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Detailed notes on Chemical Reactions for Cambridge IGCSE Chemistry, covering key concepts, explanations, examples, and exam-focused revision points.
Reactions that go both ways. Dynamic equilibrium, Le Chatelier's principle, and the conditions chosen for the Haber and Contact processes.
Mapped to the Cambridge IGCSE 0620 syllabus (2026-2028).
Both directions happen at the same time. At equilibrium, forward rate = reverse rate.
Reversible reaction. A reaction in which the products can react to re-form the reactants. Written with ⇌ instead of →.
Worked example. N2(g)+3H2(g)⇌2NH3(g). Both directions occur at the same time.
Dynamic equilibrium. When a reversible reaction is in a CLOSED SYSTEM, eventually:
That's why it's called DYNAMIC, not static. The molecules don't stop reacting; they just keep up.
Conditions for equilibrium.
Worked qualitative. A reaction A+B⇌C+D is set up. After 5 minutes, all four concentrations stop changing. The reaction is at equilibrium. Are A, B, C, D still reacting? YES — but as fast in each direction. Net change is zero, but molecular activity continues.
When you disturb an equilibrium, it shifts to oppose the disturbance.
Le Chatelier's principle. When an equilibrium system is disturbed, it shifts to OPPOSE the change and re-establish equilibrium.
Disturbance 1: Concentration. Adding more of a substance → equilibrium shifts AWAY from it (to consume the excess).
Disturbance 2: Pressure (gas-phase reactions only). Increasing pressure → equilibrium shifts to the side with FEWER MOLES of gas (to reduce pressure).
Disturbance 3: Temperature. Increasing temperature → shifts in the ENDOTHERMIC direction (which absorbs heat to oppose the change).
Disturbance 4: Catalyst. A catalyst speeds up BOTH directions EQUALLY. Equilibrium is reached FASTER but its POSITION is unchanged. Catalysts don't change the YIELD.
| Disturbance | Change made | Equilibrium shifts |
|---|---|---|
| Concentration | Add more of a substance | Away from that substance, to consume the excess |
| Pressure (gases only) | Increase pressure | Towards the side with fewer moles of gas |
| Temperature | Increase temperature | In the endothermic direction, to absorb the heat |
| Catalyst | Add a catalyst | No shift — equilibrium is reached faster, position and yield unchanged |
Worked. Haber process: N2+3H2⇌2NH3, exothermic.
Industry chooses conditions that balance YIELD against RATE and COST.
Haber process: making ammonia. N2(g)+3H2(g)⇌2NH3(g)(ΔH<0)
| Condition | Choice | Reason |
|---|---|---|
| Pressure | 200atm | High pressure favours forward (4 mol → 2 mol). Higher pressures used to be used but cost more. |
| Temperature | ∼450°C | Compromise: low T favours yield (exothermic); but reaction too slow at low T. |
| Catalyst | Iron | Speeds up reaching equilibrium without affecting position. |
| NH₃ removal | Continuous | Removes NH₃ as it forms → forward direction favoured (Le Chatelier). |
Resulting yield: ∼15% per pass. Unreacted N₂ and H₂ recycled.
Contact process: making sulfuric acid. 2SO2(g)+O2(g)⇌2SO3(g)(ΔH<0)
| Condition | Choice | Reason |
|---|---|---|
| Pressure | ∼2atm | Forward direction favoured at higher P (3 mol → 2 mol), but yield is already >95% at 2atm — no need to spend on higher pressure. |
| Temperature | ∼450°C | Compromise as in Haber. |
| Catalyst | Vanadium(V) oxide (V2O5) | Speeds up reaching equilibrium. |
After SO₃ is formed, it's dissolved in concentrated H₂SO₄ to make oleum, which is then diluted to make more H₂SO₄.
Worked qualitative. Why isn't the Haber process run at 1000°C with iron catalyst? Higher T speeds up the reaction but DECREASES yield (forward is exothermic). Iron catalyst loses effectiveness above ∼500°C. Energy costs would be huge. The chosen 450°C is the engineering sweet spot.
Verbatim phrases and definitions Cambridge mark schemes credit.
Reversible reactions appear every Paper 4 (8-12 marks) — predict equilibrium shifts (Le Chatelier), explain Haber/Contact conditions. Examiner reports flag students saying a catalyst increases yield (it doesn't), and confusing 'shifts to oppose' with 'reverses the reaction'.
Sources: Cambridge IGCSE Chemistry 0620 syllabus 2026-2028 (8.1); 0620/42 Oct/Nov 2024 — Q12 (Le Chatelier, Haber); 0620 Examiner Reports 2022-2024. Last reviewed 2026-05-07.
Step-by-step solutions to past-paper-style questions on reversible reactions and equilibrium, written exactly the way a tutor would explain them at the board.
Question
Blue hydrated copper(II) sulfate is heated until it turns white. Water is then added to the white solid. Describe the colour changes, write the reversible equation, and explain how this is used to test for water.
Step-by-step solution
Step 1
Heating drives off the water of crystallisation: blue CuSO4⋅5H2O becomes white anhydrous CuSO4.
Step 2
Adding water rehydrates the white solid back to blue, so the reaction is reversible.
CuSO4⋅5H2O⇌CuSO4+5H2O
Step 3
Because anhydrous (white) copper(II) sulfate turns blue when water is added, it is used as a chemical test for water.
Answer
Blue → white on heating; white → blue when water is added. CuSO4⋅5H2O⇌CuSO4+5H2O. White anhydrous CuSO4 turning blue is the test for water.
Examiner tip
The test only confirms water is PRESENT — it does not show the water is pure. Use the boiling point (100°C) to confirm purity.
Question
State the colour change seen when blue anhydrous cobalt(II) chloride is exposed to water, and the change seen when the resulting solid is heated. Explain why these changes show a reversible reaction.
Step-by-step solution
Step 1
Blue anhydrous cobalt(II) chloride turns PINK when water is added (forming the hydrated form).
Step 2
Heating the pink hydrated solid drives off the water and it turns BLUE again.
Step 3
The same two substances interconvert when the conditions are changed (adding water vs heating), so the reaction is reversible and is shown by ⇌.
Answer
Blue (anhydrous) ⇌ pink (hydrated): blue → pink when water is added; pink → blue on heating. The interconversion shows it is reversible.
Examiner tip
Watch the direction: ANHYDROUS cobalt(II) chloride is blue, HYDRATED is pink — the opposite colour code to copper(II) sulfate. Cobalt(II) chloride paper is the standard test for water (blue → pink).
Question
A reversible reaction reaches equilibrium in a closed container. Explain what is meant by dynamic equilibrium and why the concentrations of reactants and products stay constant.
Step-by-step solution
Step 1
At the start the forward reaction is fast (lots of reactant); as reactant is used up the forward rate falls. The reverse rate rises as product builds up.
Step 2
Equilibrium is reached when the forward and reverse reactions occur at EQUAL RATES.
Step 3
Because product forms at the same rate as it is broken down, the concentrations of all substances stay constant — but both reactions are still happening, so it is DYNAMIC, not stopped.
Step 4
This only holds in a CLOSED system (nothing enters or leaves), so no reactant or product can escape.
Answer
Dynamic equilibrium is when the forward and reverse reactions occur at equal rates in a closed system, so the concentrations of reactants and products remain constant even though both reactions continue.
Examiner tip
Equal RATES, not equal AMOUNTS. The concentrations are constant but are usually NOT the same as each other. The reaction has not stopped.
Question
For 2NO2⇌N2O4 the forward reaction is exothermic (ΔH<0). Predict and explain the effect of (a) increasing the temperature and (b) decreasing the temperature on the position of equilibrium.
Step-by-step solution
Step 1
Le Chatelier: the system shifts to oppose the change. Adding heat is opposed by absorbing heat — i.e. the system favours the ENDOTHERMIC direction.
Step 2
(a) Increasing temperature favours the endothermic (reverse) direction → equilibrium shifts LEFT → less N2O4, more NO2.
Step 3
(b) Decreasing temperature favours the exothermic (forward) direction (releasing heat to oppose the cooling) → equilibrium shifts RIGHT → more N2O4.
Answer
(a) Higher T shifts the equilibrium to the LEFT (endothermic direction) → less N2O4. (b) Lower T shifts it to the RIGHT (exothermic direction) → more N2O4.
Examiner tip
Always decide the endothermic direction first, then 'raise T → endothermic favoured'. For an exothermic forward reaction, a LOWER temperature increases the yield of product.
Question
N2(g)+3H2(g)⇌2NH3(g) (ΔH<0). The industrial conditions are about 450°C, 200atm and an iron catalyst. Explain, using Le Chatelier's principle, why these conditions are chosen.
Step-by-step solution
Step 1
Pressure: there are 4 moles of gas on the left and 2 on the right. High pressure shifts the equilibrium to the side with FEWER gas moles → toward NH3, so 200atm gives a good yield. Even higher pressure is avoided because the plant and pipework would be too expensive and dangerous.
Step 2
Temperature: the forward reaction is exothermic, so a LOW temperature would favour more NH3. But a low temperature makes the rate too slow.
Step 3
450°C is a COMPROMISE: high enough for a fast rate (and to reach equilibrium quickly), low enough that the yield is still acceptable.
Step 4
Iron catalyst: speeds up BOTH the forward and reverse reactions equally, so equilibrium is reached faster. It does NOT change the position of equilibrium or the yield.
Answer
High pressure (∼200atm) shifts equilibrium toward the 2 moles of NH3 for higher yield (limited by cost/safety). ∼450°C is a compromise between a high yield (favoured by low T for an exothermic reaction) and a fast rate (favoured by high T). The iron catalyst speeds up reaching equilibrium but does not change the yield.
Examiner tip
Three separate ideas must appear: pressure → fewer gas moles → more NH3; temperature is a COMPROMISE between yield and rate; catalyst affects RATE only, not position. Unreacted N2 and H2 are recycled.
Question
2SO2(g)+O2(g)⇌2SO3(g) (ΔH<0). The conditions used are about 450°C, 2atm and a vanadium(V) oxide (V2O5) catalyst. Explain why a relatively low pressure of only 2atm is sufficient and why 450°C is used.
Step-by-step solution
Step 1
Moles of gas: 3 on the left (2SO2+O2), 2 on the right (2SO3). Higher pressure would shift right, but the yield at low pressure is already very high (~95%+).
Step 2
Because the yield is already so high, increasing the pressure gives little extra product but adds great cost and danger — so only a low pressure of about 2atm is used.
Step 3
The forward reaction is exothermic, so a low temperature favours SO3, but too low and the rate is too slow. 450°C is the COMPROMISE between yield and rate.
Step 4
The V2O5 catalyst speeds up the attainment of equilibrium but does not change the position of equilibrium.
Answer
The equilibrium yield of SO3 is already very high at low pressure, so only ∼2atm is used — higher pressure would add cost for little extra yield. 450°C is a compromise between a high yield (low T, exothermic) and a fast rate (high T); the V2O5 catalyst only speeds up reaching equilibrium.
Examiner tip
Contrast with the Haber process: both use ∼450°C as a compromise, but the Contact process needs only LOW pressure because the yield is already near-complete. Don't claim pressure 'has no effect' — it does shift right, but the gain is not worth the cost.
High-scoring sample answers for reversible reactions and equilibrium on the Cambridge IGCSE 0620 paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.
State the symbol used to show that a reaction is reversible. (1 mark)
Model answer
The reversible-reaction arrow ⇌ (two half-arrows pointing in opposite directions).
Why this scores
One mark for ⇌. A normal single arrow (→) is wrong — it implies the reaction goes only one way.
Describe a chemical test, including the colour change, that shows a liquid contains water. (2 marks)
Model answer
Add the liquid to anhydrous copper(II) sulfate. If water is present, the white solid turns blue (CuSO4+5H2O→CuSO4⋅5H2O). (Alternatively, cobalt(II) chloride paper turns from blue to pink.)
Why this scores
Two marks: (1) name a correct reagent (anhydrous copper(II) sulfate OR cobalt(II) chloride); (2) the correct colour change (white → blue, or blue → pink). This test shows water is present but not that it is pure.
Explain what is meant by the term dynamic equilibrium. (3 marks)
Model answer
Dynamic equilibrium is reached when the forward and reverse reactions occur at the same (equal) rate, so the concentrations of reactants and products remain constant. It is dynamic because both reactions are still taking place (the reaction has not stopped); it can only be reached in a closed system.
Why this scores
Three marks: (1) forward and reverse rates are equal; (2) concentrations stay constant; (3) reactions continue / closed system. Do NOT say 'the reaction stops' or 'the concentrations become equal'.
For the equilibrium H2(g)+I2(g)⇌2HI(g), predict and explain the effect on the position of equilibrium of adding more hydrogen, and of removing some hydrogen iodide. (4 marks)
Model answer
Adding more hydrogen: by Le Chatelier's principle the system shifts to oppose the increase, so it shifts to the right (forwards) to use up the added hydrogen, producing more HI. Removing some HI: the system shifts to oppose the decrease, so it shifts to the right (forwards) to replace the HI that was removed, again producing more HI.
Why this scores
Four marks: (1) adding H2 shifts right; (2) because the system opposes the increase / uses up the added reactant; (3) removing HI shifts right; (4) because the system opposes the decrease / replaces the removed product. Each prediction must be linked to Le Chatelier's principle.
In the Haber process, N2(g)+3H2(g)⇌2NH3(g) (ΔH<0). Explain how the yield of ammonia would change if (a) the pressure were increased and (b) the temperature were decreased, and explain why the conditions actually used (∼450°C, ∼200atm) are a compromise. (5 marks)
Model answer
(a) Increasing the pressure shifts the equilibrium toward the side with fewer moles of gas — 2 moles of NH3 on the right versus 4 moles on the left — so the yield of ammonia increases. (b) The forward reaction is exothermic, so decreasing the temperature shifts the equilibrium to the right (the exothermic direction), increasing the yield of ammonia. However, the conditions are a compromise: a very low temperature would give a high yield but the rate would be too slow, so ∼450°C balances an acceptable yield against a fast enough rate; and a very high pressure would give a higher yield but is too expensive and dangerous, so ∼200atm is used.
Why this scores
Five marks: (1) higher pressure → fewer gas moles → more NH3; (2) lower temperature favours exothermic forward reaction → more NH3; (3) low temperature is too slow (rate); (4) 450°C is a temperature compromise; (5) very high pressure is too costly/dangerous so 200atm is a pressure compromise.
The Contact process makes sulfur trioxide: 2SO2(g)+O2(g)⇌2SO3(g) (ΔH<0). The conditions used are about 450°C, 2atm and a vanadium(V) oxide catalyst. Explain the choice of temperature, pressure and catalyst, referring to both yield and rate. (6 marks)
Model answer
Temperature: the forward reaction is exothermic, so a low temperature would increase the yield of SO3 (Le Chatelier favours the exothermic direction). But a low temperature makes the rate too slow, so ∼450°C is a compromise that gives a high enough yield with a fast enough rate. Pressure: there are 3 moles of gas on the left and 2 moles on the right, so higher pressure would shift the equilibrium right; however, even at a low pressure the yield is already very high (around 95-99%), so increasing the pressure is not worth the extra cost and danger — only about 2atm is needed. Catalyst: vanadium(V) oxide (V2O5) speeds up the rate of both the forward and reverse reactions equally, so equilibrium is reached faster, but it does not change the position of equilibrium or the yield.
Why this scores
Six marks: (1) exothermic → low T favours yield; (2) low T too slow, so 450°C is a compromise; (3) fewer gas moles on the right → higher pressure would raise yield; (4) yield already very high so low pressure (~2atm) is used to save cost; (5) catalyst speeds up reaching equilibrium; (6) catalyst does NOT change the position/yield. The classic error is claiming the catalyst increases the yield.
The formulae you need to memorise for reversible reactions and equilibrium on the Cambridge IGCSE 0620 paper, with every variable defined in plain English and a note on when to use it.
System shifts to oppose any change imposed on it.
When to use
Predicting how a change in temperature, pressure or concentration moves the position of equilibrium.
Definitions to memorise and the exact keywords mark schemes credit for reversible reactions and equilibrium answers — sharpened from recent examiner reports for the 2026 0620 sitting.
A reaction in which the products can react together to re-form the original reactants under suitable conditions. Shown by the symbol ⇌.
The state, reached in a closed system, in which the forward and reverse reactions occur at EQUAL RATES so the concentrations of reactants and products remain constant (though both reactions continue).
A system in which no reactants or products can enter or leave — a requirement for a reaction to reach dynamic equilibrium.
If a change (in temperature, pressure or concentration) is made to a system at equilibrium, the position of equilibrium shifts to oppose that change.
Industrial conditions chosen to balance a reasonable yield against a fast enough reaction rate (and acceptable cost/safety), e.g. ∼450°C in the Haber and Contact processes.
The traps other students keep falling into on reversible reactions and equilibrium questions — taken from recent Cambridge IGCSE 0620 examiner reports and mark schemes — and how to avoid them.
0620/42 — recurring
Why it happens
The concentrations stop changing, so it looks as if the reaction has finished.
How to avoid it
Equilibrium is DYNAMIC — the forward and reverse reactions continue at equal rates; only the NET change is zero.
Why it happens
Confusing equal RATES with equal AMOUNTS.
How to avoid it
At equilibrium the RATES are equal, but the concentrations are constant and usually NOT equal to each other.
Why it happens
Confusing rate with yield.
How to avoid it
Raising T favours the ENDOTHERMIC direction. For an exothermic forward reaction, a LOWER temperature gives a higher yield — but a compromise temperature is used so the rate is not too slow.
Why it happens
Over-generalising the rules.
How to avoid it
Pressure only shifts the position if the sides have DIFFERENT numbers of gas moles (equal moles → no shift). A catalyst speeds up BOTH directions equally → faster equilibrium but NO change in position or yield.
The things students keep getting wrong in this sub-topic, answered.