit (turbidity) of sulphur forms in the mixture. Either of these observations is acceptable: the colour change from yellow-brown to pale green, or the appearance of the yellow solid sulphur.

Bayanin Amsa

Bayanin Amsa

(a) Two uses of sodium trioxocarbonate(IV), \(Na_2CO_3\) (washing soda): in the manufacture of glass and in softening hard water. (It is also used in the manufacture of soap, detergents and paper.)

(b)(i) The phenomenon responsible for the alkalinity of sodium trioxocarbonate(IV) solution is salt hydrolysis. Being the salt of a strong base (NaOH) and a weak acid (trioxocarbonate(IV) acid), the carbonate ion reacts with water to release hydroxide ions, making the solution alkaline:

\[CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-\]

(ii) When carbon(IV) oxide is passed into a saturated sodium trioxocarbonate(IV) solution, the product obtained is sodium hydrogen trioxocarbonate(IV) (sodium hydrogencarbonate), \(NaHCO_3\):

\[Na_2CO_3 + CO_2 + H_2O \rightarrow 2NaHCO_3\]

Bayanin Amsa

(a)(i) Type of reaction in each conversion (I to V)

(a)(ii) One isomer of glucose: fructose (galactose is also acceptable), each of formula C₆H₁₂O₆.

(a)(iii) Why palm wine becomes sour on prolonged exposure to air. Palm wine contains ethanol. On exposure to air, bacterial/atmospheric oxidation converts the ethanol to ethanoic (acetic) acid, and it is this acid that gives the drink its sour taste:

\[ CH_3CH_2OH + O_2 \rightarrow CH_3COOH + H_2O \]

(b)(i) Reagents and conditions for ethanoic acid to form an alkanoate

• An alkanol (e.g. ethanol).
• Concentrated H₂SO₄ (or dry HCl) as catalyst/dehydrating agent.
• Heat, about \(40^\circ\text{C}\) to \(110^\circ\text{C}\), or refluxing.

\[ CH_3COOH + C_2H_5OH \underset{\Delta}{\overset{conc.\,H_2SO_4}{\rightleftharpoons}} CH_3COOC_2H_5 + H_2O \]

(b)(ii) Two uses of alkanoates (esters)

• In the manufacture of perfumes, flavours and cosmetics.
• As solvents in the paint, pharmaceutical and food industries; also for making polyester fibres.

(c)(i) IUPAC name of the compound: but-2-yne, \( CH_3-C{\equiv}C-CH_3 \).

(c)(ii) Chemical test to distinguish methane from but-2-yne. Pass each gas in turn into bromine water (or acidified KMnO₄). But-2-yne, being unsaturated, rapidly decolourizes the reagent; methane, being saturated, gives no visible reaction and the colour remains.

(c)(iii) Combustion of ethene in excess oxygen

\[ C_2H_{4(g)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 2H_2O_{(g)} \]

Bayanin Amsa

(a)(i) The general formula for alkanoic acids is \(C_nH_{2n+1}COOH\) (equivalently \(C_nH_{2n}O_2\)).

(ii) Two chemical properties of ethanoic acid, \(CH_3COOH\):

• It reacts with reactive metals to liberate hydrogen, e.g. \(2CH_3COOH + Mg \rightarrow (CH_3COO)_2Mg + H_2\).
• It reacts with alkanols in the presence of concentrated \(H_2SO_4\) to form an ester (esterification). (It also reacts with bases and trioxocarbonate(IV) salts, turning blue litmus red as a weak acid.)

(b) Of propene, butane and pentane:

• (i) The one that decolorizes acidified \(KMnO_4\) solution is propene (it is unsaturated, containing a C=C double bond).
• (ii) The one that can be easily polymerized is propene (alkenes polymerize; propene gives polypropene).
• (iii) The isomer of methylpropane is butane (both have the molecular formula \(C_4H_{10}\)).
• (iv) The one that can be obtained from an alkanol by dehydration is propene (dehydration of propanol gives propene).

Bayanin Amsa

(a)(i) An electrolyte is a substance which, in the molten state or in aqueous solution, conducts electricity and is chemically decomposed by it because it contains free mobile ions.

(ii)

• Potassium chloride - strong electrolyte
• Sodium ethanoate - strong electrolyte
• Aqueous ammonia - weak electrolyte
• Cane sugar - non-electrolyte

(b)(i) Daniell cell half-cell reactions

Anode (oxidation): \[Zn \rightarrow Zn^{2+} + 2e^-\]

Cathode (reduction): \[Cu^{2+} + 2e^- \rightarrow Cu\]

(ii) It is an electrochemical cell because it converts chemical energy (from the spontaneous redox reaction) into electrical energy.

(iii) Other examples: the dry (Leclanche) cell and the lead-acid accumulator.

(c)(i) In graphite, each carbon atom bonds to only three others, using three of its four valence electrons. The fourth electron of every atom is delocalized between the layers and is free to move, so graphite conducts electricity.

(ii) With platinum electrodes, Cu²⁺ ions are discharged at the cathode and removed from solution, so the blue colour fades:
\[Cu^{2+} + 2e^- \rightarrow Cu\]
at the anode oxygen is evolved: \[2H_2O \rightarrow O_2 + 4H^+ + 4e^-\]
With copper electrodes, the copper anode dissolves and replaces the Cu²⁺ removed at the cathode, so the concentration and blue colour stay the same:
\[Cu \rightarrow Cu^{2+} + 2e^-\]

(iii) In methylbenzene, hydrogen chloride exists as un-ionized covalent molecules, so there are no mobile ions and it does not conduct. In water, HCl ionizes into mobile H⁺ and Cl^- ions, so hydrochloric acid conducts electricity.

Bayanin Amsa

(a)(i) Heat of neutralization is the heat change when one mole of hydrogen ions (H⁺) from an acid reacts completely with one mole of hydroxide ions (OH^-) from a base to form one mole of water under standard conditions.

(ii) Copper(II) chloride is soluble in water, so it can be made in solution by neutralizing an acid with a base and then crystallized. Lead(II) chloride is insoluble in cold water, so a neutralization in solution would simply precipitate it; it is therefore prepared by precipitation (double decomposition), not by neutralization.

(b)(i) Manufacture of trioxonitrate(V) acid (Ostwald process)

• Ammonia is mixed with excess air and passed over a platinum-rhodium gauze catalyst at about 900 °C:
  \[4NH_3 + 5O_2 \rightarrow 4NO + 6H_2O\]
• The nitrogen(II) oxide is cooled and oxidized by more air:
  \[2NO + O_2 \rightarrow 2NO_2\]
• The nitrogen(IV) oxide is dissolved in water in the presence of excess air:
  \[4NO_2 + O_2 + 2H_2O \rightarrow 4HNO_3\]

(ii) Sodium trioxonitrate(V) on strong heating gives sodium trioxonitrate(III) and oxygen:
\[2NaNO_3 \rightarrow 2NaNO_2 + O_2\]

(c)(i) \[3Mg + N_2 \rightarrow Mg_3N_2\]

(ii) Amount of Mg \(= \dfrac{3.0}{24} = 0.125\ \text{mol}\).
From the equation, 3 mol Mg give 1 mol Mg₃N₂, so
amount of Mg₃N₂ \(= \dfrac{0.125}{3} = 0.0417\ \text{mol} \approx 4.17 \times 10^{-2}\ \text{mol}\).

Bayanin Amsa

(a) One method (apart from heating) by which the reaction can be made faster: increase the surface area of the iron(II) sulphide by using it in a finely powdered form (using a more concentrated acid is also acceptable).

(b) Calculating X. Equation: \[FeS_{(s)} + 2HCl_{(aq)} \rightarrow FeCl_{2(aq)} + H_2S_{(g)}\]

Molar mass of \(FeCl_2 = 56 + (2 \times 35.5) = 56 + 71 = 127\,g\,mol^{-1}\)

\[\text{moles of } FeCl_2 = \frac{3.20}{127} = 0.0252\,mol\]

From the equation, 1 mole \(FeS\) gives 1 mole \(FeCl_2\), so moles of \(FeS = 0.0252\,mol\).

\[X = \text{mass of } FeS = 0.0252 \times 88 = 2.22\,g\]

So X = 2.22 g (approximately).

Bayanin Amsa

(a)(i) In increasing order of reactivity:

Lead < Iron < Aluminium < Magnesium.

(ii) Aluminium reacts with sodium hydroxide solution to give hydrogen gas.

(b) Tetraoxosulphate(VI) acid behaves as:

(i) an oxidising agent in its reaction with sulphur.

(ii) a typical acid in the neutralisation of magnesium oxide.

(iii) a dehydrating agent in its reaction with sugar, \(C_{12}H_{22}O_{11}\).

Bayanin Amsa

(a) To electroplate a steel spoon with silver:

• (i) Anode: a bar (plate) of pure silver metal (which dissolves to replenish the silver ions).
• (ii) Cathode: the steel spoon itself (the article to be plated).
• (iii) Electrolyte: a solution of a soluble silver salt, e.g. silver trioxonitrate(V) solution, \(AgNO_3\) (a solution of potassium argentocyanide is used industrially).

(b)(i) An equation for one reaction in the purification of bauxite (dissolving the amphoteric aluminium oxide in hot sodium hydroxide, leaving impurities behind):

\[Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2 + H_2O\]

(ii) The carbon anodes must be changed (replaced) at intervals because the oxygen liberated at the anode reacts with the hot carbon to form carbon(IV) oxide (and carbon(II) oxide), so the carbon anodes gradually burn away (are oxidized) and become too small to use.

Bayanin Amsa

(a)(i) Quantum numbers assigned to an electron in an atom are:

• Principal quantum number, n
• Azimuthal (angular momentum) quantum number, l
• Magnetic quantum number, m_l
• Spin quantum number, m_s

(ii) The maximum number of electrons that can occupy the 3d subshell is 10.

(b) The configuration is taken as \(1s^2\,2s^2\,2p^6\,3s^2\).

(i) P loses two electrons to form P²⁺. Its electronic configuration is:

\(1s^2\,2s^2\,2p^6\)

(ii) The likely formula of the chloride of P is PCl₂.

(iii) P is a good reducing agent because it readily loses two outer electrons to attain a stable noble-gas configuration. Thus, it donates electrons to other substances.

(c)(i) Electron affinity is the energy evolved or absorbed when one mole of electrons is added to one mole of gaseous atoms to form one mole of gaseous negative ions.

(ii) Ammonia can participate in dative bonding because the nitrogen atom has a lone pair of electrons which it can donate to an electron-deficient atom or ion for sharing.

(d)(i) On losing an alpha particle, the atomic number decreases by 2. Therefore, an element in Group IV will give a product in Group II.

(ii) X, with a half-life of 8 minutes, constitutes the greater health hazard. Its shorter half-life means that it decays more rapidly and releases radiation at a higher rate than Y within a given period.

Bayanin Amsa

(a)(i) The activation energy of a reaction is the minimum amount of energy that the reacting particles must possess before a collision between them can lead to a reaction (i.e. the minimum energy needed for an effective collision).

(ii) A catalyst lowers (reduces) the activation energy of the reaction by providing an alternative reaction pathway of lower activation energy, so more colliding particles have the required energy and the reaction proceeds faster.

(b) Catalyst in each case:

• (i) Hydrogenation of oils: nickel (finely divided nickel).
• (ii) Biochemical reactions: enzymes (biological catalysts).

Bayanin Amsa

(a) Reactions of chlorine:

• (i) With potassium iodide solution (chlorine displaces iodine): \[Cl_2 + 2KI \rightarrow 2KCl + I_2\]
• (ii) With zinc on heating: \[Zn + Cl_2 \rightarrow ZnCl_2\]

(b) In \(Cl_2 + 2NaOH \rightarrow NaCl + H_2O + Q\) (cold, dilute sodium hydroxide), the product Q is sodium chlorate(I) (sodium oxochlorate(I) / sodium hypochlorite), \(NaOCl\).

The balanced equation is: \[Cl_2 + 2NaOH \rightarrow NaCl + NaOCl + H_2O\]