May/June 2025 Paper 42 Worked Answers (IGCSE Chemistry 0620 Extended)

52 questions · 80 marks · 75 minutes

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1. Question 1(a)

   1 marksPercentage of nitrogen in clean dry air
   Step 1: Clean, dry air is about 78% nitrogen and 21% oxygen, with small amounts of other gases. Step 2: The percentage of nitrogen is therefore about 78%. Step 3: 21 is the oxygen figure, 0.04 is carbon dioxide and 1 is roughly the noble gases, so the nitrogen value is 78.
   Method:

   Recall the composition of clean, dry air and give the nitrogen percentage.
   Examiner tips

   • Remember the 78% nitrogen / 21% oxygen split of clean, dry air.

2. Question 1(b)

   1 marksOperating temperature of the Contact process
   Step 1: The Contact process converts sulfur dioxide and oxygen to sulfur trioxide over a vanadium(V) oxide catalyst. Step 2: The typical operating temperature is about 450 °C, a compromise that gives a reasonable yield at a reasonable rate. Step 3: Room temperature (25 °C) is too slow and very high temperatures lower the yield of this exothermic reaction, so 450 °C is correct.
   Method:

   Recall the standard conditions for the Contact process and give the temperature.
   Examiner tips

   • Contact process: about 450 °C, 2 atm, vanadium(V) oxide catalyst.

3. Question 1(c)

   1 marksNumber of metals in Period 3
   Step 1: Period 3 contains sodium, magnesium, aluminium, silicon, phosphorus, sulfur, chlorine and argon. Step 2: The metals are sodium, magnesium and aluminium, which lie on the left of the period. Step 3: Silicon is a metalloid and the rest are non-metals, so the number of metals is 3.
   Method:

   List the Period 3 elements and count those to the left of the metal/non-metal line.
   Examiner tips

   • The metal/non-metal boundary in Period 3 falls after aluminium.

4. Question 1(d)

   1 marksNumber of halogens that are gases at r.t.p.
   Step 1: Going down Group VII the physical state changes from gas to liquid to solid. Step 2: Fluorine and chlorine are gases at r.t.p., bromine is a liquid and iodine is a solid. Step 3: So the number of halogens that are gases at r.t.p. is 2.
   Method:

   Recall the states of the halogens down the group and count the gases.
   Examiner tips

   • Down Group VII: chlorine gas, bromine liquid, iodine solid.

5. Question 1(e)

   1 marksTemperature for fermentation of glucose
   Step 1: Fermentation uses enzymes in yeast, which work best at warm body-like temperatures. Step 2: A typical fermentation temperature is around 30 °C (the mark scheme accepts any value from 25 to 35 °C). Step 3: At 5 °C the yeast is too slow and at 100 °C the enzymes are denatured, so about 30 °C is correct.
   Method:

   Recall the temperature range at which yeast enzymes ferment glucose.
   Examiner tips

   • Yeast enzymes need a warm temperature, about 30 °C, for fermentation.

6. Question 1(f)

   1 marksUnbranched isomers of C4H8 that decolourise bromine
   Step 1: Decolourising aqueous bromine needs a carbon-carbon double bond, so the isomers must be alkenes. Step 2: The unbranched alkene isomers of $\text{C}_4\text{H}_8$ are but-1-ene and but-2-ene. Step 3: Methylpropene is branched and cyclobutane has no double bond, so the number of unbranched alkene isomers is 2.
   Method:

   Draw the straight-chain alkenes of C4H8 and count those with a double bond.
   Examiner tips

   • Move the C=C along a 4-carbon chain: positions 1 and 2 give two unbranched alkenes.

7. Question 1(g)

   1 marksNumber of covalent bonds in ethanol
   Step 1: Count each single covalent bond in $\text{CH}_3\text{CH}_2\text{OH}$. Step 2: There are 5 carbon-hydrogen bonds, 1 carbon-carbon bond, 1 carbon-oxygen bond and 1 oxygen-hydrogen bond. Step 3: That gives 5 + 1 + 1 + 1 = 8 covalent bonds.
   Method:

   Draw out the displayed formula of ethanol and count all the bonds.
   Examiner tips

   • Sketch the displayed formula and count every line as one bond.

8. Question 1(h)

   1 marksNumber of carbon atoms in propyl butanoate
   Step 1: An ester name gives the alcohol part first and the acid part second. Step 2: The propyl group comes from propanol and has 3 carbon atoms; the butanoate group comes from butanoic acid and has 4 carbon atoms. Step 3: Adding these gives 3 + 4 = 7 carbon atoms.
   Method:

   Split the ester name into its alcohol and acid parts and add the carbon counts.
   Examiner tips

   • Ester name = alcohol part + acid part; count carbons in each.

9. Question 2(a)

   2 marksMeaning of a covalent bond
   Step 1: A covalent bond forms between non-metal atoms. Step 2: It is a shared pair of electrons held between the two bonded atoms. Step 3: Transfer of electrons describes ionic bonding and a sea of electrons describes metallic bonding, so a covalent bond is a shared pair of electrons between two atoms.
   Method:

   Define a covalent bond by stating that a pair of electrons is shared between two atoms.
   Examiner tips

   • Covalent = shared electrons; ionic = transferred electrons.

10. Question 2(b)(i)

    2 marksMeaning of (I) in chlorine(I) oxide
    Step 1: A Roman numeral in a name gives the oxidation number of the element it follows. Step 2: Here (I) follows chlorine, so the oxidation number of chlorine is +1. Step 3: It does not give the number of atoms or the oxidation number of oxygen, so (I) means the oxidation number of chlorine is +1.
    Method:

    Interpret the Roman numeral as the oxidation number of chlorine.
    Examiner tips

    • Roman numeral in a name = oxidation number of that element.

11. Question 2(b)(ii)

    3 marksDot-and-cross diagram of chlorine(I) oxide
    Step 1: $\text{Cl}_2\text{O}$ is covalent, so the atoms share electrons. Oxygen forms one single bond (one shared pair) to each chlorine. Step 2: Oxygen has 6 outer electrons; two are used in the two bonds, leaving 4 non-bonding electrons. Step 3: Each chlorine has 7 outer electrons; one is used in its bond, leaving 6 non-bonding electrons, so the arrangement is one shared pair per bond with 4 and 6 non-bonding electrons respectively.
    Method:

    Form one shared pair per O-Cl bond, then place the remaining outer electrons as non-bonding pairs.
    Examiner tips

    • Single bond = one shared pair; remaining outer electrons become non-bonding pairs.

12. Question 2(b)(iii)

    3 marksLow boiling point and thermal stability of Cl2O
    Step 1: When $\text{Cl}_2\text{O}$ boils, only the weak intermolecular attractions between separate molecules are overcome, which needs little energy, so the boiling point is low. Step 2: Thermal decomposition into $\text{Cl}_2$ and $\text{O}_2$ would mean breaking the strong covalent bonds inside the molecules, which needs much more energy. Step 3: Because gentle warming cannot supply enough energy to break those strong covalent bonds, the substance boils at a low temperature without decomposing.
    Method:

    Distinguish the weak forces broken on boiling from the strong covalent bonds broken on decomposition.
    Examiner tips

    • Separate the weak forces between molecules from the strong bonds within them.

13. Question 2(b)(iv)

    2 marksWhy liquid Cl2O does not conduct electricity
    Step 1: Electrical conduction needs charged particles that are free to move, either ions or mobile electrons. Step 2: $\text{Cl}_2\text{O}$ is made of neutral molecules, so there are no ions present. Step 3: It also has no mobile (delocalised) electrons, so with neither charge carrier available it cannot conduct, making it a poor conductor.
    Method:

    Explain that molecular Cl2O lacks both free ions and mobile electrons.
    Examiner tips

    • Conduction requires mobile charge carriers; molecules provide neither ions nor free electrons.

14. Question 2(c)(i)

    1 marksGiant covalent form of carbon that conducts
    Step 1: In a giant covalent structure of carbon, each atom is bonded to others in a network. Step 2: In graphite each carbon bonds to only three others, leaving one delocalised electron per atom that is free to move and conduct electricity. Step 3: Diamond uses all four electrons in bonds (no free electrons) and silicon(IV) oxide and methane are not forms of carbon, so the conducting form is graphite.
    Method:

    Pick the carbon allotrope with delocalised electrons.
    Examiner tips

    • Graphite conducts because each carbon has one delocalised electron.

15. Question 2(c)(ii)

    1 marksParticles responsible for conduction in graphite
    Step 1: In graphite each carbon atom bonds to three others, leaving one outer electron that is not held in a bond. Step 2: These delocalised electrons are free to move through the structure. Step 3: Moving electrons carry the current, so the particles responsible are electrons, not protons, ions or neutrons.
    Method:

    Identify the mobile charged particles in graphite.
    Examiner tips

    • Conduction in graphite = movement of delocalised electrons.

16. Question 2(c)(iii)

    2 marksAtoms in the giant structure of silicon(IV) oxide
    Step 1: In silicon(IV) oxide each silicon atom is bonded to four oxygen atoms and each oxygen bridges two silicon atoms. Step 2: For this particular 9-atom portion, the correct completion has 5 silicon atoms and 4 oxygen atoms. Step 3: The other combinations break the bonding pattern of the lattice, so the answer is 5 silicon and 4 oxygen atoms.
    Method:

    Use the Si-O bonding pattern to assign the symbols to the 9 atoms.
    Examiner tips

    • Each Si bonds to 4 O; each O bridges 2 Si in the SiO2 lattice.

17. Question 3(a)

    3 marksDefinition of electrolysis
    Step 1: Electrolysis breaks down compounds using electricity, and it needs ions that are free to move. Step 2: Only ionic compounds can be electrolysed, and only when they are molten or dissolved in water (aqueous solution) so the ions can move. Step 3: Covalent compounds have no ions and a solid ionic compound has fixed ions, so the definition is ionic compounds, molten or in aqueous solution.
    Method:

    Recall that electrolysis needs ionic compounds with free-moving ions.
    Examiner tips

    • Mobile ions are needed: molten or aqueous ionic compounds.

18. Question 3(b)(i)

    1 marksCathode mass with platinum electrodes
    Step 1: During electrolysis of copper(II) sulfate, copper ions, $\text{Cu}^{2+}$, are attracted to the negative cathode. Step 2: They gain electrons and are deposited as copper metal on the cathode. Step 3: Adding copper to the cathode increases its mass, so the cathode mass increases.
    Method:

    Decide which ion is discharged at the cathode and whether it adds mass.
    Examiner tips

    • Cu2+ ions are discharged at the cathode, plating copper and adding mass.

19. Question 3(b)(ii)

    1 marksColour change of electrolyte with platinum electrodes
    Step 1: The blue colour of the electrolyte comes from copper(II) ions, $\text{Cu}^{2+}$, in solution. Step 2: As electrolysis removes these ions by depositing copper at the cathode, their concentration falls. Step 3: With fewer copper(II) ions the blue colour becomes paler, fading towards colourless.
    Method:

    Link the blue colour to copper ions and to their removal during electrolysis.
    Examiner tips

    • Blue is due to Cu2+; depositing copper removes the colour.

20. Question 3(b)(iii)

    1 marksObservation at the anode with platinum electrodes
    Step 1: With inert electrodes and a dilute aqueous solution, the anode discharges hydroxide ions from the water. Step 2: This produces oxygen gas at the anode. Step 3: The oxygen is seen as bubbles of gas; copper is deposited at the cathode, not the anode, so the observation at the anode is bubbles of gas.
    Method:

    Work out the anode product with inert electrodes and describe what is seen.
    Examiner tips

    • Inert anode in dilute solution gives oxygen bubbles.

21. Question 3(b)(iv)

    3 marksIonic half-equation at the anode
    Step 1: At the anode hydroxide ions lose electrons (oxidation). Step 2: Four hydroxide ions release four electrons to form two water molecules and one oxygen molecule. Step 3: This balances for atoms and charge as $4\text{OH}^- \rightarrow 2\text{H}_2\text{O} + \text{O}_2 + 4e^-$, which is the correct anode half-equation.
    Method:

    Write an oxidation half-equation for hydroxide ions and balance atoms and charge.
    Examiner tips

    • Anode = oxidation = electrons released on the right-hand side.

22. Question 3(c)(i)

    1 marksCathode mass with copper electrodes
    Step 1: With copper electrodes, copper(II) ions are still attracted to the cathode. Step 2: They gain electrons and are deposited as copper metal on the cathode. Step 3: Copper builds up on the cathode, so its mass increases (while the copper anode dissolves to keep the solution concentration roughly constant).
    Method:

    Decide what is discharged at the cathode with copper electrodes.
    Examiner tips

    • Cathode always gains copper; with copper electrodes the anode dissolves.

23. Question 3(c)(ii)

    1 marksElectrolyte colour with copper electrodes
    Step 1: At the cathode copper(II) ions are removed as copper is deposited. Step 2: At the copper anode, copper atoms dissolve to form copper(II) ions, replacing those lost. Step 3: Because ions are added at the same rate they are removed, the copper(II) ion concentration stays constant and the blue colour does not change.
    Method:

    Compare the rate of copper ion removal and replacement at the two electrodes.
    Examiner tips

    • Copper added at the anode balances copper removed at the cathode.

24. Question 3(d)(i)

    1 marksName of the ore of aluminium
    Step 1: Aluminium is extracted from purified aluminium oxide obtained from its ore. Step 2: The main ore of aluminium is bauxite, which is purified to aluminium oxide. Step 3: Haematite is an iron ore, limestone is calcium carbonate and cryolite is an additive, so the aluminium ore is bauxite.
    Method:

    Recall the name of the aluminium-containing ore.
    Examiner tips

    • Aluminium ore = bauxite; iron ore = haematite.

25. Question 3(d)(ii)

    1 marksSubstance added to lower the operating temperature
    Step 1: Pure aluminium oxide has a very high melting point, which would make electrolysis costly. Step 2: Cryolite is added so the aluminium oxide dissolves in molten cryolite at a much lower temperature. Step 3: This lowers the operating temperature and saves energy, so the substance added is cryolite.
    Method:

    Recall the additive used to reduce energy costs in aluminium electrolysis.
    Examiner tips

    • Cryolite lowers the melting/operating temperature in aluminium extraction.

26. Question 3(d)(iii)

    2 marksWhy aluminium anodes are continually replaced
    Step 1: Oxygen is produced at the positive carbon anodes during the electrolysis. Step 2: At the high operating temperature, the hot carbon reacts with this oxygen to form carbon dioxide gas. Step 3: This reaction gradually burns away the carbon, so the anodes get smaller and must be replaced regularly.
    Method:

    Explain that oxygen at the anode reacts with the hot carbon to make carbon dioxide.
    Examiner tips

    • Anode oxygen + hot carbon → carbon dioxide, so the anode is consumed.

27. Question 4(a)(i)

    1 marksBalancing the chromite-carbon equation
    Step 1: Count the atoms already balanced: Fe and Cr match on both sides, and there are 4 carbon atoms on the left. Step 2: There are 4 oxygen atoms in $\text{FeCr}_2\text{O}_4$ to account for, and 4 carbon atoms to use. Step 3: Combining 4 carbon with 4 oxygen gives $4\text{CO}$, which balances both carbon and oxygen, so the missing product is $4\text{CO}$.
    Method:

    Balance carbon and oxygen across the equation to find the missing product.
    Examiner tips

    • Match the 4 C and 4 O atoms: they combine as 4CO.

28. Question 4(a)(ii)

    1 marksDisadvantage of extracting chromium with carbon
    Step 1: The equation shows that both iron and chromium are formed from the chromite. Step 2: This means the chromium is not pure; it is mixed with the iron. Step 3: Extra work is then needed to separate the chromium from the iron, which is the main disadvantage of this method.
    Method:

    Read the products of the equation and identify the purity problem.
    Examiner tips

    • If the ore contains iron, the product metal is a chromium-iron mixture.

29. Question 4(b)

    1 marksSubstance formed when a metal is mixed with other elements
    Step 1: A mixture of a metal with one or more other elements is given a special name. Step 2: This mixture is called an alloy, for example stainless steel. Step 3: A compound has its elements chemically combined in fixed ratios, an isotope is a form of an atom and an electrolyte conducts during electrolysis, so the correct term is alloy.
    Method:

    Recall the term for a mixture based on a metal.
    Examiner tips

    • Alloy = mixture of a metal with other elements.

30. Question 4(c)(i)

    1 marksSum of oxidation numbers in a polyatomic ion
    Step 1: The sum of the oxidation numbers of all atoms in an ion equals the overall charge on the ion. Step 2: The ion $\text{Cr}_2\text{O}_7^{2-}$ carries a charge of $2-$. Step 3: Therefore the sum of the oxidation numbers is $-2$.
    Method:

    Set the sum of oxidation numbers equal to the charge on the ion.
    Examiner tips

    • For an ion, total oxidation number = the ion charge.

31. Question 4(c)(ii)

    2 marksOxidation number of chromium in dichromate
    Step 1: The seven oxygen atoms contribute $7 \times (-2) = -14$ to the total. Step 2: The overall charge is $-2$, so the two chromium atoms together must total $-2 - (-14) = +12$. Step 3: Sharing $+12$ between the two chromium atoms gives $+12 \div 2 = +6$ for each chromium.
    Method:

    Balance the oxidation numbers: oxygens + chromiums = ion charge, then divide.
    Examiner tips

    • Total Cr = (ion charge) - (oxygen total); then divide by number of Cr atoms.

32. Question 4(c)(iii)

    3 marksVolume of gas from a mole calculation
    Step 1: Moles of $(\text{NH}_4)_2\text{Cr}_2\text{O}_7 = 1.26 \div 252 = 0.00500\ \text{mol}$. Step 2: One mole of compound gives one mole of nitrogen, so moles of $\text{N}_2 = 0.00500\ \text{mol}$. Step 3: Volume $= 0.00500 \times 24\,000 = 120\ \text{cm}^3$ at r.t.p.
    Method:

    Convert mass to moles, use the 1:1 ratio, then multiply by the molar gas volume.
    Examiner tips

    • Volume (cm³) at r.t.p. = moles × 24 000.

33. Question 5(a)

    1 marksGroup number of the alkali metals
    Step 1: The alkali metals are the very reactive metals on the far left of the Periodic Table. Step 2: They each have one electron in their outer shell, which places them in Group I. Step 3: Group II is the alkaline earth metals, Group VII the halogens and Group 0 the noble gases, so the alkali metals are in Group I.
    Method:

    Use the number of outer electrons to place the alkali metals in their group.
    Examiner tips

    • One outer electron = Group I = alkali metals.

34. Question 5(b)(i)

    1 marksAlkali metal with the highest melting point
    Step 1: In Group I the melting point decreases as you go down the group. Step 2: Lithium is at the top of the group, so it has the highest melting point. Step 3: Sodium, potassium and caesium are lower down and have progressively lower melting points, so lithium has the highest.
    Method:

    Use the down-group trend in melting point to pick the top metal.
    Examiner tips

    • Top of Group I = highest melting point.

35. Question 5(b)(ii)

    1 marksAlkali metal with the highest density
    Step 1: Density of the Group I metals generally increases going down the group. Step 2: Francium is the lowest member of Group I, so it has the highest density. Step 3: Lithium, sodium and potassium are higher up the group and have lower densities, so francium has the highest density.
    Method:

    Apply the down-group density trend to find the densest metal.
    Examiner tips

    • Bottom of Group I = highest density.

36. Question 5(b)(iii)

    1 marksAlkali metal with the lowest reactivity
    Step 1: In Group I reactivity increases going down the group, because the outer electron is lost more easily. Step 2: Lithium is at the top of the group, so it loses its outer electron least easily and is the least reactive. Step 3: Sodium, potassium and caesium are more reactive, so lithium has the lowest reactivity.
    Method:

    Use the down-group reactivity trend to pick the least reactive metal.
    Examiner tips

    • Top of Group I = least reactive; reactivity increases downwards.

37. Question 5(b)(iv)

    1 marksAlkali metal with a lilac flame colour
    Step 1: Different metal ions give characteristic flame colours. Step 2: Potassium gives a lilac (light purple) flame. Step 3: Sodium gives a yellow flame and lithium gives a red flame, so the metal with a lilac flame is potassium.
    Method:

    Match the lilac flame colour to the correct metal ion.
    Examiner tips

    • Lilac flame is the signature of potassium ions.

38. Question 5(b)(v)

    1 marksAlkali metal found in fertilisers
    Step 1: The main nutrients in NPK fertilisers are nitrogen, phosphorus and potassium. Step 2: Of the alkali metals, potassium is the one plants need and that fertilisers supply. Step 3: Lithium, caesium and rubidium are not plant nutrients, so the alkali metal in fertilisers is potassium.
    Method:

    Recall the plant nutrients in fertilisers and pick the alkali metal.
    Examiner tips

    • NPK fertiliser: N, P and potassium (K).

39. Question 5(c)

    3 marksElectronic configurations of ions in sodium sulfide
    Step 1: A sodium atom (2,8,1) loses its single outer electron to form $\text{Na}^+$ with configuration 2,8 and a 1+ charge. Step 2: A sulfur atom (2,8,6) gains two electrons to form $\text{S}^{2-}$ with configuration 2,8,8 and a 2- charge. Step 3: Two sodium ions balance one sulfide ion in $\text{Na}_2\text{S}$, so the ions are $\text{Na}^+$ (2,8) and $\text{S}^{2-}$ (2,8,8).
    Method:

    Form each ion by transferring electrons, then state the configuration and charge.
    Examiner tips

    • Metals lose electrons (positive ions); non-metals gain electrons (negative ions).

40. Question 5(d)(i)

    1 marksTerm for atoms of the same element with different masses
    Step 1: $^{85}\text{Rb}$ and $^{87}\text{Rb}$ are atoms of the same element (same protons) with different numbers of neutrons. Step 2: Atoms of the same element with different numbers of neutrons are called isotopes. Step 3: Ions are charged atoms, allotropes are forms of an element and compounds contain different elements, so the correct term is isotopes.
    Method:

    Recall the term for same-element atoms with different masses.
    Examiner tips

    • Isotopes: same proton number, different neutron (and mass) numbers.

41. Question 5(d)(ii)

    3 marksSubatomic particles in an atom and an ion
    Step 1: Protons equal the proton number 37 for both species. Step 2: Neutrons = mass number - protons, so $^{85}\text{Rb}$ has $85-37=48$ neutrons and $^{87}\text{Rb}$ has $87-37=50$ neutrons. Step 3: A neutral atom has 37 electrons; the $1+$ ion has lost one electron, leaving 36, so $^{87}\text{Rb}^+$ has 36 electrons.
    Method:

    Use proton number and mass number, then adjust electrons for the ion charge.
    Examiner tips

    • Protons = proton number; neutrons = mass - protons; ion charge changes electrons.

42. Question 5(d)(iii)

    1 marksRelative abundance from relative atomic mass
    Step 1: Let the abundance of $^{85}\text{Rb}$ be $x\%$, so $^{87}\text{Rb}$ is $(100-x)\%$. Step 2: The relative atomic mass is $\frac{85x + 87(100-x)}{100} = 85.5$, which gives $85x + 8700 - 87x = 8550$. Step 3: Solving $-2x = -150$ gives $x = 75$, so $^{85}\text{Rb}$ is 75% abundant.
    Method:

    Write the relative atomic mass as a weighted average and solve for the abundance.
    Examiner tips

    • Relative atomic mass is the weighted average of the isotope masses.

43. Question 6(a)

    1 marksTerm for a family with the same functional group
    Step 1: Organic compounds with the same functional group and general formula form a family. Step 2: This family is called a homologous series, for example the alkenes. Step 3: Allotropes are forms of an element, isotopes are atoms of one element and a mixture is not a chemical family, so the term is homologous series.
    Method:

    Recall the term for an organic family sharing a functional group.
    Examiner tips

    • Homologous series = same functional group + same general formula.

44. Question 6(b)

    1 marksMr difference between consecutive members
    Step 1: Consecutive members of a homologous series differ by one $\text{CH}_2$ group. Step 2: The relative molecular mass of $\text{CH}_2$ is $12 + (2 \times 1) = 14$. Step 3: So the next alkene ($\text{C}_3\text{H}_6$) has a relative molecular mass 14 greater than $\text{C}_2\text{H}_4$, giving a difference of 14.
    Method:

    Find the mass of the CH2 unit that separates consecutive members.
    Examiner tips

    • CH2 = 12 + 2 = 14; that is the step between members.

45. Question 6(c)

    2 marksSymbol equation for complete combustion of ethene
    Step 1: Complete combustion of a hydrocarbon gives carbon dioxide and water only. Step 2: Ethene has 2 carbon and 4 hydrogen atoms, giving $2\text{CO}_2$ and $2\text{H}_2\text{O}$. Step 3: Balancing oxygen: products need $4 + 2 = 6$ oxygen atoms, so $3\text{O}_2$ is required, giving $\text{C}_2\text{H}_4 + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O}$.
    Method:

    Write CO2 and water as products, balance carbon and hydrogen, then oxygen.
    Examiner tips

    • Balance C and H first, then count oxygen atoms in the products.

46. Question 6(d)(i)

    4 marksEffect of conditions on an equilibrium concentration
    Step 1: Lowering temperature favours the exothermic forward reaction, using up $\text{C}_2\text{H}_4$, so its concentration decreases. Step 2: Removing the product $\text{C}_2\text{H}_5\text{OH}$ shifts the position forward, so $\text{C}_2\text{H}_4$ decreases; increasing pressure favours the side with fewer gas moles (the product side), so $\text{C}_2\text{H}_4$ decreases. Step 3: A catalyst speeds both directions equally and does not change the equilibrium position, so the concentration shows no change.
    Method:

    Use Le Chatelier for each change and remember a catalyst does not move the position.
    Examiner tips

    • Catalyst = no shift; everything else follows Le Chatelier toward the product side here.

47. Question 6(d)(ii)

    3 marksCollision theory and increasing temperature
    Step 1: Raising the temperature gives the particles more kinetic energy, so they move faster. Step 2: Faster particles collide more frequently. Step 3: A greater proportion of these collisions now have energy equal to or above the activation energy, so more collisions are successful and the rate increases.
    Method:

    Link higher temperature to energy, collision frequency and the activation-energy threshold.
    Examiner tips

    • Three ideas: more kinetic energy, more frequent collisions, more collisions above activation energy.

48. Question 6(e)(i)

    1 marksMolecular formula from a displayed formula
    Step 1: Count each type of atom in compound B: there are 5 carbon atoms, 6 hydrogen atoms and 4 oxygen atoms. Step 2: Writing these in a molecular formula gives $\text{C}_5\text{H}_6\text{O}_4$. Step 3: The other options miscount the hydrogen, oxygen or carbon atoms, so the molecular formula is $\text{C}_5\text{H}_6\text{O}_4$.
    Method:

    Count each element in the displayed structure and write the molecular formula.
    Examiner tips

    • Carefully tally every atom, including those hidden in -COOH groups.

49. Question 6(e)(ii)

    1 marksWhy a compound is unsaturated
    Step 1: An unsaturated organic compound contains at least one carbon-carbon double (or triple) bond. Step 2: Compound B has a carbon-carbon bond that is not a single bond, that is a C=C double bond. Step 3: The O-H bond, single bonds and atom ratios do not define unsaturation, so it is unsaturated because of its C=C double bond.
    Method:

    Identify the carbon-carbon double bond that makes the compound unsaturated.
    Examiner tips

    • Unsaturated = contains a carbon-carbon double bond.

50. Question 6(e)(iii)

    2 marksRepeat unit of an addition polymer
    Step 1: In addition polymerisation the carbon-carbon double bond opens up. Step 2: This forms a single carbon-carbon bond, and the repeat unit is drawn with a continuation bond extending from each end carbon to show the chain continues. Step 3: No small molecule is lost in addition polymerisation, so the repeat unit keeps all of compound B and shows the opened single bond with continuation bonds.
    Method:

    Open the double bond to a single bond and draw continuation bonds on the repeat unit.
    Examiner tips

    • Addition polymer repeat unit: open the C=C, add continuation bonds, lose nothing.

51. Question 6(e)(iv)

    1 marksWhy 1 mol of compound B reacts with 2 mol NaOH
    Step 1: Sodium hydroxide neutralises carboxylic acid groups in a 1:1 ratio (one -COOH per NaOH). Step 2: If one mole of compound B reacts with two moles of NaOH, it must contain two acidic groups. Step 3: So compound B has two carboxylic acid (-COOH) groups, each reacting with one NaOH.
    Method:

    Use the 1:1 acid-to-NaOH ratio to deduce the number of -COOH groups.
    Examiner tips

    • Mole ratio with NaOH counts the number of acid groups.

52. Question 6(e)(v)

    2 marksVolume of solution from a mole calculation
    Step 1: Compound B reacts with NaOH in a 1:2 ratio, so 0.100 mol of B needs $0.100 \times 2 = 0.200\ \text{mol}$ of NaOH. Step 2: Volume in dm³ = moles ÷ concentration = $0.200 \div 0.250 = 0.800\ \text{dm}^3$. Step 3: Converting to cm³: $0.800 \times 1000 = 800\ \text{cm}^3$.
    Method:

    Use the mole ratio to find moles of NaOH, then convert to a volume in cm³.
    Examiner tips

    • Volume (cm³) = (moles ÷ concentration) × 1000.