JAMB UTME - Chemistry - 2025

Maelezo ya Majibu

Nitrogen gas (N₂) makes up about 78% of the atmosphere and has several important industrial uses. One of the most significant is the production of nitric acid (HNO₃).

The industrial production of HNO₃ occurs in two stages:

1. Haber process: Nitrogen reacts with hydrogen at high temperature and pressure in the presence of an iron catalyst to produce ammonia:
   \[\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)\]
2. Ostwald process: Ammonia is then oxidised over a platinum-rhodium catalyst to produce nitrogen monoxide, which is further oxidised and dissolved in water to form nitric acid:
   \[4\text{NH}_3 + 5\text{O}_2 \rightarrow 4\text{NO} + 6\text{H}_2\text{O}\]
   \[2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2\]
   \[3\text{NO}_2 + \text{H}_2\text{O} \rightarrow 2\text{HNO}_3 + \text{NO}\]

The other options are incorrect: margarine production uses hydrogen (not nitrogen) for hydrogenation of vegetable oils; nitrogen is not used in manufacturing oil; and while liquid nitrogen can serve as a coolant, the question refers to nitrogen as a component of air used in chemical production.

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Tristearin is a fat (triglyceride) formed from glycerol and three molecules of stearic acid. Alkaline hydrolysis of a triglyceride is the reaction of the fat with a strong alkali such as sodium hydroxide. This reaction is known as saponification and produces soap (the sodium salt of the fatty acid) and glycerol as a by-product:

\[\text{(C}_{17}\text{H}_{35}\text{COO)}_3\text{C}_3\text{H}_5 + 3\text{NaOH} \rightarrow 3\text{C}_{17}\text{H}_{35}\text{COONa} + \text{C}_3\text{H}_5\text{(OH)}_3\]

The by-product is glycerol, also known as propane-1,2,3-triol. Its structural formula shows three hydroxyl (-OH) groups, one on each of the three carbon atoms. An alcohol with three -OH groups is classified as a trihydric alkanol.

A dihydric alkanol has two -OH groups (e.g. ethane-1,2-diol). A secondary alkanol has the -OH group on a carbon bonded to two other carbon atoms. A tertiary alkanol has the -OH on a carbon bonded to three other carbons. Glycerol has two primary -OH groups and one secondary -OH group, but its defining classification is that it is trihydric, since it carries three hydroxyl groups in total.

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The energy profile diagram shows the energy changes during a chemical reaction. The reactants (A+B) start at an energy level of approximately 30 units, while the products (C+D) end at approximately 50 units. The activation energy peak reaches about 80 units.

Since the products have a higher energy level than the reactants, the reaction has absorbed energy from the surroundings. This net gain in energy by the reacting system is the defining characteristic of an endothermic reaction. The energy difference between products and reactants (\ (\Delta H\)) is positive, confirming that heat was taken in rather than released.

An exothermic reaction would show products at a lower energy level than reactants, indicating a release of energy. Here, the upward shift from reactants to products clearly indicates energy absorption.

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Tetraoxosulphate(VI) acid is the IUPAC name for sulphuric acid, \(\text{H}_2\text{SO}_4\). Its large-scale industrial manufacture uses the Contact process.

The Contact process involves three main stages:

1. Sulphur is burned in air to produce sulphur(IV) oxide: \(\text{S} + \text{O}_2 \rightarrow \text{SO}_2\)
2. Sulphur(IV) oxide is oxidised to sulphur(VI) oxide using a vanadium(V) oxide (\(\text{V}_2\text{O}_5\)) catalyst at about 450 \(^\circ\text{C}\): \(2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3\)
3. The \(\text{SO}_3\) is absorbed in concentrated \(\text{H}_2\text{SO}_4\) to form oleum (\(\text{H}_2\text{S}_2\text{O}_7\)), which is then carefully diluted with water to give concentrated sulphuric acid.

The other named processes serve different purposes. The Haber process manufactures ammonia from nitrogen and hydrogen. The Frasch process is used for mining sulphur deposits underground using superheated water. The Bosch process (or water-gas shift reaction) produces hydrogen from carbon monoxide and steam. None of these produces sulphuric acid.

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Freons are a group of chlorofluorocarbons (CFCs) - synthetic compounds containing chlorine, fluorine, and carbon. They were widely used as propellants in aerosol cans, as refrigerants in air conditioners and refrigerators, and as solvents in industrial cleaning.

When released into the atmosphere from these sources, freons rise to the stratosphere where ultraviolet radiation breaks them down, releasing chlorine atoms. These chlorine atoms catalytically destroy ozone molecules, contributing to the depletion of the ozone layer.

Fossil fuel combustion releases carbon dioxide, sulphur dioxide, and nitrogen oxides, but not freons. Photosynthesis is a biological process that produces oxygen and consumes carbon dioxide. Organic decay releases methane and carbon dioxide. None of these processes involve freons.

The Montreal Protocol (1987) restricted the production and use of CFCs, leading to a gradual recovery of the ozone layer.

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Aluminium (Al) is the metal widely used as a packaging material. It is used to make drink cans, food containers, and aluminium foil for wrapping food.

Aluminium is ideal for packaging because of several key properties:

• Low density - it is lightweight, making it easy to transport and handle.
• Malleability - it can be rolled into very thin sheets (foil) or pressed into various container shapes.
• Corrosion resistance - it forms a tough, protective oxide layer (\(\text{Al}_2\text{O}_3\)) on its surface that prevents further reaction with air or moisture, keeping the contents safe.
• Non-toxicity - it does not contaminate food or beverages.
• Recyclability - aluminium can be recycled repeatedly without losing its properties.

The other metals are unsuitable for packaging:

• Copper (Cu) is too heavy and expensive for everyday packaging, and it can impart a metallic taste to food.
• Iron (Fe) rusts readily when exposed to moisture and air, which would contaminate and spoil packaged goods.
• Sodium (Na) is an extremely reactive alkali metal that reacts violently with water and catches fire in air, making it entirely impractical for packaging.

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The cell notation (also called line notation) for an electrochemical cell follows the convention:

Anode | Anode ion || Cathode ion | Cathode

where the single vertical line (|) represents a phase boundary, and the double vertical line (||) represents the salt bridge separating the two half-cells.

For the reaction Mg + Pb²⁺ → Mg²⁺ + Pb:

• Anode (oxidation): Mg → Mg²⁺ + 2e^-. Magnesium is oxidized, so it is the anode.
• Cathode (reduction): Pb²⁺ + 2e^- → Pb. Lead ions are reduced, so lead is the cathode.

Applying the convention:

Mg | Mg²⁺ || Pb²⁺ | Pb

Using the notation in the options (where I = | and II = ||), this is written as Mg|Mg²⁺||Pb²⁺|Pb.

The anode always appears on the left and the cathode on the right. Within each half-cell, the metal (solid phase) is written adjacent to the outer edge, and the ion (aqueous phase) is written adjacent to the salt bridge.

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Dilute nitric acid (HNO\(_3\)) is an oxidising acid, which means it usually oxidises the metal and is itself reduced to nitrogen oxides (such as NO or NO\(_2\)) rather than producing hydrogen gas. This is different from non-oxidising acids like dilute HCl or dilute H\(_2\)SO\(_4\), which readily liberate H\(_2\) with reactive metals.

However, magnesium (Mg) is an exception. Because magnesium is extremely reactive (high up in the electrochemical series), it reacts so vigorously with very dilute HNO\(_3\) that the reaction proceeds faster than the acid can act as an oxidising agent. The result is that hydrogen gas is liberated:

\[\text{Mg} + 2\text{HNO}_3\text{(very dilute)} \rightarrow \text{Mg(NO}_3\text{)}_2 + \text{H}_2\uparrow\]

Copper (Cu) is below hydrogen in the activity series and cannot displace hydrogen from any acid under normal conditions. Zinc (Zn) reacts with dilute HNO\(_3\) but produces NO gas rather than H\(_2\), because it is not reactive enough to overcome the oxidising nature of the acid. Calcium (Ca) is very reactive but reacts explosively with water itself and, in practice with dilute HNO\(_3\), produces nitrogen oxides or ammonia rather than clean H\(_2\) liberation; the standard examination answer for this question is magnesium.

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Brine is a concentrated solution of sodium chloride (NaCl) in water. When brine is electrolysed using inert (neutral) electrodes such as carbon or platinum, the ions present in solution are:

• Cations: Na⁺ and H⁺ (from water)
• Anions: Cl^- and OH^- (from water)

At the anode (positive electrode), anions migrate and are discharged. Both Cl^- and OH^- are present, but Cl^- is preferentially discharged because it is present in much higher concentration in the brine solution. Despite OH^- having a lower discharge potential, the high concentration of Cl^- gives it priority at the anode.

The half-equation at the anode is:

\[2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-\]

Chlorine gas (Cl₂) is released at the anode.

Na⁺ and H⁺ are cations and migrate to the cathode, not the anode. At the cathode, H⁺ is discharged (since Na⁺ has a very high discharge potential), producing hydrogen gas.

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A covalent bond is formed when two atoms share a pair of electrons. When the two bonded atoms have different electronegativities, they do not attract the shared electron pair equally. The more electronegative atom pulls the electron pair closer to itself, creating a partial negative charge on that atom and a partial positive charge on the other. This results in a polar covalent bond.

Therefore, a polar covalent bond is characterised by the unequal sharing of an electron pair between atoms.

By contrast, when two atoms of the same element (or very similar electronegativity) bond, the electrons are shared equally, producing a non-polar covalent bond (e.g., H₂, Cl₂, O₂).

The options mentioning "lone pair" are incorrect in this context. A lone pair is an unshared pair of electrons that belongs entirely to one atom and is not involved in bonding between the two atoms. Covalent bonds (whether polar or non-polar) involve bonding pairs, not lone pairs.

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In the electrolysis of concentrated sodium chloride solution (brine) using inert graphite electrodes, the products depend on the concentration of the solution and the electrode positions.

At the anode (positive electrode), negatively charged ions migrate and are discharged. In concentrated NaCl solution, both chloride ions (Cl^-) and hydroxide ions (OH^-) from water are present. However, because the chloride ion concentration is very high, chloride ions are preferentially discharged at the anode:

\[2\text{Cl}^{-}(aq) \rightarrow \text{Cl}_2(g) + 2e^{-}\]

This produces chlorine gas, which can be identified by its greenish-yellow colour and its ability to bleach damp litmus paper.

At the cathode, hydrogen gas is produced from the reduction of water (since Na⁺ ions are too electropositive to be discharged). Oxygen gas would be the anode product only in the electrolysis of dilute sodium chloride or dilute sulphuric acid, where hydroxide ions are discharged instead of chloride ions. Water vapour and hydrogen gas are not anode products in this process.

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The fountain experiment demonstrates the very high solubility of certain gases in water. A round-bottom flask is filled with the gas and inverted over a trough of water (often containing an indicator). When a small amount of water enters the flask and dissolves the gas, the pressure inside drops dramatically. Atmospheric pressure then forces water up into the flask in a spectacular fountain.

For this experiment to work, the gas must be extremely soluble in water so that it dissolves almost instantly on contact, creating a near-vacuum inside the flask.

Hydrogen chloride (HCl) is the classic gas used. It is one of the most soluble gases in water: about 450 volumes of HCl dissolve in one volume of water at room temperature, forming hydrochloric acid. Ammonia (NH₃) is also commonly used for the same experiment, but it is not among the given options.

Hydrogen sulphide (H₂S) is only moderately soluble and is extremely toxic, making it unsuitable. Dinitrogen(I) oxide (N₂O, nitrous oxide) and nitrogen(II) oxide (NO, nitric oxide) are both poorly soluble in water and would not produce the dramatic pressure drop needed for the fountain effect.

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The electrochemical series (also called the activity series or reactivity series) is a table that ranks metals in order of their tendency to lose their valence electrons and form positive ions. Metals at the top of the series (e.g., potassium, sodium, calcium) lose electrons most readily, while those at the bottom (e.g., gold, platinum) have very little tendency to give up electrons.

The series is determined by measuring the standard electrode potentials of metals. A more negative electrode potential indicates a greater tendency to lose electrons (be oxidised), placing the metal higher in the series.

The other options are not correct:

• The periodic table arranges elements by atomic number and groups them by similar chemical properties, but it does not rank them specifically by their tendency to lose valence electrons in a sequential series.
• "Table of elements" is a vague term that does not refer to a specific scientific arrangement by electron-loss tendency.
• "Periodicals" refers to magazines or journals and has no chemical meaning.

Maelezo ya Majibu

A lone pair is a pair of valence electrons on an atom that is not shared in a bond. To find which molecule has the highest number of lone pairs, draw the Lewis structure of each molecule and count all lone pairs on every atom.

CH₄: Carbon has four bonding pairs (one to each hydrogen) and no lone pairs. Each hydrogen also has no lone pairs. Total lone pairs: 0.

NH₃: Nitrogen has three bonding pairs (one to each hydrogen) and one lone pair. Total lone pairs: 1.

H₂O: Oxygen has two bonding pairs (one to each hydrogen) and two lone pairs. Total lone pairs: 2.

CO₂: Carbon forms two double bonds (one to each oxygen) and has no lone pairs. Each oxygen in a double bond with carbon retains two lone pairs. Total lone pairs: 2 + 2 = 4.

CO₂ has the highest total number of lone pairs (four), making it the correct answer.

Exam tip: When counting lone pairs, remember to include those on every atom in the molecule, not just the central atom.

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For a reaction that follows first-order kinetics, the half-life is related to the rate constant by the formula:

\[t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}\]

Given \(k_1 = 3.85 \times 10^{-3}\, \text{s}^{-1}\):

\[t_{1/2} = \frac{0.693}{3.85 \times 10^{-3}}\]

\[t_{1/2} = \frac{0.693}{0.00385}\]

\[t_{1/2} = 180\, \text{s}\]

The half-life of radioactive phosphorus is 180 s.

An important feature of first-order kinetics is that the half-life is independent of the initial concentration - it depends only on the rate constant. This is why radioactive decay (which always follows first-order kinetics) has a constant half-life regardless of how much of the substance remains.

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The boiling point of a substance depends on the strength of its intermolecular forces and, to a lesser extent, its molecular mass. The key intermolecular forces in order of strength are: hydrogen bonding > dipole-dipole > van der Waals (London dispersion).

Consider the four compounds:

• CH₃CH₂OH (ethanol, M_r = 46): contains an -OH group, so it forms hydrogen bonds. Boiling point approximately 78 °C.
• CH₃CH₂CH₂Cl (1-chloropropane, M_r = 78.5): polar C-Cl bond gives dipole-dipole interactions but no hydrogen bonding. Boiling point approximately 47 °C.
• CH₃CH₂CH₂CH₂CH₂CH₃ (hexane, M_r = 86): non-polar, only van der Waals forces. Boiling point approximately 69 °C.
• CH₃CH₂CH₂OH (propan-1-ol, M_r = 60): contains an -OH group, so it forms hydrogen bonds, and has a higher molecular mass than ethanol. Boiling point approximately 97 °C.

Propan-1-ol (CH₃CH₂CH₂OH) has the highest boiling point. It combines hydrogen bonding (the strongest intermolecular force among these molecules) with a greater molecular mass than ethanol, giving it stronger overall intermolecular attractions.

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An alkaline gas is a gas that dissolves in water to produce a solution with a pH greater than 7 (a basic solution).

NH₃ (ammonia) is the classic example. When ammonia dissolves in water, it reacts to form ammonium hydroxide:

\[\text{NH}_3(g) + \text{H}_2\text{O}(l) \rightleftharpoons \text{NH}_4^+(aq) + \text{OH}^-(aq)\]

The production of hydroxide ions (OH^-) makes the solution alkaline.

The other gases are not alkaline:

• HCl (hydrogen chloride) dissolves in water to form hydrochloric acid, which is strongly acidic.
• N₂ (nitrogen) is an inert, unreactive gas that does not dissolve appreciably in water and produces neither acidic nor basic solutions.
• NO₂ (nitrogen dioxide) dissolves in water to form a mixture of nitric acid and nitrous acid, both of which are acidic.

Exam tip: Ammonia is the only common alkaline gas encountered at this level. Its characteristic pungent smell and ability to turn moist red litmus paper blue are standard identification tests.

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The equation Na\(_2\)X \(\rightleftharpoons\) 2Na\(^+\) + X\(^{2-}\) shows the compound Na\(_2\)X dissociating into its constituent ions: sodium ions (Na\(^+\)) and an anion X\(^{2-}\).

This dissociation into oppositely charged ions is the hallmark of an ionic bond. In ionic bonding, one or more electrons are transferred from a metal atom (here, sodium) to a non-metal atom (here, X). Sodium loses one electron to form Na\(^+\), while X gains two electrons to form X\(^{2-}\). Two sodium atoms are needed to supply the two electrons that X requires.

The other bond types do not fit:

• A covalent bond involves the sharing of electron pairs between atoms, not transfer and complete dissociation into ions.
• A dative (coordinate) bond is a special covalent bond where both shared electrons come from one atom.
• A metallic bond involves a "sea" of delocalised electrons shared among metal atoms in a lattice - it does not produce discrete anions like X\(^{2-}\).

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Amines are classified based on the number of carbon-containing groups (alkyl or aryl groups) directly bonded to the nitrogen atom:

• Primary amine: nitrogen is bonded to one carbon group and two hydrogen atoms (R-NH₂).
• Secondary amine: nitrogen is bonded to two carbon groups and one hydrogen atom (R₂-NH).
• Tertiary amine: nitrogen is bonded to three carbon groups and no hydrogen atoms (R₃-N).

In CH₃CH(NH₂)CH₂CH₂CH₃, the nitrogen atom in the -NH₂ group is bonded to one carbon atom (the CH group in the chain) and two hydrogen atoms. This fits the definition of a primary amine.

The fact that the nitrogen is attached to a secondary carbon (a carbon bonded to two other carbons) does not change the amine classification. The classification depends only on how many carbons are bonded directly to nitrogen, not on the type of carbon.

Exam tip: Do not confuse amine classification (based on bonds to nitrogen) with alcohol classification (based on the type of carbon bearing the -OH group). A primary amine simply means nitrogen has one C-N bond.

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The equation is:

\[2\text{X} + 2\text{HCl} \rightarrow 2\text{XCl} + \text{H}_2\]

The product formed is XCl, which tells us that element X combines with chlorine in a 1:1 ratio. This means X has a valency of +1 and forms a monovalent chloride.

Examining the options:

• K (potassium): Group 1 element with a valency of +1. It forms KCl. The equation becomes \(2\text{K} + 2\text{HCl} \rightarrow 2\text{KCl} + \text{H}_2\), which balances correctly.
• Mg (magnesium): Group 2 element with a valency of +2. It forms MgCl₂, not MgCl. Does not match.
• Ca (calcium): Group 2 element with a valency of +2. It forms CaCl₂, not CaCl. Does not match.
• Ba (barium): Group 2 element with a valency of +2. It forms BaCl₂, not BaCl. Does not match.

Only potassium (K) has a valency of +1 and forms a chloride with the formula XCl, making it the correct identity of X.

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Reduction is the gain of electrons (or a decrease in oxidation state). To identify which species is reduced, track the oxidation states of each element.

In the reaction \(2\text{Na} + \text{Cl}_2 \rightarrow 2\text{NaCl}\):

• Sodium (Na): Oxidation state changes from 0 (in elemental Na) to +1 (in NaCl). This is an increase in oxidation state, meaning sodium is oxidised (it loses one electron per atom).
• Chlorine (Cl₂): Oxidation state changes from 0 (in elemental Cl₂) to -1 (in NaCl as Cl^-). This is a decrease in oxidation state, meaning chlorine is reduced (each Cl atom gains one electron).

The species that undergoes reduction is Cl₂, because it is the substance that accepts electrons and has its oxidation state lowered from 0 to -1.

Note that Cl^- is the product of the reduction, not the species that undergoes it. The question asks for the species that undergoes reduction, which is the reactant Cl₂.

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The equation shown is:

2XY₃(g) ⇌ X₂(g) + 3Y₂(g)

The equilibrium constant \(K_c\) is defined as the ratio of the product concentrations raised to their stoichiometric coefficients divided by the reactant concentrations raised to their stoichiometric coefficients.

From the balanced equation, the products are X₂ (coefficient 1) and Y₂ (coefficient 3), while the reactant is XY₃ (coefficient 2). Therefore:

\[K_c = \frac{[X_2][Y_2]^3}{[XY_3]^2}\]

The coefficients become exponents in the equilibrium expression, not multipliers placed in front of the concentration brackets. This is a fundamental distinction: writing \([2XY_3]\) or \([3Y_2]\) treats the coefficient as part of the concentration term, which is incorrect.

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The molecular mass (molar mass) of a substance is defined as the mass of one mole of that substance. The relationship is:

\[\text{Molar mass} = \frac{\text{Mass}}{\text{Number of moles}}\]

Given that 0.5 mole of the alkanoic acid weighs 44 g:

\[\text{Molar mass} = \frac{44\,\text{g}}{0.5\,\text{mol}} = 88\,\text{g/mol}\]

The molecular mass of the alkanoic acid is therefore 88. This corresponds to butanoic acid (CH₃CH₂CH₂COOH), which has the molecular formula C₄H₈O₂: (4 x 12) + (8 x 1) + (2 x 16) = 48 + 8 + 32 = 88.

A common error is to multiply mass by moles instead of dividing. Remember: if a fraction of a mole has a certain mass, the full mole must weigh proportionally more.

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A diagonal relationship in the periodic table refers to the similarity in properties between an element in Period 2 and the element diagonally below and to its right in Period 3. This occurs because moving one period down increases size and metallic character, while moving one group to the right decreases them, so the two effects partially cancel out.

The well-established diagonal pairs are:

• Li (Period 2, Group 1) and Mg (Period 3, Group 2)
• Be (Period 2, Group 2) and Al (Period 3, Group 13)
• B (Period 2, Group 13) and Si (Period 3, Group 14)

From the given options, B and Si is one of the classic diagonal pairs. They share similar properties such as forming covalent compounds, acting as semiconductors, and forming acidic oxides.

The other options are not diagonal pairs:

• Be and Si are two groups apart diagonally, not one.
• Na and Be are in adjacent periods but Na is below and to the left of Be, not diagonal in the correct direction.
• Li and Al are separated by two groups in the diagonal direction.

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To determine the group and period of an element from its electron configuration, two pieces of information are needed:

• The period is the highest principal quantum number (the largest shell number) occupied by electrons.
• The group (for s-block and p-block elements) is determined by the number of electrons in the outermost shell.

The electron configuration given is \(1s^2\,2s^2\,2p^6\,3s^2\). The total number of electrons is \(2 + 2 + 6 + 2 = 12\), which identifies the element as magnesium (Mg).

The highest principal quantum number is 3 (from the \(3s^2\) subshell), so the element is in Period 3.

The outermost shell (\(n = 3\)) contains only 2 electrons (both in the \(3s\) subshell). Since these are s-block electrons, the element is in Group 2.

Therefore, the element belongs to Group 2 and Period 3.

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The melting and boiling points of a substance depend on the strength of the forces that must be overcome to change its state. For simple covalent molecules, there are two types of forces to consider:

1. Intramolecular forces (covalent bonds within the molecule): These are strong but are not broken during melting or boiling.
2. Intermolecular forces (forces between molecules): These are the forces that must be overcome during phase changes. In simple covalent molecules, these are typically weak van der Waals forces or dipole-dipole interactions.

Because the intermolecular forces are weak, relatively little energy is needed to separate the molecules from one another. This is why simple covalent substances such as water, methane, and carbon dioxide have low melting and boiling points compared to ionic or metallic substances.

The other options do not explain the low melting and boiling points:

• Having definite shapes describes molecular geometry but does not determine the strength of forces between molecules.
• Being three-dimensional structures is a feature of giant covalent substances (like diamond), which actually have very high melting points.
• Possessing shared electron pairs is what makes them covalent, but the shared pairs (covalent bonds) are not broken during melting or boiling, so they do not account for the low melting/boiling points.

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The reaction is:

\[\text{NH}_3(g) + \text{HCl}(g) \rightarrow \text{NH}_4\text{Cl}(s)\]

On the reactant side, there are 2 moles of gas (1 mole of NH₃ + 1 mole of HCl). On the product side, NH₄Cl is a solid, so there are effectively 0 moles of gas.

According to Le Chatelier's principle, when the pressure of a gaseous system at equilibrium is increased, the equilibrium shifts towards the side with fewer moles of gas to reduce the pressure.

Since the product side has fewer gaseous moles than the reactant side, increasing the pressure will shift the equilibrium to the right, favouring the product (NH₄Cl).

Note that changing pressure shifts the position of equilibrium but does not change the equilibrium constant (K). The equilibrium constant is only affected by changes in temperature, not pressure or concentration.

Exam tip: When applying Le Chatelier's principle to pressure changes, count only the moles of gaseous species on each side. Solids and liquids are not affected by pressure changes.

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Fermentation is the biochemical process in which enzymes (particularly zymase, found in yeast) convert glucose into ethanol and carbon dioxide. The overall equation is:

\[\text{C}_6\text{H}_{12}\text{O}_6 \xrightarrow{\text{zymase}} 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2\]

This is an anaerobic process, meaning it occurs without oxygen. The key word in the question is enzymatic, which points directly to fermentation, since it is the only process among the given options that is enzyme-catalysed.

Polymerization is the joining of small monomer molecules into a large polymer chain. Hydrogenation is the addition of hydrogen gas across unsaturated bonds, typically using a metal catalyst such as nickel. Saponification is the alkaline hydrolysis of fats or oils to produce soap and glycerol. None of these processes involves the enzymatic breakdown of glucose to ethanol.

Whenever a question mentions the biological or enzymatic conversion of sugars to alcohol, the answer is fermentation.

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Crude oil (petroleum) is a complex mixture of hydrocarbons with different boiling points. To separate it into useful fractions (such as petrol/gasoline, kerosene, diesel, lubricating oil, and bitumen), fractional distillation is used.

In fractional distillation, crude oil is heated in a furnace until most of it vaporises. The vapour enters a tall fractionating column that is hot at the bottom and cool at the top. As the vapour rises through the column:

• Hydrocarbons with higher boiling points condense lower in the column (e.g., bitumen, heavy fuel oil).
• Hydrocarbons with lower boiling points rise higher before condensing (e.g., petrol, refinery gas).

The column contains trays at different heights where each fraction is collected.

The other separation methods are not suitable:

• Steam distillation is used for heat-sensitive compounds (like essential oils) that would decompose at their normal boiling points.
• Fractional crystallisation separates solids from solution based on differences in solubility at different temperatures.
• Precipitation is used to separate insoluble solids from a solution by adding a reagent.

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The structural formula shows H₃C-CH₂-C(=O)-O-CH₂-CH₃, which is an ester. To name an ester using IUPAC nomenclature, identify two parts:

The acid component (to the left of the ester linkage -C(=O)-O-): There are three carbon atoms (CH₃-CH₂-C=O), which corresponds to propanoic acid. In the ester name, this becomes propanoate.

The alkyl component (to the right of the ester oxygen): There are two carbon atoms (-O-CH₂-CH₃), which is an ethyl group.

Combining both parts, the ester is named ethyl propanoate. The alkyl group name comes first, followed by the name derived from the parent carboxylic acid with the -ic acid suffix replaced by -ate.

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The structure shown is R-C(=O)-NH-H, which contains a carbonyl group (C=O) directly bonded to a nitrogen atom bearing hydrogen atoms. This is the defining arrangement of the amide functional group (-CONH₂).

An alkanamide (also called an amide) has the general formula R-CONH₂, where R is an alkyl group. The key feature distinguishing it from the other options is the simultaneous presence of both the C=O and the N-H bonds on the same carbon.

An alkylamine (R-NH₂) has nitrogen bonded to an alkyl group but no carbonyl. An alkanone (R-CO-R') has a carbonyl flanked by two carbon groups with no nitrogen. An amino acid would require both an amine group (-NH₂) and a carboxyl group (-COOH) on the same molecule, which is not the case here.

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The diagram shows the equation 2NaCl(l) → 2Na(l) + Cl₂(g) with electricity as the energy source. This is the electrolysis of molten sodium chloride to produce metallic sodium and chlorine gas.

This industrial process is known as the Downs process, named after J.C. Downs who patented the Downs cell in 1924. In the Downs cell, molten NaCl (often mixed with CaCl₂ to lower the melting point from 801°C to about 600°C) is electrolysed. At the cathode, Na⁺ ions are reduced to liquid sodium metal, while at the anode, Cl⁻ ions are oxidised to produce chlorine gas.

The Bosch process produces hydrogen gas from water gas. The Chlor-alkali process electrolyses aqueous (not molten) NaCl to give NaOH, Cl₂, and H₂. The Browning process is not a standard industrial chemistry term in this context.

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The electron configuration 1s² 2s² 2p⁶ 3s² 3p⁵ has a total of 2 + 2 + 6 + 2 + 5 = 17 electrons, which identifies the element as chlorine (Cl, atomic number 17).

The group number of an element in the periodic table is determined by the number of electrons in its outermost (valence) shell. For chlorine, the outermost shell is the third shell (n = 3), which contains:

\[3s^2\,3p^5 = 2 + 5 = 7 \text{ electrons}\]

Therefore, chlorine belongs to Group 7 (also called Group VII or Group 17 in modern IUPAC numbering). Group 7 elements are the halogens: fluorine, chlorine, bromine, iodine, and astatine. They all have seven electrons in their outermost shell, giving them the general outer-shell configuration ns² np⁵.

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Ammonium trioxonitrate(V) is the IUPAC name for ammonium nitrate, NH₄NO₃. When heated gently (thermal decomposition), it breaks down as follows:

\[\text{NH}_4\text{NO}_3 \xrightarrow{\text{heat}} \text{N}_2\text{O} + 2\text{H}_2\text{O}\]

The products are dinitrogen monoxide (N₂O, also known as nitrous oxide or laughing gas) and water (H₂O).

To verify, check that the equation is balanced:

• N: 2 on each side (1 from NH₄⁺ and 1 from NO₃^- on the left; 2 in N₂O on the right).
• H: 4 on each side (4 in NH₄⁺ on the left; 4 in 2H₂O on the right).
• O: 3 on each side (3 in NO₃^- on the left; 1 in N₂O + 2 in 2H₂O on the right).

The other options are incorrect: NO₂ and H₂O would not balance correctly with the given reactant; N₂O and O₂ would leave hydrogen unaccounted for; and NO₃ is not a stable molecular product of thermal decomposition.

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This question tests the ability to read and interpret a solubility-temperature graph. The graph plots solubility (y-axis) against temperature in °C (x-axis) for four substances: X, Y, Z, and Q.

Examining each curve on the graph:

• X shows a steep, curved increase in solubility with temperature. Its solubility rises slowly at lower temperatures but accelerates sharply at higher temperatures, giving a concave-upward curve.
• Y follows a straight line from the origin, rising at a constant rate as temperature increases. This is a steady, linear increase in solubility.
• Z appears as a nearly horizontal line at about 45 units of solubility, meaning its solubility barely changes regardless of temperature. Z is essentially temperature-independent.
• Q rises to a peak around 40 to 50 °C and then decreases, meaning its solubility first increases and then drops with further heating.

The correct inference is that the solubility of Y increases steadily as temperature increases. The word "steadily" is key here: Y's straight-line graph means its solubility rises at a uniform, constant rate per degree of temperature increase. X also increases with temperature, but its increase is not steady; it accelerates (curves upward), so the rate of increase itself changes.

The claim that the solubility of X and Y is the same at all temperatures is incorrect because the two curves only intersect at a single point; at all other temperatures, their solubilities differ. The claim that the solubility of X, Y, and Z is temperature dependent is incorrect because Z is nearly flat, showing its solubility is essentially independent of temperature. The claim that the solubility of Z increases as temperature increases is directly contradicted by Z's horizontal line on the graph.

Exam tip: When a question uses the word "steadily," look for a straight-line relationship on the graph. A curve that bends upward or downward represents a changing rate of increase, not a steady one.

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The equation shows propane (\(C_3H_8\)) reacting with chlorine gas (\(Cl_2\)) in the presence of ultraviolet light to produce chloropropane (\(C_3H_7Cl\)) and hydrogen chloride (\(HCl\)).

In this reaction, a hydrogen atom on the propane molecule is replaced by a chlorine atom. This is the hallmark of a substitution reaction, specifically a free-radical substitution. The UV light provides the energy needed to break the \(Cl-Cl\) bond homolytically, generating chlorine free radicals that then attack the alkane.

It is not neutralization (no acid-base reaction), not polymerization (no repeating monomer units are joined), and not oxidation in the classical sense used here. The defining feature is the direct replacement of one atom (H) by another (Cl) in the organic molecule.

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To name an organic compound using IUPAC nomenclature, follow these steps:

Step 1: Identify the structure. The compound is CH₃C≡CCH(CH₃)₂. Writing it out carbon by carbon:

• C-1: CH₃
• C-2: C (triple bond to C-3)
• C-3: C (triple bond to C-2)
• C-4: CH with two CH₃ groups attached

Step 2: Find the longest carbon chain containing the triple bond. The four carbons above give a chain of 4. However, one of the methyl groups on C-4 can extend the chain to 5 carbons: C-1, C-2, C-3, C-4, C-5 (incorporating one methyl into the main chain). The remaining methyl group on C-4 becomes a branch.

Step 3: Number the chain to give the triple bond the lowest possible locants. Numbering from the CH₃ end: the triple bond is at positions 2-3. This gives pent-2-yne.

Step 4: Name the substituent. The methyl branch is on C-4.

The complete IUPAC name is 4-methylpent-2-yne.

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Benzene (\(\text{C}_6\text{H}_6\)) is a cyclic aromatic hydrocarbon with a six-membered ring containing three alternating double bonds (or, more precisely, delocalised electrons). When benzene undergoes catalytic hydrogenation, three molecules of hydrogen add across the ring, saturating all the double bonds while preserving the ring structure:

\[\text{C}_6\text{H}_6 + 3\text{H}_2 \xrightarrow{\text{Ni, heat/pressure}} \text{C}_6\text{H}_{12}\]

The product is cyclohexane, a six-membered saturated ring. The key point is that hydrogenation adds hydrogen to the double bonds but does not break open the ring. Hexane (\(\text{C}_6\text{H}_{14}\)) is a straight-chain alkane, which would require ring-opening and further hydrogen addition; that is not what happens here.

Margarine is produced by the catalytic hydrogenation of unsaturated vegetable oils (fats), not benzene. Hexene is an unsaturated six-carbon compound that would result from incomplete hydrogenation of a different starting material, not from benzene.

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In qualitative analysis, ions are classified as either acid radicals (anions) or basic radicals (cations).

• Acid radicals are the negative ions (anions) derived from acids. Examples include CO₃^2- (from H₂CO₃), SO₄^2- (from H₂SO₄), NO₃^- (from HNO₃), PO₄^3- (from H₃PO₄), and Cl^- (from HCl).
• Basic radicals are the positive ions (cations) derived from bases. Examples include Na⁺, K⁺, Mg²⁺, Ca²⁺, and Fe²⁺.

The question asks which group contains only acid radicals. Examining each option:

• CO₃^2-, PO₄^3-, Mg²⁺: contains Mg²⁺, which is a basic radical (cation).
• CO₃^2-, SO₄^2-, Na⁺: contains Na⁺, which is a basic radical (cation).
• CO₃^2-, SO₄^2-, NO₃^-: all three are anions derived from acids. This group contains only acid radicals.
• CO₃^2-, SO₄^2-, K⁺: contains K⁺, which is a basic radical (cation).

The correct answer is the group containing CO₃^2-, SO₄^2-, and NO₃^-, as all three are acid radicals.

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Carbohydrates follow the general formula \(\text{C}_x(\text{H}_2\text{O})_y\). To find the value of x for fructose, you need to know the molecular formula of fructose.

Fructose has the molecular formula C₆H₁₂O₆. Rewriting this in the carbohydrate general formula:

\[\text{C}_6\text{H}_{12}\text{O}_6 = \text{C}_6(\text{H}_2\text{O})_6\]

Comparing with \(\text{C}_x(\text{H}_2\text{O})_y\), the value of x = 6 (and y = 6 as well).

Fructose is a monosaccharide (simple sugar) that is an isomer of glucose. Both have the same molecular formula C₆H₁₂O₆, but they differ in structural arrangement: glucose is an aldose (contains an aldehyde group) while fructose is a ketose (contains a ketone group).