Chemical Reactions

• Core 6.4.2 Rate answers “how fast”. Equilibrium answers “how far”. Redox answers “who gave, who took”. Before you write, check which of the three the question actually asked — answering a different one is the easiest way to write something true that earns nothing.
• Twelve Core statements. Notice what is not among them: collision theory, activation energy, the position of equilibrium, the Haber process, the Contact process, electrons, oxidation number rules and oxidising agents. A Core candidate is not examined on any of those.
• Twenty-one Supplement statements, on top of the twelve Core ones. Thirty-three in all for an Extended candidate.
• Those twelve rows are the whole of the Core course for Topic 6. If you are a Core candidate, the sections they point to — 6.1, 6.2A, 6.2B, the two investigation sections, the rate graph studio, 6.3A, 6.4A and 6.4B — are the whole of your reading, and they are written to stand on their own.
• Has at least one new chemical substance been formed? Yes → chemical change. No → physical change.
• A catalyst increases the rate of a reaction and is chemically unchanged at the end of the reaction.
• A catalyst decreases the activation energy, \(E_\mathrm{a}\), of a reaction — not by pushing the existing barrier down, but by offering a different route that has a lower barrier.
• Every evaluation answer has the same skeleton: this measurement is imperfect because …, which makes the readings too high / too low / too coarse, and the fix is … because it removes that specific cause.

• Core 6.1.1 Identify physical and chemical changes, and describe the differences between them.
• Core 6.2.1 Describe the effect on the rate of reaction of changing the concentration of solutions, changing the pressure of gases, changing the surface area of solids, changing the temperature, and adding or removing a catalyst, including enzymes.
• Core 6.2.2 State that a catalyst increases the rate of a reaction and is unchanged at the end of a reaction.
• Core 6.2.3 Describe practical methods for investigating the rate of a reaction, including change in mass of a reactant or a product, and the formation of a gas.
• Core 6.2.4 Interpret data, including graphs, from rate of reaction experiments.
• Core 6.3.1 State that some chemical reactions are reversible, as shown by the symbol \(\rightleftharpoons\).
• Core 6.3.2 Describe how changing the conditions can change the direction of a reversible reaction, for the effect of heat on hydrated compounds and the addition of water to anhydrous compounds, limited to copper(II) sulfate and cobalt(II) chloride.
• Core 6.4.1 Use a Roman numeral to indicate the oxidation number of an element in a compound.
• Core 6.4.2 Define redox reactions as involving simultaneous oxidation and reduction.
• Core 6.4.3 Define oxidation as gain of oxygen and reduction as loss of oxygen.
• Core 6.4.4 Identify redox reactions as reactions involving gain and loss of oxygen.
• Core 6.4.5 Identify oxidation and reduction in redox reactions.
• Supplement 6.2.5 Describe collision theory in terms of the number of particles per unit volume, the frequency of collisions between particles, the kinetic energy of particles, and activation energy, \(E_\mathrm{a}\).
• Supplement 6.2.6 Describe and explain, using collision theory, the effect on the rate of reaction of all five factors named in 6.2.1.
• Supplement 6.2.7 State that a catalyst decreases the activation energy, \(E_\mathrm{a}\), of a reaction.
• Supplement 6.2.8 Evaluate practical methods for investigating the rate of a reaction, including change in mass of a reactant or a product, and the formation of a gas.
• Supplement 6.3.3 State that a reversible reaction in a closed system is at equilibrium when the rate of the forward reaction equals the rate of the reverse reaction and the concentrations of reactants and products are no longer changing.
• Supplement 6.3.4 Predict and explain, for a reversible reaction, how the position of equilibrium is affected by changing temperature, changing pressure, changing concentration and using a catalyst, using information provided.
• Supplement 6.3.5 State the symbol equation for the production of ammonia in the Haber process.
• Supplement 6.3.6 State the sources of the hydrogen (methane) and the nitrogen (air) in the Haber process.
• Supplement 6.3.7 State the typical conditions in the Haber process as \(450\,{}^\circ\mathrm{C}\), \(20\,000\ \mathrm{kPa}\) (\(200\) atm) and an iron catalyst.
• Supplement 6.3.8 State the symbol equation for the conversion of sulfur dioxide to sulfur trioxide in the Contact process.
• Supplement 6.3.9 State the sources of the sulfur dioxide (burning sulfur or roasting sulfide ores) and the oxygen (air) in the Contact process.
• Supplement 6.3.10 State the typical conditions for that conversion as \(450\,{}^\circ\mathrm{C}\), \(200\ \mathrm{kPa}\) (\(2\) atm) and a vanadium(V) oxide catalyst.
• Supplement 6.3.11 Explain, in terms of rate of reaction and position of equilibrium, why the typical conditions stated are used in the Haber process and in the Contact process, including safety considerations and economics.
• Supplement 6.4.6 Define oxidation in terms of loss of electrons and of an increase in oxidation number.
• Supplement 6.4.7 Define reduction in terms of gain of electrons and of a decrease in oxidation number.
• Supplement 6.4.8 Identify redox reactions as reactions involving gain and loss of electrons.
• Supplement 6.4.9 Identify redox reactions by changes in oxidation number, using the four rules: an uncombined element is zero; a monatomic ion equals its charge; the sum in a compound is zero; the sum in an ion equals the charge on the ion.
• Supplement 6.4.10 Identify redox reactions by the colour changes involved when using acidified aqueous potassium manganate(VII) or aqueous potassium iodide.
• Supplement 6.4.11 Define an oxidising agent as a substance that oxidises another substance and is itself reduced.
• Supplement 6.4.12 Define a reducing agent as a substance that reduces another substance and is itself oxidised.
• Supplement 6.4.13 Identify oxidising agents and reducing agents in redox reactions.

Why it matters. Roman numerals are already all over this chapter: manganese(IV) oxide catalysed the hydrogen peroxide in Section 6.2B, cobalt(II) chloride and copper(II) sulfate carried the reversible reactions in Section 6.3A, and vanadium(V) oxide catalyses the Contact process. Every one of those names is telling you an oxidation number. Reading them is a Core skill in its own right, and it is worth a mark on its own.

• 1. “It is a physical change because you can reverse it.” Core 6.1.1 Why temptingMost physical changes are easy to reverse, so the correlation is real. RepairReversibility is a correlation, not a definition. Heating hydrated copper(II) sulfate is chemical and reverses in seconds; cutting paper is physical and cannot be undone. Ask instead whether a new substance has formed. TransferClassify “anhydrous cobalt(II) chloride turning pink in damp air” and justify it. Chemical: a different compound, the hydrate, has formed.
• 2. “There was a colour change, so a chemical reaction happened.” Core 6.1.1 Why temptingColour change is on every list of signs of a reaction. RepairThose signs are supporting evidence, not proof. Iodine sublimes from grey solid to purple vapour with no new substance at all. TransferGive one colour change that is not a chemical reaction and one that is. Iodine subliming; copper(II) sulfate crystals turning white on heating.
• 3. “More collisions, so the reaction is faster.” Supplement 6.2.5 Why temptingIt is half of the right answer, and it sounds like a full one. RepairInsert the word successful: collisions in which the particles have combined energy at least equal to the activation energy. Almost all collisions achieve nothing. TransferRewrite “higher concentration means more collisions so it is faster” as a full-mark sentence.
• 4. “Increasing the concentration makes the particles move faster.” Supplement 6.2.6 Why tempting“Faster reaction” slides into “faster particles”. RepairAverage particle speed depends on temperature alone. Concentration changes how crowded the particles are, so they meet more often at the same speed. TransferWhich of the five rate factors do change particle speed? Only temperature.
• 5. “Grinding the solid increases its concentration.” Core 6.2.1 Why temptingBoth changes make a reaction faster, so the words get swapped. RepairSolids do not have a concentration; concentration is defined for a solute in a solution. Grinding increases the surface area, exposing more particles. TransferState what is unchanged when a lump is ground to powder. The mass, the number of particles present and the final amount of product.
• 6. “Powdered reactant gives more gas in total.” Core 6.2.4 Why temptingThe powder curve is above the lump curve for most of the graph. RepairIt is above only until the lump catches up. Both plateau at the same height, because the plateau is fixed by the limiting reactant, and grinding changes no amounts. TransferTwo curves plateau at different heights. What must have been changed? The amount of limiting reactant.
• 7. “Heating lowers the activation energy.” Supplement 6.2.6 Why temptingHeating and catalysis both make more collisions succeed. Repair\(E_\mathrm{a}\) is a fixed property of the reaction pathway. Heating raises the particles towards the barrier; only a catalyst lowers the barrier, by offering a different pathway. TransferOn an energy profile, which of the two changes would you be able to see? The catalyst; heating changes nothing on the diagram.
• 8. “Heating gives all the particles enough energy to react.” Supplement 6.2.6 Why temptingIt sounds like a stronger version of the right answer. RepairIt raises the fraction of particles with energy at or above \(E_\mathrm{a}\). Many still fall short — which is why reactions have a rate rather than finishing instantly. TransferIf every particle had enough energy, what would the graph look like? Vertical, then flat: the reaction would be instantaneous.
• 9. “The catalyst is used up, so you have to keep adding it.” Core 6.2.2 Why temptingIt clearly takes part, so it feels like a reactant. RepairIt is regenerated. Filter it off at the end, dry it and weigh it: the mass is unchanged. That is the experimental evidence, and it is worth quoting. TransferWhy does a catalyst not appear in the equation? It is neither consumed nor produced, so it cancels from both sides.
• 10. “A catalyst increases the yield.” Supplement 6.3.4 Why temptingMore product appears sooner, which looks like more product. RepairIt lowers the activation energy of forward and reverse reactions equally, so both rates rise by the same factor. Equilibrium arrives sooner at exactly the same composition. TransferExplain why iron is used in the Haber process. To reach equilibrium fast enough at \(450\,{}^\circ\mathrm{C}\) — not to raise the yield.
• 11. “The flat part of the graph shows equilibrium.” Core 6.2.4 Supplement 6.3.3 Why temptingNothing is changing, which is what equilibrium looks like. RepairIn an open flask the curve flattens because a reactant has run out and the reaction has stopped. Equilibrium needs a closed system and two reactions still running at equal rates. TransferWhat two things does a vessel need before it can reach equilibrium? A reversible reaction, and a seal so that no product escapes. Marble and acid fails the first test — carbon dioxide, calcium chloride and water do not react back — so sealing that flask alone would achieve nothing.
• 12. “At equilibrium the amounts of reactants and products are equal.” Supplement 6.3.3 Why tempting“Equilibrium” and “balance” suggest equality. RepairThe rates are equal; the concentrations are constant. The Contact process sits at equilibrium with about \(98\%\) product. TransferFill the gap: “the concentrations remain ______.” Constant.
• 13. “At equilibrium the reaction has stopped.” Supplement 6.3.3 Why temptingNothing observable changes. RepairBoth reactions continue at equal, non-zero rates. That is exactly what the word dynamic is doing in “dynamic equilibrium”. TransferWhat would happen to the colour of a sealed \(\mathrm{NO_2}\)/\(\mathrm{N_2O_4}\) tube if the forward reaction really stopped? It would keep changing until one substance ran out.
• 14. “Raising the pressure always moves an equilibrium to the right.” Supplement 6.3.4 Why temptingIt does in both industrial examples in this chapter. RepairIt moves towards the side with fewer gas moles, whichever side that is — and does nothing at all if the counts are equal. TransferPredict the effect of higher pressure on \(\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}\). No shift: two moles of gas on each side.
• 15. “Adding an inert gas at constant volume raises the pressure, so the equilibrium shifts.” Supplement 6.3.4 Why temptingThe pressure gauge really does read higher. RepairWhat matters is the number of reacting molecules per unit volume, and that is unchanged. No shift. TransferWhat would shift it? Reducing the volume, which genuinely crowds the reacting molecules.
• 16. “Heat acts as a catalyst.” Supplement 6.3.4 Why temptingBoth speed reactions up. RepairA catalyst lowers \(E_\mathrm{a}\) and leaves the position of equilibrium alone. Heat leaves \(E_\mathrm{a}\) alone and does move the position. They are opposites in both respects. TransferWhich of the two changes the yield? Heat.
• 17. “The Contact process runs at 200 atmospheres.” Supplement 6.3.10 Why temptingThe two processes share a temperature, so the pressures get shared too. RepairContact runs at \(200\ \mathrm{kPa}\), which is \(2\) atmospheres. Haber runs at \(20\,000\ \mathrm{kPa}\), which is \(200\) atmospheres. Both numbers contain “200”, and the unit is the whole difference. TransferGive both pressures with units, then say why they differ.
• 18. “Industrial conditions are chosen to give the highest possible yield.” Supplement 6.3.11 Why temptingYield is what the chemistry is about. RepairThey are chosen to give the most product per day at an acceptable cost and risk. \(450\,{}^\circ\mathrm{C}\) is deliberately not the yield-maximising temperature. TransferWhat temperature would maximise the Haber yield, and why is it not used? As low as possible; the rate would be far too slow.
• 19. “The species oxidised is the oxidising agent.” Supplement 6.4.13 Why temptingThe two phrases share a word and differ by three letters. RepairThe species oxidised gave electrons away, causing reduction elsewhere, so it is the reducing agent. An oxidising agent is itself reduced. TransferIn \(\mathrm{2Mg + O_2 \rightarrow 2MgO}\), name both agents. Magnesium reducing; oxygen oxidising.
• 20. “The Roman numeral is the charge on the ion.” Core 6.4.1 Why temptingFor simple ions such as \(\mathrm{Fe^{3+}}\) in iron(III) chloride, the two do coincide. RepairThe Roman numeral is the oxidation number of one element. In manganate(VII) manganese is \(+7\) while the whole \(\mathrm{MnO_4^{\,-}}\) ion carries only \(1-\). TransferState the oxidation number of chromium in dichromate(VI) and the charge on the ion. \(+6\) per chromium atom; the ion is \(2-\).

• Describe against explain is the tier line in miniature. Core statement 6.2.1 says describe the effect of each rate factor; Supplement statement 6.2.6 says describe and explain the same five effects using collision theory. A Core candidate who writes a collision-theory explanation has not earned extra marks, and has spent time they needed elsewhere. An Extended candidate who writes only the effect has answered half the question.
• Say “surface area”, never “concentration”. A solid does not have a concentration; concentration is defined for a solute in a solution. “Grinding the solid increases its concentration” names the wrong quantity, so the sentence is wrong however sound the reasoning around it is. In the same way, a pressure change only affects the rate if at least one reactant is a gas — check for a \((\mathrm{g})\) state symbol before you write anything about pressure.
• Why the plug is cotton wool and not a bung. The reaction is vigorous and throws fine droplets of acid upwards. If those droplets escaped, the mass loss would be too large and would not be due to gas alone. Loose cotton wool lets the carbon dioxide through while trapping the spray — it makes the measurement valid without sealing the system.
• Comparing two curves in one sentence each. Say something about the steepness and something about the plateau, and say what each one tells you. “Curve A is steeper at the start, so its initial rate is higher; both curves level off at the same volume, so both experiments contained the same amount of limiting reactant.” That single sentence pattern answers most graph-comparison questions in this topic.
• The word that carries the explanation. A rate explanation is only complete when it contains the word successful, or an equivalent such as “collisions with energy greater than or equal to the activation energy”. “More collisions, so faster” leaves out the entire energy condition, which is the half of collision theory that does the explaining.
• Where pressure applies. A pressure change only affects the rate if at least one reactant is a gas. Increasing the pressure above a solution of hydrochloric acid does essentially nothing to the rate, because liquids are almost incompressible — the particles were already touching. If a question mentions pressure, check for a \((\mathrm{g})\) state symbol before you write anything.
• Drawing the diagram. An answer asked to add a catalysed pathway to a reaction-pathway diagram must keep the reactant level and the product level exactly where they were, and draw a second, lower hump between them. Moving either level changes the energy of the reaction, which a catalyst cannot do; drawing only one lowered hump loses the fact that the uncatalysed route still exists.
• The rule that links the last two lines. An improvement is only worth writing if it fixes the limitation you named. “Repeat the experiment” does not fix a systematic error such as a leak or a soluble gas — repeating it just gives the same wrong answer three times. Match the repair to the fault.
• The blue trap. “Blue” appears twice in that table, on opposite sides. For copper(II) sulfate, blue is the hydrated form. For cobalt(II) chloride, blue is the anhydrous form. A useful anchor: cobalt chloride paper is used to test for water, and a test must start in a state that visibly changes, so the dry paper is blue and turns pink when water arrives.
• Read the last two columns together. Rows one and two are the reason industrial chemists cannot simply pick the temperature that gives the best yield: for an exothermic forward reaction, the temperature that maximises yield is also the temperature at which you would wait forever. That tension is the whole of the compromise argument in Sections 6.3F and 6.3G.
• The row worth memorising is the third. Whenever the gas counts are equal, the correct answer to “what happens to the position of equilibrium?” is nothing. Students who have learned “pressure pushes it right” as a slogan get this wrong every time. Pressure still speeds the reaction up — but it does not move the position.
• Read the last column. All four changes shorten the time to equilibrium, so “it gets there faster” never distinguishes them and never answers a question about yield. The column that discriminates is the second one — and only one row in it says “no change”.
• Give the pressure in the unit the question uses. \(20\,000\ \mathrm{kPa}\) and \(200\) atmospheres are the same pressure, and the syllabus states both. Quoting both costs nothing and covers you whichever unit a question is set in.
• The pressure is the discriminating fact. Both processes run at \(450\,{}^\circ\mathrm{C}\), and both use a catalyst, so the temperature and the idea of a catalyst do not tell the two apart. The pressures differ by a factor of a hundred: \(200\) atmospheres for Haber, \(2\) atmospheres for Contact. Both numbers contain the digits “200”, so check the unit every time before you write one of them down.
• The sentence that finishes an evaluation. “The chemistry favours this change, but the cost and safety consequences outweigh the gain, so the proposal should be rejected.” Or the reverse. An evaluation is the judgement plus the reasons that support it; a list of effects with no verdict has stopped one step short.
• Read the numeral off the name, and read the charge off the formula. They are two different sources of information and they answer two different questions. If a question asks for the oxidation number of an element and the name contains a Roman numeral, the answer is already written down for you — add the sign and you are finished.
• Two values you will use constantly. In its compounds hydrogen is almost always \(+1\) and oxygen almost always \(-2\). These are not extra rules to memorise — they follow from rules 2 and 3 applied to the simple compounds you already know — but having them ready makes every ledger a one-line calculation.
• Evidence first, name second. “Identify the oxidising agent” is asking for two things: which species it is, and how you know. Write the oxidation number change, or the electron transfer, and then name the agent. A bare name is an assertion; the evidence is what makes it an answer — and, since the inversion is so easy to get backwards, writing the evidence first is also the best way to catch your own error.
• Using this list under exam conditions. Nine of the twenty errors above (numbers 3, 4, 7, 8, 10, 12, 14, 16, 19) are triggered by a single word slipping out of place. If you have thirty seconds at the end of a paper, re-read your rate and redox sentences hunting for those words: successful, fraction, constant, fewer, and the direction of the agent inversion.
• How to use these across a revision week. Do not work through all eighty-six in one sitting. Take one heading per day, answer every prompt cold, and mark only whether you produced the answer without help. Return the following day to the ones you missed. Spaced, effortful retrieval is what moves this topic into long-term memory; re-reading the chapter does not.
• Marking yourself honestly. Award a mark only where your answer contains the idea in the marking point, not merely a related word. If you wrote “more collisions” where the point says “more successful collisions”, that is not a mark. These marking points are original to this chapter and are a guide to completeness, not a Cambridge mark scheme, so treat your score as a map of what to revise rather than as a prediction. Whatever the total, the useful step is the same: list the sections that produced the losses and return to those, rather than repeating the paper.
• If you have only one hour before the exam. Every candidate should read the seven danger zones and the nine comparison matrices. A Core candidate should then re-read Figure 6.2, the five factors, and the four colours in Section 6.3A. An Extended candidate should instead re-read the five-line specifications of the Haber and Contact processes and the four oxidation number rules. Those are the densest summaries of the topic for each route: the recall facts, the distinctions that are easiest to merge, and the errors that turn correct chemistry into a wrong answer.

• Cambridge IGCSE Chemistry 0620 is assessed on two routes. Which papers you sit is decided by the route you are entered for, and it decides which half of this chapter you are examined on.
• The practical component is taken by both routes. Every candidate sits either Paper 5 or Paper 6, whichever route they are entered for, so experimental skills are never Supplement-only. What differs is the demand: describing a rate method is Core, and evaluating one is Supplement.
• Extended candidates are examined on Core content as well. Papers 2 and 4 assess Core plus Supplement together, so nothing in the Core half of this chapter is optional for them.
• The same content is worth different marks depending on the verb, and in this topic the verb is often exactly what separates the routes.
• Describe against explain is the tier line in miniature. Core statement 6.2.1 says describe the effect of each rate factor; Supplement statement 6.2.6 says describe and explain the same five effects using collision theory. A Core candidate who writes a collision-theory explanation has not earned extra marks, and has spent time they needed elsewhere. An Extended candidate who writes only the effect has answered half the question.
• Core 6.2.1 Rate descriptions name the change and its effect: “the rate increases”. That is the whole answer, and it is complete.