Question 1
A list of gases is shown.
Answer the following questions about these gases. Each gas may be used once, more than once or not at all.
(a) State one gas which:
(i) is the main constituent of natural gas
(ii) is responsible for both photochemical smog and acid rain
(iii) is unsaturated
(iv) has monatomic particles
(v) reduces iron(III) oxide in a blast furnace.
(b) Nitrogen dioxide, \(NO_2\), and carbon monoxide are removed from a car exhaust by a catalytic converter. Write the symbol equation for this reaction.
Most-appropriate topic codes (Cambridge IGCSE Chemistry 0620):
• Topic 11.3 — Fuels (a)(i)
• Topic 10.3 — Air quality and climate (a)(ii)
• Topic 11.5 — Alkenes (a)(iii)
• Topic 8.5 — Noble gases (a)(iv)
• Topic 9.6 — Extraction of metals (a)(v)
• Topic 10.3 — Air quality and climate (b)
▶️ Answer/Explanation
Detailed Solution:
(a)(i) Methane: According to the syllabus (Topic 11.3), natural gas is a fossil fuel, and methane (CH₄) is named as its main constituent.
(a)(ii) Nitrogen dioxide: The syllabus (Topic 10.3) states that oxides of nitrogen from car engines cause photochemical smog and, as they dissolve in water to form acids, are also responsible for acid rain.
(a)(iii) Propene: Unsaturated hydrocarbons contain at least one carbon-carbon double bond (C=C). According to the diagram, propene (C₃H₆) fits the general formula CₙH₂ₙ for alkenes, which are unsaturated (Topic 11.5).
(a)(iv) Helium: Noble gases (Group VIII) exist as single, separate atoms (monatomic) due to their full outer electron shell, which makes them unreactive (Topic 8.5).
(a)(v) Carbon monoxide: In the blast furnace extraction of iron (Topic 9.6), carbon monoxide (CO) acts as the reducing agent. It reacts with and removes oxygen from iron(III) oxide (Fe₂O₃).
(b) 2NO₂ + 4CO → N₂ + 4CO₂: The syllabus (Topic 10.3 Supplement) requires knowledge of how catalytic converters remove oxides of nitrogen. They convert harmful NO and NO₂ into harmless nitrogen gas (N₂) while also converting toxic carbon monoxide (CO) into carbon dioxide (CO₂).
Question 2
A list of five metals is shown.
(a) All metals form positive ions.
(i) Describe how atoms form positive ions.
(ii) State which of the five metals in the list has the greatest tendency to form positive ions.
(iii) Suggest one of the five metals in the list which is not likely to show catalytic properties.
(iv) State which of the five metals in the list is a major component of stainless steel.
(b) A student adds a sample of a metal to an aqueous metal salt in a beaker to see if a displacement reaction takes place. Complete Table 2.1 to show the colour of the solution in the beaker at the start and at the end of the experiment. 
(c) Most Group II metals form a gas when placed into cold water. An alkaline solution is also formed.
(i) Name the gas formed when strontium is added to cold water.
(ii) Name the alkaline solution formed when strontium is added to cold water.
(iii) One Group II metal reacts very slowly when placed in cold water. When heated, the metal reacts with steam to form a white solid. Identify this metal and name the white solid formed.
(d) Under certain conditions, iron will react with steam to form an oxide of iron with the formula \(Fe_3O_4\). \(Fe_3O_4\) reacts with dilute hydrochloric acid to form a mixture of iron(II) and iron(III) salts and water. Deduce the symbol equation for the reaction between \(Fe_3O_4\) and dilute hydrochloric acid.
Most-appropriate topic codes (Cambridge IGCSE Chemistry 0620):
• Topic 2.2 — Atomic structure and the Periodic Table (Part (a)(i))
• Topic 9.4 — Reactivity series (Part (a)(ii))
• Topic 8.4 — Transition elements (Part (a)(iii))
• Topic 9.3 — Alloys and their properties (Part (a)(iv))
• Topic 9.4 — Displacement reactions (Part (b))
• Topic 9.1 — Properties of metals / Group II properties (Part (c))
• Topic 6.4 — Redox and equations (Part (d))
▶️ Answer/Explanation
(a)(i) Atoms form positive ions by losing one or more electrons from their outer shell. This loss reduces the number of negative charges, resulting in a net positive charge.
(a)(ii) Potassium has the greatest tendency to form positive ions because it is the most reactive metal in the list, lying at the top of the reactivity series.
(a)(iii) Magnesium (or potassium) is not likely to show catalytic properties, as these metals are not transition elements; catalytic activity is a common characteristic of transition metals like iron and copper.
(a)(iv) Iron is a major component of stainless steel, which is an alloy of iron with other elements such as chromium, nickel, and carbon.
(b) Completed Table 2.1:
Explanation: Zinc displaces iron from iron(II) sulfate, removing the pale green Fe²⁺(aq) ions. Copper displaces silver from silver nitrate, forming blue Cu²⁺(aq) ions.
(c)(i) Hydrogen is the gas formed when strontium is added to cold water, as Group II metals react to produce hydrogen gas and the metal hydroxide.
(c)(ii) Strontium hydroxide is the alkaline solution formed when strontium reacts with cold water (Sr + 2H₂O → Sr(OH)₂ + H₂).
(c)(iii) M1: Magnesium (Mg). M2: Magnesium oxide (MgO) is the white solid formed when magnesium reacts with steam (Mg + H₂O → MgO + H₂).
(d) The balanced symbol equation is: Fe₃O₄ + 8HCl → 2FeCl₃ + FeCl₂ + 4H₂O
Explanation: Fe₃O₄ contains iron in both +2 and +3 oxidation states. It reacts with 8 moles of HCl to produce one mole of iron(II) chloride, two moles of iron(III) chloride, and four moles of water, balancing all atoms and charges.
Question 3
The symbol equation for the industrial production of ammonia is shown.
(a) Name this industrial process.
(b) State the meaning of ∆H.
(c) State the typical conditions and name the catalyst used in the industrial production of ammonia.
temperature and units …………………………….
pressure and units ……………………………
catalyst used ………………………………
(d) State two methods of increasing the rate of this reaction.
(e) The symbol equation for the reaction can be represented as shown in Fig. 3.1. 
Table 3.1 shows some bond energies. 
Use the bond energies in Table 3.1 and ∆H to calculate the bond energy of an N–H bond, in kJ/mol. Use the following steps.
- Calculate the energy needed to break bonds in the reactants.
- Calculate the energy released when bonds form in the products.
- Calculate the energy of an N–H bond.
(f) An incomplete symbol equation for the preparation of ammonia in the laboratory is shown.
\(CaO + 2NH_4Cl → CaCl_2 + ……………………. + 2NH_3\)
(i) Complete the symbol equation.
(ii) Name \(NH_4Cl\).
(iii) Calculate the volume of ammonia, \(NH_3\), measured at room temperature and pressure, which forms when 1.12g of CaO is heated with excess \(NH_4Cl\).
[\(M_r\): CaO, 56]
Most-appropriate topic codes (Cambridge IGCSE Chemistry 0620):
• Topic 6.3 — Reversible reactions and equilibrium / Haber process (Parts a, c, d)
• Topic 5.1 — Exothermic and endothermic reactions / Enthalpy change (Parts b, e)
• Topic 6.2 — Rate of reaction (Part d)
• Topic 3.3 — The mole and the Avogadro constant (Part f-iii)
• Topic 7.1 — Characteristic properties of acids and bases (Part f-i, f-ii, context)
▶️ Answer/Explanation
(a) Haber process
(b) enthalpy change (of reaction)
(c) 450 and °C
200 and atm or 20 000 and kPa
iron
(d) M1 increase temperature
M2 increase pressure
(e) M1 bond energy in breaking bonds = [945 + (3×435)] = (+)2250 kJ/mol
M2 = 2250 + 90 = 2340 kJ/mol
M3 = 2340 / 6 = 390 kJ/mol
(f)(i) H₂O
(f)(ii) Ammonium chloride
(f)(iii) Moles of CaO = 1.12 / 56 = 0.02 mol
Mole ratio CaO : NH₃ = 1 : 2 → moles NH₃ = 0.04 mol
Volume NH₃ = 0.04 × 24 = 0.96 dm³
Question 4
A carboxylic acid reacts with an alcohol to produce an ester and water. Under certain conditions, this reaction can be reversed so an ester reacts with water to produce a carboxylic acid X and an alcohol Y. The reaction reaches an equilibrium.
The forward reaction is endothermic.
(a) Deduce the empirical formula of the ester.
(b) Name the ester.
(c) Name carboxylic acid X and draw its displayed formula.
(d) Name alcohol Y and give its structural formula.
(e) Complete Table 4.1 to show the effect, if any, for each change of condition. 
(f) At the beginning of the reaction between the ester and water, no carboxylic acid is present in the reaction mixture.
(i) Suggest how the pH of the reaction mixture changes from the start of the reaction until equilibrium is reached. Assume alcohols and esters are neutral.
pH at start of reaction ……………………………
pH at equilibrium ……………………………..
(ii) Identify the ion that causes the change in pH.
(iii) Name an indicator which can be used to follow the change in pH.
Most-appropriate topic codes (Cambridge IGCSE Chemistry 0620):
• Topic 3.3 — Stoichiometry: The mole and empirical formula (Part a)
• Topic 11.2 & 11.7 — Organic chemistry: Naming esters, alcohols, and carboxylic acids (Parts b, c, d)
• Topic 6.3 — Reversible reactions and equilibrium: Le Chatelier’s principle (Part e)
• Topic 7.1 — Acids, bases and salts: pH and H+ ions (Part f)
▶️ Answer/Explanation
(a) C2H4O
From the given structure, the molecular formula of the ester is C4H8O2. To find the empirical formula, divide each subscript by the greatest common divisor (2), giving C2H4O.
(b) methyl propanoate
The ester is formed from propanoic acid and methanol. In naming an ester, the first part (methyl) comes from the alcohol (methanol), and the second part (propanoate) comes from the carboxylic acid (propanoic acid).
(c) propanoic acid
Hydrolysis of this ester splits it into propanoic acid (C2H5COOH) and methanol. The displayed formula shows all atoms and bonds in the propanoic acid molecule (CH3CH2COOH).
Displayed formula of propanoic acid:
H – C – C – C = O
| | |
H H O – H
(d) methanol, CH3OH
The other product of the hydrolysis is alcohol Y, which is methanol. Its structural formula shows a methyl group (CH3</sub-) bonded to a hydroxyl group (-OH).
(e) Table completion:
- Decrease temperature: The equilibrium shifts to oppose the change. Since the reverse reaction (hydrolysis) is endothermic, decreasing temperature favours the exothermic forward reaction, so the yield of carboxylic acid decreases.
- Add a catalyst: A catalyst lowers activation energy for both forward and reverse reactions equally, speeding up attainment of equilibrium without shifting the position. Therefore, the yield of carboxylic acid has no effect.
(f)(i) pH at start of reaction: 7 (neutral). pH at equilibrium: Between 3 and 7 (acidic).
Initially, only neutral ester and water are present (pH 7). As the reaction proceeds, propanoic acid (a weak acid) is produced, releasing H+ ions into the solution, causing the pH to drop below 7 at equilibrium.
(f)(ii) H+(aq) / hydrogen ion
The carboxylic acid produced dissociates slightly in water to release hydrogen ions (H+), which are responsible for the decrease in pH.
(f)(iii) universal indicator
Universal indicator gives a broad range of colour changes across different pH values, making it suitable for following the gradual change in pH from neutral to acidic during the reaction.
Question 5
Sulfur is a Group VI element.
(a) A sample of sulfur contains two isotopes, \(^{32}S\) and \(^{34}S\).
(i) Complete Table 5.1 to show the number of protons and neutrons in one atom of each isotope of sulfur.
(ii) State why these isotopes have identical chemical properties.
(iii) State the mass of \(6.02 × 10^{23}\) atoms of \(^{34}S\). Include units in your answer.
(iv) State the name of the amount of substance which contains \(6.02 × 10^{23}\) atoms.
(v) Table 5.2 shows the relative abundance of these isotopes of sulfur in the sample. 
Calculate the relative atomic mass of sulfur in this sample to one decimal place.
(b) Sulfur reacts with magnesium to form magnesium sulfide, MgS, an ionic compound.
(i) Complete the dot-and-cross diagram in Fig. 5.1 of the ions in magnesium sulfide. Give the charges on the ions.
(ii) State why MgS has a high melting point.
(iii) State why molten MgS conducts electricity.
(c) An acid containing sulfur reacts with sodium hydroxide, NaOH, to form a salt and water. The salt has the formula \(Na_2SO_3\).
(i) Deduce the formula of this acid.
(ii) Deduce the formula of the anion in \(Na_2SO_3\).
(d) \(Na_2SO_3\) is oxidised by acidified aqueous potassium manganate(VII).
(i) State what VII refers to in the name potassium manganate(VII).
(ii) State the colour change when this reaction happens.
Most-appropriate topic codes (Cambridge IGCSE Chemistry 0620):
• Topic 2.3 — Isotopes (Parts (a)(i)-(v))
• Topic 2.4 — Ions and ionic bonds (Parts (b)(i)-(iii))
• Topic 7.1 — Characteristic properties of acids (Part (c)(i)-(ii))
• Topic 6.4 — Redox (Part (d)(i)-(ii))
▶️ Answer/Explanation
(a)(i) The completed table shows that both isotopes have 16 protons, while \(^{32}S\) has 16 neutrons and \(^{34}S\) has 18 neutrons. The atomic number of sulfur is 16.
(a)(ii) Isotopes have identical chemical properties because they have the same electronic configuration (same number and arrangement of electrons). The number of protons and electrons determines chemical behavior, not neutrons.
(a)(iii) The mass is 34 g. \(6.02 \times 10^{23}\) atoms is one mole of the isotope, and the mass of one mole in grams is equal to the mass number (for a specific isotope).
(a)(iv) The amount of substance which contains \(6.02 \times 10^{23}\) particles (atoms, molecules, ions) is called one mole. The number itself is the Avogadro constant.
(a)(v) Relative atomic mass = \( \frac{(32 \times 95) + (34 \times 5)}{100} = \frac{3040 + 170}{100} = \frac{3210}{100} = 32.1 \). This weighted average accounts for the much higher abundance of the lighter isotope.
(b)(i) The dot-and-cross diagram should show: Mg atom losing two electrons to become Mg\(^{2+}\) with an empty outer shell (8 electrons in the second shell). S atom gaining two electrons to become S\(^{2-}\) with a full outer shell (8 electrons in the third shell).
Magnesium atom (Mg, electron config 2,8,2) loses the two outer electrons. It now has 12 protons and 10 electrons, giving a charge of 2+ (shown on the answer line). The sulfur atom (S, electron config 2,8,6) gains two electrons to complete its outer shell. It now has 16 protons and 18 electrons, giving a charge of 2-.
(b)(ii) MgS has a high melting point because it is an ionic compound held together by strong electrostatic forces of attraction (ionic bonds) between the Mg\(^{2+}\) and S\(^{2-}\) ions. A large amount of thermal energy is needed to overcome these strong bonds.
(b)(iii) Molten MgS conducts electricity because the ions (Mg\(^{2+}\) and S\(^{2-}\)) are mobile (free to move) when the lattice is melted, allowing them to carry charge through the liquid.
(c)(i) The acid formula is \(H_2SO_3\) (sulfurous acid). Neutralization involves hydrogen ions from the acid reacting with hydroxide ions from the base. Since the salt contains two sodium ions (Na\(^+\)), the acid must provide two hydrogen ions (H\(^+\)), and the anion is SO\(_3^{2-}\).
(c)(ii) The formula of the anion is \(SO_3^{2-}\) (sulfite ion). In the salt Na\(_2\)SO\(_3\), each sodium ion has a charge of 1+. To balance the 2+ total charge from two sodium ions, the remaining group (SO\(_3\)) must have a charge of 2-.
(d)(i) VII refers to the oxidation number (or oxidation state) of manganese (Mn) in the compound, which is +7. This Roman numeral notation is used in naming transition metal compounds to indicate the metal’s oxidation state.
(d)(ii) The colour change is from purple to colourless. The purple MnO\(_4^-\) ion (Mn in +7 state) is reduced by the sulfite ion (which is oxidized to sulfate) to the nearly colourless Mn\(^{2+}\) ion. This is a common test for a reducing agent.
Question 6
Glucose is involved in two processes.
(a) Glucose, \(\mathrm{C_6H_{12}O_6}\), is made in plants from carbon dioxide and water.
(i) Name this process.
(ii) Write the symbol equation for this process.
(iii) State two essential conditions needed for this process to happen.
(b) Glucose is converted to ethanol.
(i) Name this process.
(ii) Name the other product formed when glucose is converted to ethanol.
(c) Ethanol is made by reacting ethene with steam in an industrial process.
(i) State the conditions and type of catalyst used in this industrial production of ethanol.
temperature and units …………………………………….
pressure and units …………………………………………
type of catalyst used ………………………………………
(ii) Explain why this reaction is an addition reaction.
(iii) Complete the dot-and-cross diagram in Fig. 6.1 of a molecule of ethanol.
Most-appropriate topic codes (Cambridge IGCSE Chemistry 0620):
• Topic 10.3 — Air quality and climate / Photosynthesis (Part (a))
• Topic 11.6 — Alcohols (Parts (b), (c))
• Topic 11.2 — Naming organic compounds (Part (c)(ii))
• Topic 2.5 — Simple molecules and covalent bonds (Part (c)(iii))
▶️ Answer/Explanation
(a)(i) Photosynthesis
This is the biological process by which plants convert light energy into chemical energy, using carbon dioxide and water to produce glucose and oxygen.
(a)(ii) \(6\mathrm{CO_2} + 6\mathrm{H_2O} \rightarrow \mathrm{C_6H_{12}O_6} + 6\mathrm{O_2}\)
The balanced symbol equation shows that six molecules of carbon dioxide react with six molecules of water to produce one molecule of glucose and six molecules of oxygen.
(a)(iii) Two essential conditions: Presence of chlorophyll and energy from light (sunlight).
Chlorophyll is the green pigment that absorbs light energy, which is then used to drive the endothermic reaction of photosynthesis.
(b)(i) Fermentation
This is an anaerobic (in the absence of oxygen) biological process where enzymes in yeast break down glucose into simpler substances.
(b)(ii) Carbon dioxide (\(\mathrm{CO_2}\))
The balanced equation for fermentation is \(\mathrm{C_6H_{12}O_6} \rightarrow 2\mathrm{C_2H_5OH} + 2\mathrm{CO_2}\), identifying carbon dioxide as the other product alongside ethanol.
(c)(i) M1: 300 °C (or 300 degrees Celsius); M2: 60 atm (or 6000 kPa); M3: Acid catalyst (e.g., phosphoric acid)
These conditions are chosen to achieve a reasonable rate of reaction and a good yield of ethanol in the industrial hydration of ethene.
(c)(ii) It is an addition reaction because only one product is formed (no other products).
Ethene (\(\mathrm{C_2H_4}\)) and water (\(\mathrm{H_2O}\)) combine directly to form a single molecule of ethanol (\(\mathrm{C_2H_5OH}\)), which is characteristic of an addition reaction.
(c)(iii) The completed dot-and-cross diagram should show:
• A C–C single bond (one dot from one carbon, one cross from the other).
• All C–H bonds and the O–H bond as shared pairs (dot-and-cross).
• Two lone pairs (four non-bonding electrons) on the oxygen atom to complete its octet, and no non-bonding electrons on carbon or hydrogen atoms.