← Topic 6
⚡ Challenge Paper Preparation

Challenge Prep: Chemical Reactions

IGCSE Chemistry 0620 — Topic 6

Topic 6 looks generous — four short subtopics, plenty of recall marks. Then the challenge paper arrives and asks you to read a gradient rather than a plateau, to predict which way an equilibrium shifts when a plant is heated rather than cooled, and to say whether carbon monoxide is the oxidising agent or the substance oxidised. This guide hunts down every one of those traps: rate versus amount, catalysts versus yield, dynamic versus finished, and agent versus victim in redox. Work through it and Topic 6 turns from a minefield into a scoring opportunity.

⚠️ Common Traps & Misconceptions

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Twelve traps that cost students marks on Topic 6 questions. Every one of them appears on challenge papers regularly.

⚠️ TRAP
Trap 1: Confusing the RATE of a reaction with the AMOUNT of product
The TrapStudents see a steeper curve and write "so more gas is produced". Or they double the concentration of an acid that is already in excess and claim the final volume of gas will double.
The TruthRate = how fast, shown by the gradient. Amount = how much, shown by the height of the plateau. Rate factors (concentration, temperature, surface area, catalyst, pressure) change only the gradient. The plateau height is decided solely by the moles of the limiting reactant.
Why It MattersAlmost every challenge-paper rate question includes a sketch-the-second-curve instruction. Draw it steeper AND finishing higher when only a rate factor changed, and you lose the mark even though the shape looks impressive.
Example Question"0.10 g of magnesium reacts with excess 1.0 mol/dm3 HCl. Sketch the curve obtained if the experiment is repeated with 2.0 mol/dm3 HCl, the same mass of magnesium."
⚠️ TRAP
Trap 2: Misreading what a flat section of a rate graph means
The Trap"The graph is horizontal, so the reaction is going at a slow steady rate." Or: "the graph flattens because the reaction has run out of energy."
The TruthA horizontal line means zero gradient, which means zero rate — the reaction has stopped because one reactant has been completely used up. The curve flattens gradually before that because the concentration of reactants is falling, so collisions become less frequent.
Why It MattersExaminers award separate marks for "the rate decreases because the concentration of acid decreases" and "the reaction has stopped because all the magnesium has reacted". Saying "it slows down and stops" without naming the cause scores neither.
Example Question"Explain the shape of the curve between 0 and 60 s, and explain why it becomes horizontal after 60 s."
⚠️ TRAP
Trap 3: Thinking a catalyst increases the yield or changes ΔH
The Trap"Adding an iron catalyst increases the yield of ammonia." "The catalyst makes the reaction more exothermic." "The catalyst shifts the equilibrium to the right."
The TruthA catalyst provides an alternative pathway with a lower activation energy. That is the whole story. It does NOT change ΔH, does NOT change the position of equilibrium, and does NOT change the amount of product. In an equilibrium it speeds up the forward and backward reactions equally, so it changes only the time taken to reach equilibrium.
Why It MattersThis is tested in all three sub-topics: on a rate graph (same plateau), on an energy profile (same start and end levels), and in the Haber/Contact process (same yield, reached sooner). One misconception, three lost marks.
Example Question"State the effect of adding an iron catalyst on (i) the rate of reaction and (ii) the equilibrium yield of ammonia."
⚠️ TRAP
Trap 4: Getting the Le Chatelier temperature direction backwards
The Trap"The forward reaction is exothermic, so heating it gives more product." Students reason that heating "gives the reaction more energy so it goes forwards faster", and confuse rate with position.
The TruthHeating shifts the equilibrium in the endothermic direction, because the system opposes the change by absorbing the heat you supplied. If the forward reaction is exothermic, the backward reaction is endothermic, so heating shifts it BACKWARDS and lowers the yield. Cooling shifts it in the exothermic direction. Heating does increase the rate — but rate and position are different questions.
Why It MattersEvery Haber and Contact process question turns on this. If you say heating raises the yield, the whole "why is 450 °C a compromise" answer collapses, because the compromise only exists because heating LOWERS the yield.
Example Question"N2 + 3H2 ⇌ 2NH3, forward reaction exothermic. State and explain the effect of increasing the temperature on the equilibrium yield of ammonia."
⚠️ TRAP
Trap 5: Thinking "reversible" means the reaction stops at equilibrium
The Trap"At equilibrium the reaction has finished." "Nothing is happening any more." "The concentrations are constant because the particles have stopped reacting."
The TruthEquilibrium is dynamic. Both reactions continue at full speed — reactants are still becoming products and products are still becoming reactants — but the rate of the forward reaction equals the rate of the backward reaction, so the concentrations do not change. Nothing appears to happen precisely because so much is happening in both directions at once.
Why It MattersThe mark scheme phrase is almost always "rate of forward reaction = rate of backward reaction" plus "concentrations remain constant". "The reaction stops" is an outright contradiction of the word "dynamic" and scores zero.
Example Question"A sealed flask contains an equilibrium mixture and the colour no longer changes. Explain what is happening to the molecules in the flask."
⚠️ TRAP
Trap 6: Believing equilibrium means equal amounts of reactants and products
The Trap"At equilibrium there are equal concentrations of reactants and products, a 50:50 mixture."
The TruthEquilibrium means equal rates, not equal amounts. The position of equilibrium may lie far to the right (mostly products, like the Contact process at about 96% SO3) or far to the left (mostly reactants). Changing conditions moves that position without ever making the rates unequal at the new equilibrium.
Why It MattersQuestions ask you to describe the "position of equilibrium" and whether it lies to the left or right. If you think equilibrium is always 50:50, the question makes no sense to you.
Example Question"At 450 °C and 2 atm, 96% of the sulfur dioxide is converted. State where the position of equilibrium lies and explain your answer."
⚠️ TRAP
Trap 7: Confusing the oxidising agent with the substance that is oxidised
The TrapIn Fe2O3 + 3CO → 2Fe + 3CO2, students write "CO is oxidised, so CO is the oxidising agent." The word "oxidised" is right there in the sentence, so it feels correct.
The TruthAn agent does the job to something else and has the opposite done to itself. The oxidising agent is REDUCED (it gains electrons). The reducing agent is OXIDISED (it loses electrons). So CO is oxidised ⇒ CO is the reducing agent; Fe2O3 is reduced ⇒ it is the oxidising agent.
Why It MattersThis single confusion is the most common redox error in the whole IGCSE. It also corrupts the two colour tests: acidified manganate(VII) is an oxidising agent, so it detects a REDUCING agent; potassium iodide is a reducing agent, so it detects an OXIDISING agent.
Example Question"In the reaction 2Fe3+ + 2I− → 2Fe2+ + I2, identify the oxidising agent and the reducing agent, giving reasons."
⚠️ TRAP
Trap 8: Oxidation number sign and direction errors
The TrapThree separate slips: (1) writing 2+ instead of +2; (2) thinking Fe2+ → Fe3+ is reduction "because it gained a positive charge"; (3) thinking −2 → 0 is a decrease "because 0 is smaller than 2".
The TruthOxidation numbers are written sign first (+2, −1, +7); ionic charges are written the other way round (Fe2+). On the number line, −2 < −1 < 0 < +1 < +2 < +3. So −2 → 0 is an increase (oxidation), and +2 → +3 is an increase (oxidation — the ion has LOST a negative electron).
Why It MattersChallenge papers deliberately use changes that cross zero or go from negative to less negative. Sketch a quick number line in the margin and mark the two values on it — the direction becomes obvious.
Example Question"In 2Mg + CO2 → 2MgO + C, deduce the oxidation number change of carbon and state whether it is oxidised or reduced."
⚠️ TRAP
Trap 9: The incomplete temperature explanation
The Trap"Increasing the temperature makes the particles move faster so they collide more, so the rate increases." Full stop. This scores about one mark out of three.
The TruthYou need both effects, and the second is the important one: (1) particles gain kinetic energy and move faster, so they collide more frequently; (2) a greater proportion of particles now have energy equal to or greater than the activation energy, so a much larger fraction of collisions are successful. Effect (2) is why a 10 °C rise can double the rate — collision frequency alone rises by only a few percent.
Why It MattersChallenge papers ask "explain fully" or allocate 3 marks. The activation energy point is the one most students omit, and it is usually worth two of the three.
Example Question"Explain, in terms of collision theory, why the rate of reaction increases when the temperature is raised from 20 °C to 40 °C. [3]"
⚠️ TRAP
Trap 10: Saying "more particles" instead of "more particles per unit volume"
The Trap"A more concentrated solution has more particles, so there are more collisions." Also: "the particles move faster in a concentrated solution."
The TruthConcentration is about crowding: more particles in the same volume, so the particles are closer together and collide more frequently. Concentration does NOT change the speed of the particles and does NOT change the activation energy. A bigger beaker of the same solution also has "more particles" but exactly the same rate.
Why It MattersMark schemes specifically require "per unit volume" or "closer together". The bare phrase "more particles" is listed as insufficient on Cambridge mark schemes.
Example Question"Explain why increasing the concentration of hydrochloric acid increases the rate of reaction with calcium carbonate. [2]"
⚠️ TRAP
Trap 11: Using reversibility to decide physical versus chemical change
The Trap"Heating hydrated copper(II) sulfate is a physical change because adding water turns it blue again." Or: "burning is chemical because it cannot be reversed."
The TruthThe only test is: has a new substance with a different chemical formula been formed? CuSO4·5H2O ⇌ CuSO4 + 5H2O changes the formula, so it is a chemical change that happens to be reversible. Melting and dissolving are reversible physical changes. Reversibility tells you nothing.
Why It MattersThe hydrated/anhydrous salts sit in the syllabus twice — once as reversible reactions and once as tests for water — so examiners love asking students to classify them. A reversibility-based answer is marked wrong even when the classification happens to be right.
Example Question"Blue crystals are heated and turn white; adding water turns them blue again. State, with a reason, whether this is a physical or a chemical change."
⚠️ TRAP
Trap 12: Applying the pressure rule without counting moles of gas
The Trap"Increasing the pressure always shifts the equilibrium to the right." Or students count solids and liquids as part of the mole comparison, or forget to check whether any gas is involved at all.
The TruthCount only the moles of GAS on each side. Increasing pressure shifts the equilibrium towards the side with fewer moles of gas. If both sides have the same number of moles of gas (as in H2 + I2 ⇌ 2HI), pressure has no effect at all on the position. If no gases are present, pressure is irrelevant.
Why It MattersChallenge papers include at least one "equal moles" equilibrium precisely to catch students applying the rule blindly. Always write the mole counts above each side before answering.
Example Question"For H2(g) + I2(g) ⇌ 2HI(g), state and explain the effect of increasing the pressure on the position of equilibrium."

🧩 Multi-Step Reasoning Walkthroughs

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Six challenging questions broken down step by step. Try each step yourself before revealing the next.

Walkthrough 1 — Sketching a Second Rate Curve0.24 g of magnesium ribbon reacts with 50 cm3 of excess 1.0 mol/dm3 hydrochloric acid at 20 °C. The curve of volume of hydrogen against time plateaus at 240 cm3 after 90 s. On the same axes, sketch and justify the curve obtained if the experiment is repeated with 0.24 g of magnesium POWDER and 50 cm3 of 2.0 mol/dm3 acid at 30 °C.
1

Separate the variables

Three things changed: surface area (ribbon → powder), concentration (1.0 → 2.0 mol/dm3) and temperature (20 → 30 °C). One thing did NOT change: the mass of magnesium, still 0.24 g. Also note the acid is stated to be in excess in both runs.

2

All three changes speed it up

Powder gives a far larger surface area, so more magnesium atoms are exposed. Doubling the acid concentration puts more H+ ions per unit volume. Raising the temperature gives particles more kinetic energy and, crucially, a greater proportion with energy ≥ Ea. All three increase the frequency of successful collisions, so the new curve is much steeper at the start.

3

Same limiting reactant, same amount of product

Magnesium is the limiting reactant (the acid is in excess). Its mass is unchanged at 0.24 g, so the same number of moles reacts and the same number of moles of hydrogen is produced. The new curve must therefore level off at exactly 240 cm3 — the same height as the original.

4

Shape rules

Start at the origin (both curves must). Rise more steeply than the original. Curve over and become horizontal earlier — perhaps around 30–40 s. Then run horizontally at 240 cm3, meeting and lying exactly on top of the original curve's plateau. Never let the new curve cross above the 240 cm3 line.

Final AnswerA curve that is steeper than the original, becomes horizontal sooner, and plateaus at the same volume of 240 cm3. Justification: powder, higher concentration and higher temperature all increase the frequency of successful collisions (and temperature also increases the proportion of particles with energy ≥ Ea), so the rate rises; but the mass of magnesium, the limiting reactant, is unchanged, so the total volume of hydrogen is unchanged.
Examiner's NoteThe commonest error by far is drawing the new curve finishing higher. Examiners often award one mark for "steeper" and one mark for "same final volume", so a curve that is steeper but too high scores 1 out of 2. Second error: starting the new curve above the origin. Third: drawing the steeper curve crossing the original one twice.
Walkthrough 2 — The Full Haber Compromise AnswerThe Haber process, N2(g) + 3H2(g) ⇌ 2NH3(g), has an exothermic forward reaction and is run at 450 °C, 200 atm, with an iron catalyst. Explain the choice of each of the three conditions, and state the effect of each on both the rate and the yield. [6]
1

Two pieces of information, extracted

Write "exo →" and "endo ←" above the arrows. Then count the moles of gas: left = 1 + 3 = 4; right = 2. Every part of the answer follows from these two facts. Do this before writing a single sentence.

2

Two competing effects

Yield: the forward reaction is exothermic, so raising the temperature shifts the equilibrium in the endothermic direction — backwards. Higher temperature therefore lowers the yield; a low temperature would give a high yield.
Rate: a low temperature means very few particles have energy ≥ Ea, so the rate would be far too slow and equilibrium would take far too long to reach.
Conclusion: 450 °C is a compromise — a moderate yield obtained in an acceptable time.

3

Yield and rate both improve

Yield: 4 moles of gas on the left, 2 on the right. Increasing the pressure shifts the equilibrium towards the side with fewer moles of gas, i.e. to the right, so the yield increases.
Rate: higher pressure means more gas particles per unit volume, so collisions are more frequent and the rate also increases.
Limit: pressures much above 200 atm are avoided because the plant becomes prohibitively expensive and dangerous. This is an economic and safety limit, not a chemical one.

4

Rate yes, yield no

The iron catalyst provides an alternative pathway with a lower activation energy, so more collisions are successful and the rate increases. It speeds up the forward and backward reactions equally, so the position of equilibrium and the yield are completely unchanged — equilibrium is simply reached sooner. Say both halves explicitly.

5

Removal and recycling

The ammonia is cooled and condensed out as a liquid while nitrogen and hydrogen remain gaseous. Removing a product shifts the equilibrium to the right, so more ammonia forms; the unreacted gases are recycled back into the reactor, raising the overall conversion well above the single-pass equilibrium yield.

Final Answer450 °C: a lower temperature would give a higher yield because the forward reaction is exothermic, but the rate would be uneconomically slow — so 450 °C is a compromise between yield and rate. 200 atm: increases the yield (equilibrium shifts to the side with fewer moles of gas, 4 → 2) and increases the rate (more particles per unit volume); higher pressures are too costly and hazardous. Iron catalyst: increases the rate by lowering the activation energy, so equilibrium is reached sooner, but has no effect on the yield.
Examiner's NoteMark schemes for this question hand out marks in pairs: for temperature you need the yield statement AND the rate statement AND the word "compromise". For pressure you need the mole count as the reason, not just "high pressure helps". For the catalyst, "increases rate" alone is half an answer — you must also say the yield is unchanged. Writing "the catalyst increases the yield" typically costs two marks because it also contradicts the compromise reasoning.
Walkthrough 3 — Full Redox AnalysisAnalyse the reaction MnO4− + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O. Deduce all oxidation number changes, identify what is oxidised and reduced, name the oxidising and reducing agents, and state the colour change observed.
1

Work through systematically

In MnO4−: oxygen is −2 each, so x + 4(−2) = −1, giving Mn = +7.
Fe2+ is a monatomic ion, so Fe = +2.
H+ is a monatomic ion, so H = +1.

2

And compare

Mn2+: Mn = +2. Fe3+: Fe = +3. In H2O: H = +1, O = −2 — both unchanged from the left-hand side, so hydrogen and oxygen take no part in the electron transfer. Only manganese and iron have changed.

3

Direction of change

Mn: +7 → +2. That is a decrease of 5, so manganese is reduced (each Mn gains 5 electrons).
Fe: +2 → +3. That is an increase of 1, so iron is oxidised (each Fe loses 1 electron).
Check the bookkeeping: 1 Mn gaining 5 electrons balances 5 Fe each losing 1 — which is exactly why the coefficient 5 appears in the equation.

4

Agent = opposite of what happens to it

MnO4− is reduced, therefore MnO4− (the manganate(VII) ion) is the oxidising agent. Fe2+ is oxidised, therefore Fe2+ is the reducing agent. If you find yourself writing "Fe2+ is oxidised so it is the oxidising agent", stop and reread.

5

Purple disappears

MnO4− is deep purple; Mn2+ is colourless (very pale pink). The solution therefore changes from purple to colourless. Note also why the H+ ions appear in the equation — the reduction all the way to Mn2+ requires acidic conditions, which is why the reagent must be acidified potassium manganate(VII).

Final AnswerMn is reduced from +7 to +2; Fe is oxidised from +2 to +3. The oxidising agent is MnO4− (it is reduced and gains electrons); the reducing agent is Fe2+ (it is oxidised and loses electrons). Observation: the purple solution turns colourless. Hydrogen and oxygen do not change oxidation number.
Examiner's NoteQuote the actual numbers — "Mn is reduced from +7 to +2" scores where "Mn is reduced" alone often does not. Do not forget to check H and O; stating explicitly that they are unchanged shows you assigned every atom. And write +7, not 7+.
Walkthrough 4 — Reading a Rate Graph PreciselyA gas syringe records the volume of CO2 from marble chips and acid: 0 s = 0 cm3; 20 s = 44 cm3; 40 s = 68 cm3; 60 s = 78 cm3; 80 s = 80 cm3; 100 s = 80 cm3. (a) Calculate the average rate over the first 20 s and over the whole reaction. (b) Explain why the rate decreases. (c) State when the reaction finished and how you know. (d) Explain what changed if a repeat run plateaus at 40 cm3.
1

Rate = change in volume ÷ time

First 20 s: 44 cm3 ÷ 20 s = 2.2 cm3/s.
Whole reaction: the reaction is over at 80 s, and 80 cm3 was produced, so 80 ÷ 80 = 1.0 cm3/s. Use 80 s, not 100 s — nothing happened in the last 20 s.

2

Use collision theory, not vague language

As the reaction proceeds the acid is used up, so the concentration of acid falls and there are fewer H+ ions per unit volume. The marble chips also get smaller, reducing the surface area available. Both reduce the frequency of successful collisions, so the rate falls and the gradient gets shallower. Do not write "the reaction gets tired" or "it runs out of energy".

3

Read the data, not the graph shape

The volume is 80 cm3 at 80 s and still 80 cm3 at 100 s. Since no further gas is produced, the reaction finished at 80 s. It finished because one of the reactants (the limiting reactant) was completely used up, so the rate is now zero.

4

Half the gas means half the limiting reactant

A plateau of 40 cm3 instead of 80 cm3 means only half as many moles of CO2 were produced. That can only be caused by using half the amount (moles) of the limiting reactant — for example half the mass of marble, or half the volume/concentration of acid if the acid was limiting. It CANNOT be caused by temperature, surface area or a catalyst, because those change only the gradient.

Final Answer(a) 2.2 cm3/s for the first 20 s; 1.0 cm3/s overall. (b) The concentration of acid and the surface area of the chips both decrease as reactants are used up, so the frequency of successful collisions falls. (c) The reaction finished at 80 s, because no more gas was produced after that time — the limiting reactant had been completely used up. (d) A lower plateau means a smaller amount (fewer moles) of the limiting reactant was used; it is not a rate effect.
Examiner's NotePart (a) trips students who divide 80 cm3 by 100 s. Part (c) is a "state and explain" — give the time AND the evidence AND the chemical reason. Part (d) is the discriminating question on most papers: candidates who have not separated gradient from plateau in their heads reliably answer "the temperature was lower", which is worth nothing.
Walkthrough 5 — Predicting Equilibrium Shifts Under PressureFor 2SO2(g) + O2(g) ⇌ 2SO3(g), forward reaction exothermic, predict the effect on the position of equilibrium and on the rate of: (a) increasing the temperature; (b) increasing the pressure; (c) adding V2O5; (d) removing SO3 as it forms; (e) adding argon at constant volume.
1

Annotate

Forward = exothermic, backward = endothermic. Moles of gas: left = 2 + 1 = 3; right = 2. Keep the two questions separate in your head: "which way does the position move?" and "how fast does it get there?" are different questions with different answers.

2

Temperature and pressure

(a) Higher temperature: position shifts in the endothermic direction, i.e. to the left, so the yield of SO3 decreases. The rate increases (more particles with energy ≥ Ea). Two opposite-sounding answers, both correct.
(b) Higher pressure: position shifts towards fewer moles of gas, i.e. to the right (3 → 2), so the yield increases. The rate also increases (more particles per unit volume).

3

Catalyst and product removal

(c) V2O5: position unchanged — it speeds up forward and backward reactions equally. The rate increases, so equilibrium is reached sooner. Yield identical.
(d) Removing SO3: the system opposes the removal by making more, so the position shifts to the right and more SO2 is converted. (Strictly the system never rests at equilibrium while removal continues, which is exactly why continuous removal is so effective industrially.)

4

An inert gas at constant volume changes nothing

Argon takes no part in the reaction. At constant volume the concentrations of SO2, O2 and SO3 are all unchanged, so the collision frequency between reacting particles is unchanged. The position of equilibrium does not move and the rate does not change — even though the total pressure gauge reads higher. The total pressure is a red herring; what matters is the concentration of the reacting gases.

Final Answer(a) Position left, yield down; rate up. (b) Position right, yield up; rate up. (c) Position unchanged, yield unchanged; rate up. (d) Position right, more SO3 formed overall. (e) No change to position or rate — argon is inert and the concentrations of the reacting gases are unaltered at constant volume.
Examiner's NoteParts (a) and (c) are where marks are lost. In (a) candidates who confuse rate with position write "the yield increases because the reaction is faster" — a double error. In (c) they write "the catalyst increases the yield". Part (e) appears on higher-demand papers to separate students who understand why pressure matters (concentration of gas particles) from those who have simply memorised "high pressure shifts to fewer moles".
Walkthrough 6 — Designing a Fair Rate ExperimentA student wants to investigate the effect of concentration on the rate of the reaction between sodium thiosulfate and hydrochloric acid using the disappearing-cross method. Describe the method, identify the independent, dependent and control variables, explain how the results are processed, and evaluate one weakness with an improvement.
1

Practical detail earns marks

Measure 50 cm3 of sodium thiosulfate solution into a conical flask using a measuring cylinder. Place the flask on a printed paper cross. Add 5 cm3 of dilute hydrochloric acid and start the stopclock at the moment of mixing. Look down vertically through the solution and stop the clock when the cross can no longer be seen, obscured by the pale yellow sulfur precipitate. Repeat with thiosulfate solutions of different concentrations, made by diluting the stock with measured volumes of distilled water so that the total volume stays constant.

2

Name them precisely

Independent: the concentration of the sodium thiosulfate solution.
Dependent: the time taken for the cross to disappear.
Control: total volume of solution; volume and concentration of hydrochloric acid; temperature; the same paper cross; the same flask (so the same depth of liquid); the same observer judging disappearance.

3

Rate is proportional to 1/time

A shorter time means a faster reaction, so calculate 1/t for each run and use it as a measure of rate. Plot 1/t (y-axis) against concentration (x-axis). A straight line through the origin shows that the rate is directly proportional to the concentration. Plotting t against concentration instead gives a curve that is much harder to interpret.

4

One weakness, one improvement, both specific

Weakness: deciding exactly when the cross has "disappeared" is subjective — different observers, and even the same observer on different runs, will judge the end point differently, reducing the reliability of the results.
Improvement: use a light sensor or colorimeter with a data logger and record the time for the light transmission to fall to a fixed value. This removes human judgement entirely and gives a reproducible end point. A further improvement is to use a water bath to keep the temperature genuinely constant, since the reaction itself changes the temperature slightly.

Final AnswerMix measured volumes of thiosulfate and acid over a paper cross, timing until the cross disappears; vary the thiosulfate concentration by dilution while keeping the total volume constant. Independent variable = thiosulfate concentration; dependent = time for the cross to disappear; controls = total volume, acid volume and concentration, temperature, same cross, same flask, same observer. Plot 1/t against concentration; a straight line through the origin shows proportionality. Weakness: the end point is subjective — improve by using a light sensor and data logger.
Examiner's NoteTwo marks are routinely thrown away here. First, forgetting to keep the total volume constant when diluting — without water added, you change both concentration and volume and it is no longer a fair test. Second, offering "repeat the experiment" as the improvement; repeats improve reliability but do nothing about a systematically subjective end point. Name the specific instrument.

🔍 Spot the Difference

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Pairs of questions that look nearly identical but have different answers. Spot the key distinction.

Question A
0.1 g Mg + excess 1.0 mol/dm3 HCl is repeated with 2.0 mol/dm3 HCl. What happens to the final volume of hydrogen?
No change. The magnesium is limiting and its mass is unchanged, so the same moles of H2 form. Only the gradient is steeper.
Question B
0.1 g Mg + excess 1.0 mol/dm3 HCl is repeated with 0.2 g Mg. What happens to the final volume of hydrogen?
It doubles. Twice the moles of the limiting reactant means twice the moles of hydrogen. The plateau is twice as high.
Key DifferenceIn A a rate factor changed → gradient changes, plateau does not. In B the amount of limiting reactant changed → plateau changes. Ask yourself every time: "did the moles of the limiting reactant change?" If no, the plateau cannot move.
Question A
What does the gradient of a volume-time curve tell you?
The rate at that instant. Steepest at t = 0 when the concentration is highest; it decreases as reactants are used up.
Question B
What does the height of the plateau tell you?
The total amount of product formed, fixed by the moles of the limiting reactant. It says nothing at all about how fast the reaction went.
Key DifferenceGradient answers "how fast?". Height answers "how much?". Two curves can have completely different gradients and identical heights, or identical gradients and different heights. Never let one word of your answer slide from one to the other.
Question A
A catalyst is added to N2 + 3H2 ⇌ 2NH3. What happens to the rate?
It increases. The catalyst provides an alternative pathway with a lower activation energy, so a greater proportion of collisions are successful.
Question B
A catalyst is added to N2 + 3H2 ⇌ 2NH3. What happens to the yield?
Nothing. The forward and backward reactions are sped up equally, so the position of equilibrium is unchanged. Equilibrium is simply reached sooner.
Key DifferenceSame catalyst, same reaction, opposite answers — because rate and yield are different quantities. A catalyst is the one factor in the whole of Topic 6 that affects rate and only rate.
Question A
The forward reaction is exothermic. Temperature is increased. Which way does the equilibrium shift?
Backwards (to the left) — the endothermic direction — because the system absorbs the extra heat. Yield of product falls.
Question B
The forward reaction is endothermic. Temperature is increased. Which way does the equilibrium shift?
Forwards (to the right) — because now the forward direction is the endothermic one. Yield of product rises.
Key DifferenceThe rule is never "heating shifts it right" or "heating shifts it left". The rule is heating always shifts it towards the ENDOTHERMIC direction. Which direction that is depends entirely on the reaction. Write "endo" on the correct arrow before you answer.
Question A
Increase the pressure on N2(g) + 3H2(g) ⇌ 2NH3(g).
Shifts right. 4 moles of gas on the left, 2 on the right; the system moves to the side with fewer gas molecules.
Question B
Increase the pressure on H2(g) + I2(g) ⇌ 2HI(g).
No shift. 2 moles of gas on each side, so neither direction relieves the pressure. The rate increases but the position does not move.
Key DifferenceAlways count the moles of gas on each side before answering. "Increasing pressure shifts equilibrium right" is not a rule — it is only true when the right-hand side has fewer moles of gas. Equal moles = no effect.
Question A
In Fe2O3 + 3CO → 2Fe + 3CO2, which substance is oxidised?
CO — it gains oxygen and carbon goes from +2 to +4, an increase.
Question B
In Fe2O3 + 3CO → 2Fe + 3CO2, which substance is the oxidising agent?
Fe2O3 — it is the substance that is reduced, so it is the one doing the oxidising.
Key Difference"Oxidised" and "oxidising agent" refer to different substances in the same equation. The oxidising agent is always the one that is REDUCED. Read the question wording letter by letter — those two extra syllables reverse the answer.
Question A
Acidified potassium manganate(VII) turns from purple to colourless. What has been added?
A reducing agent. The Mn has been reduced from +7 to +2, so something donated electrons to it.
Question B
Potassium iodide solution turns from colourless to brown. What has been added?
An oxidising agent. Iodide has been oxidised from −1 to 0, forming brown iodine, so something accepted those electrons.
Key DifferenceEach reagent detects the opposite of what it is. Manganate(VII) is an oxidising agent and detects reducing agents. Iodide is a reducing agent and detects oxidising agents. Learn the pair together: "purple disappears, brown appears."
Question A
Fe2+ → Fe3+. Oxidation or reduction?
Oxidation. The oxidation number increases from +2 to +3 because the ion has LOST one electron.
Question B
Mn7+ in MnO4− → Mn2+. Oxidation or reduction?
Reduction. The oxidation number decreases from +7 to +2 because the ion has GAINED five electrons.
Key DifferenceBoth ions stay positive throughout, which is why students freeze. Ignore the fact that it is positive and look only at the direction of movement along the number line: up = oxidation, down = reduction. Gaining electrons always makes the number go down.
Question A
Blue crystals are heated and turn white; adding water turns them blue again. Physical or chemical?
Chemical. CuSO4·5H2O → CuSO4: the formula changes, so a new substance is formed. It just happens to be reversible.
Question B
Blue crystals dissolve in water to give a blue solution; evaporating recovers the crystals. Physical or chemical?
Physical. Dissolving separates the ions but CuSO4·5H2O is recovered unchanged — no new substance.
Key DifferenceSame compound, both reversible, opposite answers. Reversibility is never the test. Ask only: does the chemical formula of the substance change? In A it does; in B it does not.
Question A
At equilibrium, what is happening to the molecules?
Both reactions are still occurring, at equal rates. Reactants become products and products become reactants continuously.
Question B
When a rate curve becomes horizontal, what is happening to the molecules?
The reaction has genuinely stopped. A reactant has been completely used up, so the rate is zero.
Key DifferenceTwo flat lines that mean completely opposite things. A flat rate graph means the reaction has finished. A constant concentration at equilibrium means two reactions are perfectly balanced. Check which situation the question describes before choosing your words.

🔗 Chemical Reactions Concept Map

▼

Click each node to see how the subtopics connect.

⭐ CORE FRAMEWORK 1
Collision theory is the engine that drives all of 6.2
Collision Theory ▶
The rule: particles must collide with energy ≥ Ea AND in the correct orientation for a successful collision.
Rate depends on: the FREQUENCY of successful collisions per second.
Universal answer template: what changed → effect on collision frequency or proportion with E ≥ Ea → effect on successful collisions per second → effect on rate.
The Five Rate Factors ▶
Concentration: more particles per unit volume → more frequent collisions.
Pressure (gases only): the same idea as concentration — particles forced closer together.
Surface area: more of the solid exposed → more sites for collisions. Powder > chips > lump.
Temperature: faster particles (more frequent collisions) AND a greater proportion with E ≥ Ea — the second effect dominates.
Catalyst: alternative pathway with lower Ea → greater proportion of collisions successful. Never changes yield or ΔH.
Measuring and Graphing Rate ▶
Gas syringe: volume of gas / cm3. Curve rises and plateaus. Best for H2 or CO2.
Mass loss: mass / g. Curve falls and levels off. Only for dense gases — useless for hydrogen.
Disappearing cross: rate ∝ 1/t. Subjective end point — improve with a light sensor.
Reading a curve: gradient = rate; plateau height = amount of product; horizontal = reaction stopped.
Enzymes and Photochemical Reactions ▶
Enzymes: biological catalysts (proteins). Optimum temperature about 37 °C; denatured above it, so the rate-temperature curve peaks then collapses.
Photochemical: light supplies the energy instead of heat.
Photosynthesis: 6CO2 + 6H2O → C6H12O6 + 6O2, endothermic, chlorophyll absorbs light.
Photography: 2AgBr → 2Ag + Br2. Light decomposes the silver halide; the silver formed is black, giving a negative.
⭐ CORE FRAMEWORK 2
Equilibrium — where rate ideas meet yield ideas
Reversible Reactions and Dynamic Equilibrium ▶
Symbol: ⇌ means reversible. Forward = left to right; backward = right to left.
Equilibrium: rate forward = rate backward, so concentrations are constant. NOT stopped, NOT necessarily equal amounts.
Three requirements: reversible reaction, closed system, constant temperature.
Energy link: if forward is exothermic by X kJ/mol, backward is endothermic by exactly X kJ/mol.
Hydrated and Anhydrous Salts ▶
Copper: CuSO4·5H2O (blue) ⇌ CuSO4 (white) + 5H2O. Heating is endothermic; adding water is exothermic.
Cobalt: CoCl2·6H2O (pink) ⇌ CoCl2 (blue) + 6H2O.
Tests for water: white anhydrous CuSO4 → blue; blue cobalt(II) chloride paper → pink.
Link to 6.1: these are reversible CHEMICAL changes — the formula changes, so reversibility is irrelevant to classification.
Shifting the Position of Equilibrium ▶
Concentration: add a reactant or remove a product → shifts right. Add a product → shifts left.
Temperature: heat → shifts in the ENDOTHERMIC direction. Cool → shifts in the EXOTHERMIC direction.
Pressure: increase → shifts towards FEWER moles of gas. Equal moles of gas = no effect.
Catalyst: no effect on position or yield — equilibrium is reached sooner only.
Haber and Contact Processes ▶
Haber: N2 + 3H2 ⇌ 2NH3. 450 °C, 200 atm, iron catalyst. 4 → 2 moles of gas. Exothermic forwards.
Contact: 2SO2 + O2 ⇌ 2SO3. 450 °C, 2 atm, V2O5 catalyst. 3 → 2 moles of gas. Exothermic forwards.
Why 450 °C in both: a compromise — lower temperature gives a higher yield but too slow a rate.
Why the pressures differ: Contact already achieves about 96% at 2 atm, so high pressure is not worth the cost; Haber's yield is poor at low pressure, so 200 atm pays for itself.
⭐ CORE FRAMEWORK 3
Redox — follow the electrons and everything else falls out
The Three Definitions ▶
Oxygen: oxidation = gain of oxygen; reduction = loss of oxygen.
Electrons (OIL RIG): Oxidation Is Loss; Reduction Is Gain.
Oxidation number: oxidation = increase; reduction = decrease.
They never disagree. Use whichever the question makes easiest, but the electron definition always works.
Agents — the Great Reversal ▶
Oxidising agent: oxidises something else, so it is itself REDUCED and GAINS electrons. Examples: O2, Cl2, MnO4−, Fe3+.
Reducing agent: reduces something else, so it is itself OXIDISED and LOSES electrons. Examples: C, CO, H2, Zn, Mg, I−.
Test yourself: in Fe2O3 + 3CO, CO is oxidised, so CO is the REDUCING agent.
Oxidation Number Rules ▶
Element = 0 (including O2, Cl2, H2). Simple ion = its charge.
O = −2 except peroxides (−1). H = +1 except metal hydrides (−1).
Sum = 0 in a neutral compound; sum = the charge in a polyatomic ion.
Roman numerals give the oxidation number: iron(III) = +3, manganate(VII) = +7, lead(IV) = +4.
The Two Colour Tests ▶
Acidified potassium manganate(VII): purple → colourless. Detects a REDUCING agent. Mn goes +7 → +2.
Potassium iodide: colourless → brown. Detects an OXIDISING agent. I goes −1 → 0.
Why acidified: H+ ions are needed to reduce MnO4− all the way to colourless Mn2+; without acid you get brown MnO2.
Where Redox Appears Elsewhere ▶
Electrolysis: Red Cat, An Ox — reduction at the cathode, oxidation at the anode.
Metal extraction: Fe2O3 + 3CO → 2Fe + 3CO2 in the blast furnace.
Rusting and its prevention: iron oxidised 0 → +3; sacrificial zinc is oxidised instead.
NOT redox: neutralisation and precipitation — no oxidation number changes at all.

❌ "Why Is This Wrong?" Exercises

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Spot the error in each student's answer. Think before revealing.

Exercise 1: "0.20 g of magnesium reacts with excess dilute HCl. Predict the effect on the volume of hydrogen collected if the concentration of the acid is doubled."
Student's Answer"The reaction will be faster, so twice as much hydrogen will be collected. The curve will be steeper and finish higher."
The FlawThe student has fused rate and amount into one idea. Faster does not mean more. The acid was already in excess, so it was never the factor limiting how much hydrogen could form.
Correct Answer"The volume of hydrogen collected is unchanged. The magnesium is the limiting reactant and its mass is the same, so the same number of moles of hydrogen is produced. The curve is steeper and reaches the plateau sooner, but the plateau is at the same height."
Key RuleRate factors change the gradient. Only a change in the moles of the limiting reactant changes the plateau height.
Exercise 2: "Explain the shape of a volume-time graph that rises steeply, curves over, and becomes horizontal after 90 s."
Student's Answer"The reaction starts fast because the chemicals are fresh, then it gets tired and slows down, and after 90 s it is going very slowly at a constant rate."
The FlawTwo errors. "Gets tired" is not chemistry — it explains nothing in terms of particles. And a horizontal line is not a slow constant rate; a zero gradient means a zero rate.
Correct Answer"The rate is greatest at the start because the concentration of the reactants is highest, giving the greatest frequency of successful collisions. As the reactants are used up, the concentration falls, collisions become less frequent and the rate decreases, so the gradient becomes shallower. After 90 s the graph is horizontal because the reaction has stopped — the limiting reactant has been completely used up."
Key RuleExplain every graph feature with particles: concentration → collision frequency → rate. And zero gradient = zero rate = reaction finished.
Exercise 3: "Explain why increasing the temperature from 20 °C to 40 °C roughly quadruples the rate of reaction. [3]"
Student's Answer"The particles have more energy so they move faster and collide more often, which makes the reaction faster."
The FlawOnly half the explanation, and it is the less important half. Raising the temperature by 20 °C increases the average particle speed by only a few percent — nowhere near enough to quadruple the rate. The student has omitted the activation energy effect entirely.
Correct Answer"The particles gain kinetic energy and move faster, so they collide more frequently [1]. More importantly, a greater proportion of the particles now have energy equal to or greater than the activation energy [1], so a much larger fraction of the collisions are successful. The frequency of successful collisions per second therefore increases greatly and the rate rises [1]."
Key RuleTemperature always needs both points, and the activation energy point is usually worth two of the three marks. Never stop at "they collide more often".
Exercise 4: "N2 + 3H2 ⇌ 2NH3. The forward reaction is exothermic. Explain the effect of increasing the temperature on the yield of ammonia."
Student's Answer"Increasing the temperature gives the particles more energy so the forward reaction goes faster, which means the equilibrium shifts to the right and the yield of ammonia increases."
The FlawThe student has used a rate argument to answer a yield question, and has got the direction exactly backwards. Heating speeds up the forward AND backward reactions, so "the forward reaction goes faster" tells you nothing about which way the balance tips.
Correct Answer"The forward reaction is exothermic, so the backward reaction is endothermic. Increasing the temperature causes the equilibrium to shift in the endothermic direction to oppose the increase, that is, to the left. The yield of ammonia therefore decreases. (The rate of reaction does increase, but that is a separate effect.)"
Key RuleHeating shifts equilibrium towards the ENDOTHERMIC direction. Answer "which way does it move?" with equilibrium reasoning and "how fast?" with collision theory — never mix them.
Exercise 5: "State the effect of adding an iron catalyst to the Haber process."
Student's Answer"The catalyst lowers the activation energy so more ammonia is produced at equilibrium and the reaction is more exothermic."
The FlawThe first clause is right; everything after it is wrong twice over. A catalyst does not increase the equilibrium yield, and it does not change ΔH. Lowering Ea lowers the hump on the energy profile; it does not move the reactant or product energy levels.
Correct Answer"The iron catalyst provides an alternative reaction pathway with a lower activation energy, so a greater proportion of collisions are successful and the rate increases. It speeds up the forward and backward reactions equally, so the position of equilibrium and the yield of ammonia are unchanged — equilibrium is simply reached in a shorter time. ΔH is also unchanged."
Key RuleCatalyst = rate only. Never yield, never ΔH, never equilibrium position.
Exercise 6: "A sealed flask contains an equilibrium mixture whose colour has stopped changing. Explain what is happening in the flask."
Student's Answer"The reaction has finished because all the reactants have turned into products, so nothing else can happen and the colour stays the same."
The FlawTwo misconceptions in one sentence. Equilibrium does not mean the reaction has stopped, and it does not mean the reactants have been fully converted — both reactants and products are present.
Correct Answer"The system has reached dynamic equilibrium. Both the forward and the backward reactions are still taking place, but the rate of the forward reaction is equal to the rate of the backward reaction. The concentrations of reactants and products therefore remain constant, so the colour does not change. Both reactants and products are present in the flask."
Key Rule"Dynamic" is in the name for a reason. Constant concentrations, not stopped reactions. And equilibrium never means everything has been converted.
Exercise 7: "In Fe2O3 + 3CO → 2Fe + 3CO2, identify the oxidising agent and explain your choice."
Student's Answer"Carbon monoxide is the oxidising agent because it is oxidised to carbon dioxide by gaining oxygen."
The FlawThe observation is correct — CO is indeed oxidised — but the conclusion is inverted. A substance that is oxidised is the reducing agent, because it gave its electrons away to reduce something else.
Correct Answer"Fe2O3 is the oxidising agent. It loses oxygen and iron's oxidation number falls from +3 to 0, so it is reduced — and the substance that is reduced is the oxidising agent. CO gains oxygen and carbon rises from +2 to +4, so CO is oxidised and is therefore the reducing agent."
Key RuleThe agent always has the opposite done to it. Oxidising agent = reduced. Reducing agent = oxidised. Write this in the margin before every redox question.
Exercise 8: "Deduce the oxidation number of manganese in MnO4− and state the change when it becomes Mn2+."
Student's Answer"Mn is 7+ because the four oxygens are 2− each giving 8− and the ion is 1−. Going from 7+ to 2+ means it is oxidised because it is losing charge."
The FlawThe arithmetic is right but two conventions are wrong. Oxidation numbers are written sign first: +7, not 7+. And a fall from +7 to +2 is a decrease, which is reduction — caused by gaining five electrons, not "losing charge".
Correct Answer"Let Mn = x. Then x + 4(−2) = −1, so Mn = +7. Going from +7 to +2 is a decrease of 5 in oxidation number, so manganese has been reduced, having gained 5 electrons. The purple colour of MnO4− disappears because Mn2+ is colourless."
Key RuleSign first for oxidation numbers. Gaining electrons makes the number go DOWN — even when the species stays positive throughout.
Exercise 9: "Blue hydrated copper(II) sulfate is heated and turns white; adding water turns it blue again. Is this a physical or a chemical change? Give a reason."
Student's Answer"It is a physical change because it can easily be reversed by adding water, just like melting ice can be reversed by freezing it."
The FlawReversibility is not the test for a physical change. Many chemical changes are reversible — this one, the Haber process, and every equilibrium in the syllabus. The student has used a clue instead of the definition.
Correct Answer"It is a chemical change, because a new substance with a different chemical formula is formed: CuSO4·5H2O becomes CuSO4. The water of crystallisation is chemically bonded within the lattice and is driven off by heating; the colour change from blue to white confirms a new substance. The fact that it is reversible does not make it physical."
Key RuleThe only test is: has a new substance with a different formula been formed? Reversibility, temperature change, bubbles and colour changes are clues, not proof.
Exercise 10: "Explain why increasing the concentration of hydrochloric acid increases the rate of its reaction with marble chips. [2]"
Student's Answer"There are more acid particles so they move faster and collide with more energy, so more collisions are successful."
The FlawThree errors. "More particles" without "per unit volume" is insufficient. Concentration does not make particles move faster — that is temperature. And it does not change the energy of the collisions either, only how often they happen.
Correct Answer"A more concentrated acid contains more acid particles per unit volume, so the particles are closer together [1]. They therefore collide with the surface of the marble more frequently, so the frequency of successful collisions per second increases and the rate increases [1]."
Key RuleConcentration and pressure change collision frequency. Temperature changes frequency AND collision energy. A catalyst changes the energy barrier. Keep the three mechanisms separate.
Exercise 11: "For H2(g) + I2(g) ⇌ 2HI(g), state and explain the effect of increasing the pressure on the position of equilibrium."
Student's Answer"Increasing the pressure always shifts the equilibrium to the right because the particles are pushed closer together and react more, so more HI is formed."
The FlawThe student has applied a memorised rule without counting the moles of gas, and has again used a rate argument ("react more") for a position question. Both sides here have 2 moles of gas.
Correct Answer"There are 2 moles of gas on the left (1 H2 + 1 I2) and 2 moles of gas on the right (2 HI). Since the number of moles of gas is the same on both sides, neither direction relieves the increased pressure, so the position of equilibrium is unchanged. The rate of reaction does increase, so equilibrium is reached sooner, but the yield of HI is not affected."
Key RuleCount the moles of gas on each side BEFORE applying the pressure rule. Equal moles = no shift. Solids and liquids do not count.
Exercise 12: "A student adds a substance to potassium iodide solution and it turns brown. What can be deduced about the substance?"
Student's Answer"The substance is a reducing agent, because potassium iodide is a reducing agent and it has reduced the iodide ions to brown iodine."
The FlawThe student has copied the identity of the reagent onto the unknown, and has also described iodide going to iodine as "reduction". Iodine goes from −1 to 0, which is an increase — oxidation.
Correct Answer"The substance is an oxidising agent. The colourless iodide ions have been oxidised to brown iodine (2I− → I2 + 2e−), with iodine's oxidation number rising from −1 to 0. The electrons lost must have been accepted by the added substance, so it has been reduced and is therefore the oxidising agent."
Key RuleEach test reagent detects the OPPOSITE of itself. KI is a reducing agent, so it detects oxidising agents. Acidified manganate(VII) is an oxidising agent, so it detects reducing agents.

✍️ Ultra-Detailed Practice Questions

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Ten Cambridge-style challenge questions. Write your answer, then reveal the model answer with mark scheme and examiner's notes.

Question 1
[6 marks]
Priya in Pune reacts 0.12 g of magnesium ribbon with 50 cm3 of excess 1.0 mol/dm3 hydrochloric acid at 25 °C and collects 120 cm3 of hydrogen, the reaction finishing after 100 s. (a) Calculate the average rate of reaction in cm3/s. [1] (b) Explain, using collision theory, why the graph is steepest at the start and becomes horizontal at 100 s. [3] (c) She repeats the experiment with 0.12 g of magnesium powder. State two differences and one similarity in the curve obtained. [2]
Model Answer(a) rate = 120 ÷ 100 = 1.2 cm3/s [1]
(b) At the start the concentration of the acid and the surface area of the magnesium are at their greatest, so the frequency of successful collisions is highest and the rate is fastest [1]. As the reaction proceeds the acid is used up and the ribbon becomes smaller, so collisions become less frequent and the rate decreases [1]. At 100 s the graph is horizontal because the reaction has stopped — all the magnesium (the limiting reactant) has been used up, so the rate is zero [1]
(c) Differences: the curve is steeper at the start, and it becomes horizontal sooner [1]. Similarity: it levels off at the same volume, 120 cm3 [1]
Examiner's NotesPart (c) is the discriminator. The similarity mark is lost by anyone who thinks powder produces more gas — the mass of magnesium is identical, so the moles of hydrogen must be identical. In (b), "it slows down because it runs out of chemicals" is too vague; you must name the concentration and surface area, and you must say the reaction has STOPPED (not "gone very slow") at 100 s.
Question 2
[6 marks]
The Haber process is N2(g) + 3H2(g) ⇌ 2NH3(g). The forward reaction is exothermic. A plant in Billingham, England, operates at 450 °C and 200 atm with an iron catalyst. (a) State and explain the effect on the equilibrium yield of ammonia of increasing the temperature. [2] (b) State and explain the effect on the yield of increasing the pressure. [2] (c) Explain why 450 °C is used even though a lower temperature would give a greater yield. [2]
Model Answer(a) The yield decreases [1]. The forward reaction is exothermic, so the backward reaction is endothermic; increasing the temperature shifts the equilibrium in the endothermic direction, i.e. to the left, to oppose the increase [1]
(b) The yield increases [1]. There are 4 moles of gas on the left and only 2 on the right, so increasing the pressure shifts the equilibrium towards the side with fewer moles of gas, i.e. to the right [1]
(c) At a lower temperature the rate of reaction would be far too slow, because fewer particles would have energy ≥ the activation energy, so equilibrium would take too long to reach and the process would be uneconomic [1]. 450 °C is therefore a compromise between an acceptable yield and an acceptable rate [1]
Examiner's NotesEach part needs a statement plus a reason. In (a) the reason must name the endothermic direction, not just "Le Chatelier says so". In (b) quote the mole counts 4 and 2 — "fewer molecules on the right" without numbers is often not credited. In (c) the word "compromise" plus a rate justification are both needed; students who write only "it would be too slow" typically get 1 of 2.
Question 3
[7 marks]
Consider the reaction 2Fe3+(aq) + 2I−(aq) → 2Fe2+(aq) + I2(aq). (a) Deduce the oxidation number of iron and of iodine on each side, and state which species is oxidised and which is reduced. [3] (b) Identify the oxidising agent and the reducing agent, giving a reason for each. [2] (c) Write the two half-equations. [2]
Model Answer(a) Iron: +3 → +2, a decrease, so Fe3+ is reduced [1]. Iodine: −1 → 0, an increase, so I− is oxidised [1]. Both changes stated with correct numbers [1]
(b) Oxidising agent = Fe3+, because it is the species that is reduced (it gains electrons) [1]. Reducing agent = I−, because it is the species that is oxidised (it loses electrons) [1]
(c) Reduction: Fe3+ + e− → Fe2+ (or 2Fe3+ + 2e− → 2Fe2+) [1]
Oxidation: 2I− → I2 + 2e− [1]
Examiner's NotesPart (b) is where the marks vanish. Roughly half of candidates write "I− is oxidised so it is the oxidising agent." Always finish the sentence with the reversal. In (c), check the charge balances: left-hand side of the oxidation half-equation is 2−; right-hand side is 0 + 2(−1) = 2−. If the charges do not balance, the electrons are wrong. Expect a follow-up asking for the observation: the solution turns brown as iodine forms.
Question 4
[6 marks]
A student investigates 2SO2(g) + O2(g) ⇌ 2SO3(g), the key step of the Contact process. The forward reaction is exothermic and a vanadium(V) oxide catalyst is used at about 450 °C and 2 atm. (a) Explain what is meant by "dynamic equilibrium". [2] (b) State the effect of the catalyst on the rate and on the yield, with reasons. [2] (c) Explain why a pressure of only 2 atm is used when high pressure would increase the yield. [2]
Model Answer(a) Both the forward and the backward reactions are still taking place [1], but the rate of the forward reaction equals the rate of the backward reaction, so the concentrations of reactants and products remain constant [1]
(b) The rate increases, because the catalyst provides an alternative pathway with a lower activation energy so a greater proportion of collisions are successful [1]. The yield is unchanged, because the catalyst speeds up the forward and backward reactions equally, so the position of equilibrium is unaffected — equilibrium is simply reached sooner [1]
(c) At 2 atm the yield of SO3 is already about 96%, so raising the pressure could increase it only very slightly [1]. High-pressure vessels, pipework and compressors are extremely expensive and hazardous, so the small gain in yield does not justify the cost [1]
Examiner's NotesIn (a), "the reaction has stopped" scores zero and often forfeits both marks. In (b) you must give both halves — rate up, yield unchanged — and the reason for each. In (c) the expected answer is economic. Candidates who write "because there are fewer moles of gas on the right" have answered a different question; that explains why high pressure would help, not why it is not used.
Question 5
[7 marks]
Blue crystals of hydrated copper(II) sulfate are heated strongly in a test tube. A white solid remains and a colourless liquid collects at the cooler top of the tube. The tube is allowed to cool and the liquid is returned to the white solid. (a) Write an equation, using the ⇌ symbol, for the reaction. [1] (b) State the observation when the liquid is added back, and state the energy change. [2] (c) The student claims this is a physical change because it is reversible. Evaluate this claim. [2] (d) Describe how you would show the colourless liquid is pure water. [2]
Model Answer(a) CuSO4·5H2O ⇌ CuSO4 + 5H2O [1]
(b) The white solid turns blue [1] and the tube becomes warm — the backward reaction is exothermic [1]
(c) The claim is incorrect [1]. Reversibility is not the test for a physical change; the correct test is whether a new substance with a different chemical formula is formed. Here CuSO4·5H2O becomes CuSO4, a different compound with a different colour, so it is a chemical change that happens to be reversible [1]
(d) Show it turns white anhydrous copper(II) sulfate blue (or blue cobalt(II) chloride paper pink), which shows water is present [1]. Then show it boils at exactly 100 °C (and freezes at 0 °C) at standard pressure, which shows it is pure [1]
Examiner's NotesPart (d) catches almost everyone: the colour test shows water is present, but only a sharp, correct boiling or melting point shows it is pure. Both marks require both ideas. In (b), do not forget the energy change — if the forward reaction needed continuous heating (endothermic), the backward reaction must be exothermic. In (c) an evaluation needs a verdict AND the correct criterion.
Question 6
[6 marks]
A class in Nairobi investigates the reaction Na2S2O3 + 2HCl → 2NaCl + SO2 + S + H2O by timing how long a paper cross takes to disappear. Results: 20 °C, 96 s; 30 °C, 48 s; 40 °C, 24 s. (a) Explain how the results should be processed to obtain a measure of rate, and describe the trend. [2] (b) Explain the trend fully in terms of collision theory. [3] (c) Suggest one improvement to the method and justify it. [1]
Model Answer(a) Rate is proportional to 1/t, so calculate 1/96 = 0.0104, 1/48 = 0.0208 and 1/24 = 0.0417 s−1 [1]. The rate doubles for each 10 °C rise in temperature [1]
(b) At higher temperature the particles have more kinetic energy and move faster, so they collide more frequently [1]. Crucially, a greater proportion of the particles have energy equal to or greater than the activation energy [1], so a much larger fraction of collisions are successful; the frequency of successful collisions per second therefore increases and the rate rises [1]
(c) Use a light sensor and data logger to detect when a fixed amount of light is blocked, because judging by eye when the cross disappears is subjective and varies between observers [1]
Examiner's NotesPart (a) needs the actual processing (1/t), not just "the time gets shorter". Part (b) is the classic three-marker: the activation energy sentence is worth more than the "move faster" sentence, and omitting it caps you at 1 mark. In (c), "repeat the experiment" is not accepted as an improvement to a subjective end point — name the instrument and say what problem it removes.
Question 7
[6 marks]
(a) Deduce the oxidation number of the named element in each of: (i) chromium in Cr2O72−, (ii) nitrogen in NH3, (iii) oxygen in H2O2, (iv) sulfur in SO42−. [4] (b) Magnesium burns in carbon dioxide: 2Mg + CO2 → 2MgO + C. Deduce the oxidation number change of carbon and state, with a reason, which substance is the oxidising agent. [2]
Model Answer(a) (i) 2x + 7(−2) = −2, so 2x = +12 and Cr = +6 [1]
(ii) x + 3(+1) = 0, so N = −3 [1]
(iii) H2O2 is a peroxide: with H = +1, 2(+1) + 2x = 0, so O = −1 [1]
(iv) x + 4(−2) = −2, so S = +6 [1]
(b) Carbon changes from +4 in CO2 to 0 in C, a decrease, so carbon is reduced [1]. CO2 is therefore the oxidising agent, because it is the substance that is reduced (magnesium, going 0 → +2, is oxidised and is the reducing agent) [1]
Examiner's NotesIn (a)(i) the commonest slip is forgetting to divide by 2 and giving +12. In (a)(iii) students apply the −2 rule automatically; peroxides are examined precisely because of that reflex. Part (b) is deliberately counter-intuitive — carbon dioxide, usually thought of as inert, acts as the oxidising agent, and a metal burns in it. Quote the numbers +4 → 0 for the mark.
Question 8
[6 marks]
Two experiments follow the reaction of calcium carbonate with hydrochloric acid using a balance. Experiment 1: 5.0 g of large marble chips with 50 cm3 of 1.0 mol/dm3 acid; total mass loss 1.10 g after 300 s. Experiment 2: 5.0 g of powdered marble with 50 cm3 of 1.0 mol/dm3 acid; total mass loss 1.10 g after 90 s. (a) Explain why the mass decreases during the reaction. [1] (b) Sketch, in words, how the two curves of mass against time compare. [2] (c) Explain the difference in terms of collision theory. [2] (d) Explain why this method would be unsuitable for the reaction of magnesium with hydrochloric acid. [1]
Model Answer(a) Carbon dioxide gas escapes from the open flask, so the mass of the flask and contents falls [1]
(b) Both curves start at the same mass and fall, becoming shallower and then horizontal [1]. The powder curve falls more steeply and becomes horizontal sooner (about 90 s rather than 300 s), but both level off at the same total mass loss of 1.10 g [1]
(c) The powder has a much greater surface area for the same mass, so more calcium carbonate particles are exposed to the acid [1]. This gives a greater frequency of successful collisions per second, so the rate is higher [1]
(d) Hydrogen has a very low density, so the mass loss would be too small to measure accurately on a school balance; a gas syringe should be used instead [1]
Examiner's NotesPart (b) has the usual trap in disguise: because this is a mass-loss graph the curves go DOWN, so "the powder curve is steeper" means steeper downwards. The equal final mass loss of 1.10 g is the mark most often dropped — the same mass of marble must release the same mass of CO2. Part (d) needs the reason (low density / very small mass change), not just "it would not work".
Question 9
[6 marks]
A student is given three colourless solutions, P, Q and R, and tests each one. With acidified potassium manganate(VII), P turns the purple solution colourless while Q and R leave it purple. With potassium iodide solution, Q turns brown while P and R stay colourless. (a) Deduce what P and Q are, explaining each deduction in terms of electron transfer. [4] (b) State what can and cannot be concluded about R. [1] (c) Explain why the potassium manganate(VII) must be acidified. [1]
Model Answer(a) P is a reducing agent [1]. The purple MnO4− has been reduced to colourless Mn2+ (Mn from +7 to +2), so P must have donated electrons to it [1].
Q is an oxidising agent [1]. The colourless iodide ions have been oxidised to brown iodine, 2I− → I2 + 2e− (iodine from −1 to 0), so Q must have accepted those electrons [1]
(b) R gave a negative result with both reagents, so it is neither a reducing agent nor an oxidising agent strong enough to react with these reagents; no positive conclusion about its identity can be drawn [1]
(c) H+ ions are needed for MnO4− to be reduced all the way to colourless Mn2+; without acid a brown precipitate of manganese(IV) oxide forms instead and the colour change is unclear [1]
Examiner's NotesThe whole question turns on the reversal rule — each reagent detects the opposite of what it is. Part (b) tests scientific reasoning: a negative result is not proof of the opposite property, and candidates who write "R is a catalyst" or "R is neutral" invent information. In (a) the marks are split between the conclusion and the electron-transfer explanation, so a bare "P is a reducing agent" scores half.
Question 10
[8 marks]
Ethanol is manufactured by hydrating ethene: C2H4(g) + H2O(g) ⇌ C2H5OH(g). The forward reaction is exothermic. Industrial conditions are about 300 °C, 60 atm and a phosphoric acid catalyst. (a) State and explain the effect on the equilibrium yield of ethanol of (i) increasing the temperature and (ii) increasing the pressure. [4] (b) State the effect of the catalyst on the yield and on the time taken to reach equilibrium. [2] (c) Unreacted ethene and steam are separated from the ethanol and recycled. Explain how continuously removing the ethanol affects the equilibrium. [1] (d) A student says "at equilibrium the reaction has stopped, so recycling is pointless." Explain why this is wrong. [1]
Model Answer(a)(i) The yield decreases [1], because the forward reaction is exothermic so the backward reaction is endothermic; raising the temperature shifts the equilibrium in the endothermic direction, to the left, to oppose the increase [1]
(a)(ii) The yield increases [1], because there are 2 moles of gas on the left and only 1 on the right, so increasing the pressure shifts the equilibrium towards the side with fewer moles of gas, to the right [1]
(b) The yield is unchanged — the catalyst speeds up forward and backward reactions equally so the position of equilibrium is unaffected [1]. Equilibrium is reached in a shorter time, because the catalyst provides an alternative pathway with a lower activation energy [1]
(c) Removing the ethanol (a product) causes the equilibrium to shift to the right to replace it, so more ethene is converted and the overall conversion rises well above the single-pass equilibrium yield [1]
(d) At equilibrium the reaction has not stopped — it is dynamic, with the forward and backward reactions still proceeding at equal rates. Removing ethanol disturbs that balance so the forward reaction can proceed further, which is exactly why recycling works [1]
Examiner's NotesThis question combines every Topic 6 equilibrium idea and is typical of the highest-demand papers. Part (a)(ii) requires you to count the moles of gas for an unfamiliar equation — note that 2 → 1 is an even larger relative reduction than in the Haber process, which is why 60 atm is worthwhile. Part (d) awards the mark only for the word "dynamic" or an explicit statement that both reactions continue. Watch for the classic double error in (a)(i): using a rate argument to claim the yield rises.
IGCSE Chemistry 0620 — Topic 6 Challenge Prep Guide. Work through every section, and revisit the twelve traps the night before your exam.