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Question 1 -- Temperature and Collision Theory
Total: 12 marks
A class in Kolkata investigates how temperature affects the rate of reaction between sodium thiosulfate solution and dilute hydrochloric acid. A conical flask is placed on a printed paper cross and the time taken for the cross to become invisible from above is measured. The equation is Na₂S₂O₃(aq) + 2HCl(aq) → 2NaCl(aq) + SO₂(aq) + S(s) + H₂O(l)
Temperature / °C
Time for cross to disappear / s
1/time / s⁻¹
20
120
0.0083
30
75
0.0133
40
48
0.0208
50
30
0.0333
60
20
to be calculated
(a)[4]
(i) Calculate the value of 1/time at 60 °C, giving your answer to three significant figures. [2]
(ii) Explain why 1/time is used as a measure of the rate of this reaction rather than the time itself. [2]
Model Answer -- 1(a)
1/time = 1 ÷ 20 s [1]
= 0.0500 s⁻¹ (3 significant figures) [1]
A faster reaction gives a shorter time, so time is inversely related to rate; taking the reciprocal gives a quantity that increases as the rate increases [1]
1/time is therefore directly proportional to rate, so a graph of 1/time against temperature shows the trend directly [1]
⚠ If you missed marks here: Many candidates write 0.05 instead of 0.0500 and lose the significant-figures mark, or say "1/time is used because it is easier" without stating that rate is inversely proportional to time. Writing "time measures the rate" is wrong — a long time means a slow rate.
(b)[5]
Using collision theory, explain fully why increasing the temperature from 20 °C to 60 °C makes this reaction faster. Your answer must clearly separate the effect on how often particles collide from the effect on the energy of the collisions. [5]
Model Answer -- 1(b)
At the higher temperature the particles gain kinetic energy and move faster [1]
So the particles collide more frequently (more collisions per second) [1]
More importantly, a greater proportion of the colliding particles now have energy ≥ the activation energy [1]
Therefore a greater proportion of collisions are successful (result in reaction) [1]
The energy effect is the larger of the two, which is why a rise of only 40 °C cuts the time from 120 s to 20 s [1]
⚠ If you missed marks here: The commonest error is writing only "particles move faster so they collide more" — that scores at most 2 of 5. You must state that a greater proportion of particles have energy greater than or equal to the activation energy. Saying "heating lowers the activation energy" is wrong: activation energy is fixed for a given reaction and only a catalyst provides a lower-energy route.
(c)[3]
(i) State one variable, other than temperature, that must be kept constant for these results to be valid. [1]
(ii) Suggest why two students timing the same flask may record slightly different times, and state one improvement to reduce this error. [2]
Model Answer -- 1(c)
Any one of: concentration of sodium thiosulfate / concentration of hydrochloric acid / total volume of solution / depth of liquid above the cross / size and darkness of the printed cross [1]
Judging the exact moment the cross disappears is subjective, so different students choose different end points [1]
Improvement: use a light sensor and datalogger to detect a fixed drop in light transmitted, or have the same student judge every run against a reference [1]
⚠ If you missed marks here: "Use a better stopwatch" gains nothing — reaction times of 20–120 s are far longer than any stopwatch error; the uncertainty is in judging the end point, not in the timer. Also do not name temperature as the controlled variable: it is the independent variable here.
Question 2 -- Catalysts and Enzymes
Total: 12 marks
A technician in Bristol demonstrates the decomposition of hydrogen peroxide solution. In flask A the solution is left alone; in flask B a spatula of black manganese(IV) oxide powder is added and oxygen is given off vigorously. The equation is 2H₂O₂(aq) → 2H₂O(l) + O₂(g)
(a)[4]
(i) Define the term catalyst. [2]
(ii) Explain, in terms of activation energy, how manganese(IV) oxide speeds up the decomposition of hydrogen peroxide. State the effect of the catalyst on the products and on the overall energy change of the reaction. [2]
Model Answer -- 2(a)
A catalyst increases the rate of a chemical reaction [1]
and is chemically unchanged (and not used up) at the end of the reaction [1]
The manganese(IV) oxide provides an alternative reaction pathway of lower activation energy, so a greater proportion of collisions are successful [1]
The products are unchanged (still water and oxygen) and the overall energy change of the reaction is unchanged [1]
⚠ If you missed marks here: "A catalyst speeds up a reaction without taking part" is not accepted at Extended level — a catalyst does take part, it is simply regenerated. Writing "the catalyst lowers the activation energy of the reaction" is loose; the accepted wording is that it provides an alternative route with lower activation energy. Do not say the catalyst increases the yield or changes the products.
(b)[3]
The technician weighs 0.50 g of manganese(IV) oxide before adding it to flask B. When the bubbling stops, the mixture is filtered, and the residue is washed, dried and reweighed.
(i) State the mass of residue expected and explain what this shows. [2]
(ii) State one further observation from flask B that shows manganese(IV) oxide is a catalyst and not a reactant. [1]
Model Answer -- 2(b)
The residue has a mass of 0.50 g, the same as the mass added [1]
showing the catalyst is not used up / is chemically unchanged by the reaction [1]
The recovered solid is still black and, when added to a fresh portion of hydrogen peroxide, it catalyses the decomposition again (a small mass catalyses a large amount of hydrogen peroxide) [1]
⚠ If you missed marks here: A common wrong answer is "the mass will be less than 0.50 g because some was used up" — this contradicts the definition of a catalyst. Note the residue must be washed and dried first; if you answer "more than 0.50 g" you are describing wet residue, which is an experimental error and not a chemical change.
(c)[5]
Fresh liver contains the enzyme catalase, which also decomposes hydrogen peroxide.
(i) State what is meant by an enzyme. [1]
(ii) A student finds that liver boiled for five minutes produces no bubbles with hydrogen peroxide. Explain this result in terms of the structure of the enzyme. [2]
(iii) Name one industrial catalyst other than manganese(IV) oxide and state the reaction it is used in. [2]
Model Answer -- 2(c)
An enzyme is a biological catalyst, a protein that speeds up a reaction in a living organism [1]
Above its optimum temperature the enzyme is denatured: the shape of the protein (and of its active site) is permanently changed [1]
so the hydrogen peroxide molecules no longer fit the active site and the reaction is no longer catalysed [1]
Named catalyst, for example iron [1]
used in the Haber process, N₂ + 3H₂ ⇌ 2NH₃ (accept vanadium(V) oxide in the Contact process, or nickel in the hydrogenation of alkenes) [1]
⚠ If you missed marks here: Writing "the enzyme is killed" scores zero — an enzyme is a molecule, not a living thing; the required word is denatured. Also avoid "the enzyme melts". In (iii), naming "iron oxide" for the Haber process loses the mark: the catalyst is iron.
Question 3 -- Measuring Rate by Loss of Mass
Total: 12 marks
In a laboratory in Lagos, 5.00 g of marble chips (excess) is added to 50.0 cm³ of dilute hydrochloric acid in a conical flask standing on an electronic balance. A loose plug of cotton wool is pushed into the neck of the flask. The reading on the balance is recorded every 30 s. CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)
(a)[4]
(i) Explain why the reading on the balance decreases during the reaction. [2]
(ii) Explain why the cotton wool plug is needed, and why an airtight bung must not be used instead. [2]
Model Answer -- 3(a)
Carbon dioxide gas is produced [1]
and escapes from the open flask into the air, so the total mass of the flask and contents falls [1]
The cotton wool lets carbon dioxide out but stops any spray or droplets of acid being carried out of the flask, which would cause extra mass loss [1]
An airtight bung would trap the gas so no mass loss would be measured, and pressure would build up and could blow the bung out [1]
⚠ If you missed marks here: Writing "mass is lost because mass is destroyed" contradicts conservation of mass — the mass leaves as gas, it is not destroyed. Saying the cotton wool "stops the gas escaping" is self-defeating: if the gas could not escape there would be nothing to measure. The cotton wool stops liquid spray, not gas.
(b)[3]
The loss of mass was 0.88 g after 60 s and reached a final value of 1.76 g.
(i) Calculate the average rate of reaction over the first 60 s, in g/s, to two significant figures. [2]
(ii) State how the average rate over the first 60 s compares with the average rate over the whole 300 s, and justify your answer using the graph. [1]
Model Answer -- 3(b)
rate = loss of mass ÷ time = 0.88 g ÷ 60 s [1]
= 0.015 g/s (2 significant figures) [1]
Over 300 s the average rate is 1.76 ÷ 300 = 0.0059 g/s, which is smaller; the graph is steepest at the start and becomes horizontal, so the early rate is much greater than the overall average [1]
⚠ If you missed marks here: Omitting the unit g/s costs a mark even when the number 0.015 is right. Do not divide 1.76 by 60 — only 0.88 g has been lost in the first 60 s. Writing 0.0147 without rounding to two significant figures also loses the second mark.
(c)[5]
(i) Explain, in terms of particles, why the curve is steepest at the start and why it becomes horizontal after about 240 s. [3]
(ii) The experiment is repeated using the same mass of marble but ground into a fine powder, with everything else unchanged. Describe and explain the shape of the new curve compared with the original. [2]
Model Answer -- 3(c)
At the start the acid is at its highest concentration, so there are most acid particles per unit volume and the frequency of successful collisions with the marble surface is greatest [1]
As the reaction proceeds the acid is used up, its concentration falls, collisions become less frequent and the curve becomes less steep [1]
The line becomes horizontal because all the acid has been used up (the marble is in excess) so the reaction has stopped [1]
The powdered marble gives a steeper curve at the start / the reaction finishes sooner, because the powder has a larger surface area so more particles are exposed and collision frequency is greater [1]
The curve levels off at the same final loss of mass of 1.76 g, because the limiting amount of acid is unchanged [1]
⚠ If you missed marks here: The most frequently lost mark is the last one: candidates draw the powder curve levelling off higher up, but surface area changes only the rate, never the final yield. Also, "the reaction stops because the marble runs out" is wrong here — the marble is stated to be in excess, so it is the acid that is used up.
Question 4 -- Dynamic Equilibrium and Le Chatelier
Total: 12 marks
A research group in Hyderabad studies three reversible reactions in sealed vessels:
Reaction A: N₂O₄(g) ⇌ 2NO₂(g) forward reaction endothermic (pale yellow to dark brown)
Reaction B: H₂(g) + I₂(g) ⇌ 2HI(g) forward reaction exothermic
Reaction C: CoCl₄²⁻(aq) + 6H₂O(l) ⇌ Co(H₂O)₆²⁺(aq) + 4Cl⁻(aq) blue to pink
(a)[4]
(i) State what the symbol ⇌ tells you about a reaction. [1]
(ii) Reaction A is left in a sealed tube until the brown colour stops changing. Describe three features of the system at this point that show it is at dynamic equilibrium. [3]
Model Answer -- 4(a)
The reaction is reversible: it goes in both directions, products can re-form the reactants [1]
The rate of the forward reaction equals the rate of the reverse reaction [1]
The concentrations of N₂O₄ and NO₂ remain constant (so the colour stops changing) [1]
Both reactions continue to occur (the equilibrium is dynamic, not stopped) and the system must be closed so nothing enters or leaves [1]
⚠ If you missed marks here: Two answers are penalised almost every year: "the reaction has stopped" (it has not — both directions continue at equal rates) and "the amounts of reactants and products are equal" (they are constant, not equal). Constant is the key word.
(b)[5]
(i) A sealed tube of the equilibrium mixture in reaction A is placed in iced water. Predict and explain the colour change observed. [2]
(ii) The pressure on the equilibrium mixture in reaction A is increased at constant temperature. Predict and explain the effect on the position of equilibrium. [2]
(iii) Explain why increasing the pressure has no effect on the position of equilibrium in reaction B. [1]
Model Answer -- 4(b)
The mixture becomes paler / pale yellow [1]
because lowering the temperature shifts the equilibrium in the exothermic direction, which here is the reverse reaction, forming more colourless-to-pale N₂O₄ [1]
Increasing the pressure shifts the equilibrium to the left / towards N₂O₄ [1]
because the left has 1 mole of gas and the right has 2 moles, so the system shifts to the side with fewer moles of gas to oppose the increase in pressure [1]
In reaction B there are 2 moles of gas on the left (H₂ + I₂) and 2 moles of gas on the right (2HI), so there is no side with fewer gas moles and the position of equilibrium does not move [1]
⚠ If you missed marks here: A frequent error is saying cooling shifts the equilibrium "towards the endothermic side" — it is the opposite: cooling favours the exothermic direction because that releases heat to oppose the cooling. In (iii), do not answer "pressure never affects liquids" — reaction B is entirely gaseous; the reason is the equal number of gas moles.
(c)[3]
A pink solution of the cobalt equilibrium (reaction C) is prepared in a test tube.
(i) Concentrated hydrochloric acid is added dropwise. Predict the colour change and explain it. [2]
(ii) State what would be seen if water is then added to the tube. [1]
Model Answer -- 4(c)
The solution turns blue [1]
Adding concentrated hydrochloric acid increases the concentration of chloride ions, so the equilibrium shifts to the left to remove the added Cl⁻, forming more of the blue CoCl₄²⁻ ion [1]
Adding water shifts the equilibrium back to the right and the solution turns pink again [1]
⚠ If you missed marks here: Candidates often say the acid "reacts with the cobalt to make a new compound" — no new reaction occurs, the position of an existing equilibrium moves. Also do not answer "it turns green": the two species here are pink and blue, and the intermediate purple shade is a mixture, not a third compound.
Question 5 -- The Contact Process
Total: 10 marks
A sulfuric acid plant near Auckland manufactures sulfuric acid by the Contact process. The key reversible step is 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) and the forward reaction is exothermic. The plant operates at 450 °C and 2 atm with a vanadium(V) oxide catalyst.
(a)[4]
(i) Write the equation for the first stage, in which sulfur is burned in air to make sulfur dioxide. [1]
(ii) Name the catalyst used in the Contact process and give its formula. [1]
(iii) In the final stage the sulfur trioxide is absorbed in concentrated sulfuric acid to form oleum, H₂S₂O₇, which is then diluted with water to give sulfuric acid. Write the equation for the reaction of oleum with water. [1]
(iv) Suggest why sulfur trioxide is not added directly to water in the plant. [1]
Model Answer -- 5(a)
S(s) + O₂(g) → SO₂(g) [1]
Vanadium(V) oxide, V₂O₅ [1]
H₂S₂O₇(l) + H₂O(l) → 2H₂SO₄(l) [1]
The direct reaction of SO₃ with water is violently exothermic and produces a dense, corrosive mist of sulfuric acid that is difficult and dangerous to condense [1]
⚠ If you missed marks here: Writing the catalyst as VO₅ or V₅O₂ loses the mark — vanadium(V) oxide is V₂O₅. The oleum equation is often left unbalanced as H₂S₂O₇ + H₂O → H₂SO₄; two molecules of sulfuric acid are formed. In (iv), "because water is cheaper" is not a reason.
(b)[6]
(i) Use the graph to state the yield of sulfur trioxide at 300 °C, and explain in terms of Le Chatelier's principle why a lower temperature gives a higher yield. [2]
(ii) Explain why the plant nevertheless operates at 450 °C rather than 300 °C. [2]
(iii) Increasing the pressure would increase the yield, yet the plant uses only 2 atm. Explain why the pressure is kept so low. [2]
Model Answer -- 5(b)
Yield at 300 °C is about 99% (accept 98–100%) [1]
The forward reaction is exothermic, so lowering the temperature shifts the equilibrium to the right to release heat, giving a higher yield of SO₃ [1]
At 300 °C the rate of reaction would be too slow, because fewer particles have energy ≥ the activation energy [1]
450 °C is a compromise: a slightly lower yield of about 96% but reached quickly, so more SO₃ is made per hour and the process is economic [1]
The yield at 2 atm is already about 96%, so higher pressure would give very little extra product [1]
and high-pressure plant is expensive to build and needs a large amount of energy to run, so the extra cost is not justified [1]
⚠ If you missed marks here: The classic error is claiming 450 °C gives the highest yield — it does not; it gives an acceptable yield fast. Another is saying the catalyst increases the yield: vanadium(V) oxide only speeds up the attainment of equilibrium. For pressure, "2 atm is used because SO₃ is a gas" scores nothing; the argument is that yield is already high so the extra plant cost is not worth it.
Question 6 -- Redox by Electron Transfer
Total: 12 marks
In a school in Glasgow, a strip of clean zinc is placed in blue copper(II) sulfate solution. The blue colour fades, a brown solid coats the zinc and the tube becomes warm. Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)
(a)[5]
(i) Define oxidation and reduction in terms of electrons. [2]
(ii) Write the ionic half equation for the oxidation and the ionic half equation for the reduction taking place in this experiment. [2]
(iii) Write the overall ionic equation for the reaction, omitting spectator ions, and name the spectator ion. [1]
Model Answer -- 6(a)
Oxidation is the loss of electrons [1]
Reduction is the gain of electrons [1]
Oxidation: Zn → Zn²⁺ + 2e⁻ [1]
Reduction: Cu²⁺ + 2e⁻ → Cu [1]
Overall: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s); the spectator ion is the sulfate ion, SO₄²⁻ [1]
⚠ If you missed marks here: Half equations must balance for charge as well as atoms: Zn → Zn²⁺ + e⁻ is wrong because two electrons are lost. Defining oxidation as "gain of oxygen" is accepted at Core but not for this electron-transfer question. Including sulfate in the overall ionic equation loses the mark since it is a spectator.
(b)[4]
(i) Identify the oxidising agent and the reducing agent in the zinc and copper(II) sulfate reaction, and justify each choice. [2]
(ii) When chlorine gas is bubbled through colourless potassium bromide solution, the solution turns orange. Write the ionic equation and state which species is the oxidising agent. [2]
Model Answer -- 6(b)
Oxidising agent is Cu²⁺ because it accepts electrons and so oxidises the zinc [1]
Reducing agent is Zn because it donates electrons and so reduces the copper(II) ions [1]
Cl₂(aq) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(aq) [1]
Chlorine is the oxidising agent: it gains electrons and oxidises bromide ions to bromine, which gives the orange colour [1]
⚠ If you missed marks here: The two agents are constantly swapped: the species that is itself reduced is the oxidising agent. Answering "zinc is oxidised so zinc is the oxidising agent" is the classic trap. In (ii), writing Cl₂ + Br⁻ → Cl⁻ + Br₂ is unbalanced — two bromide ions are needed.
(c)[3]
(i) Deduce the oxidation number of the underlined element in each of: manganese in MnO₄⁻, chromium in Cr₂O₇²⁻, and nitrogen in NH₃. [2]
(ii) Explain what the Roman numeral in the name iron(III) chloride tells you, and give the formula of the compound. [1]
Model Answer -- 6(c)
Mn in MnO₄⁻ is +7 and Cr in Cr₂O₇²⁻ is +6 [1]
N in NH₃ is −3 [1]
The Roman numeral gives the oxidation number of the iron, +3; the formula is FeCl₃ [1]
⚠ If you missed marks here: For Cr₂O₇²⁻ remember there are TWO chromium atoms: 2x + 7(−2) = −2 gives x = +6, not +12. In NH₃ hydrogen is +1 and nitrogen is negative, so −3, not +3 — sign errors here are very common. Writing iron(III) chloride as FeCl₂ contradicts the Roman numeral.
Question 7 -- Physical and Chemical Change, Photochemical Reactions
Total: 10 marks
A museum conservator in Auckland studies old photographic plates coated with silver bromide, which darken when exposed to light. Her students first revise the difference between physical and chemical changes.
(a)[4]
For each change below, state whether it is a physical change or a chemical change and give one reason for your choice.
(i) Solid iodine warmed in a test tube forms a purple vapour which re-forms grey crystals on the cool upper glass. [2]
(ii) Blue hydrated copper(II) sulfate crystals are heated and turn to a white powder; adding water turns them blue again with a hiss and a rise in temperature. [2]
Model Answer -- 7(a)
(i) Physical change [1]
because no new substance is formed — it is still iodine, only the state has changed, and the change is easily reversed [1]
(ii) Chemical change [1]
because a new substance (anhydrous copper(II) sulfate) is formed, shown by the colour change and the energy change on adding water [1]
⚠ If you missed marks here: "It is reversible so it must be physical" is not a safe rule — the hydration of copper(II) sulfate is reversible but is still a chemical change; the test is whether a new substance with new chemical properties is made. Do not call the iodine change chemical just because a colour appears: purple iodine vapour is still iodine.
(b)[3]
A student burns 2.40 g of magnesium ribbon in a crucible and obtains 4.00 g of white magnesium oxide.
(i) Write the balanced equation for the reaction. [1]
(ii) Explain how the increase in mass is consistent with the law of conservation of mass. [2]
Model Answer -- 7(b)
2Mg(s) + O₂(g) → 2MgO(s) [1]
The magnesium combines with oxygen from the air, and the mass of oxygen gained is 4.00 − 2.40 = 1.60 g [1]
No atoms are created or destroyed: the total mass of magnesium plus oxygen used equals the mass of magnesium oxide formed, so mass is conserved once the gas is counted [1]
⚠ If you missed marks here: Do not write Mg + O → MgO — oxygen in the air is the diatomic molecule O₂, so the equation needs 2Mg. Answers such as "mass increases because heat is added" score zero; the extra 1.60 g is oxygen atoms from the air, and mass is only apparently gained because the gaseous reactant was not weighed at the start.
(c)[3]
(i) State what is meant by a photochemical reaction and write the equation for photosynthesis, naming the energy source and the substance that absorbs it. [2]
(ii) Explain, with an equation, why photographic film containing silver bromide must be handled in a darkroom. [1]
Model Answer -- 7(c)
A photochemical reaction is one in which light energy is absorbed and causes (or drives) a chemical change [1]
6CO₂ + 6H₂O → C₆H₂₂O₆ + 6O₂; the energy source is sunlight and it is absorbed by chlorophyll [1]
Light decomposes silver bromide to silver, which darkens the film: 2AgBr → 2Ag + Br₂, so any stray light would blacken the film and ruin the image [1]
⚠ If you missed marks here: Photosynthesis is endothermic and needs 6CO₂ and 6H₂O — an unbalanced version, or naming the absorber as "the leaf" rather than chlorophyll, loses the mark. For the film, saying "the silver bromide melts" or "reacts with air" is wrong: it is decomposed by light into grey silver metal and bromine.
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