What is a reversible reaction in Chemistry?

In Chemistry, a reversible reaction is a type of chemical reaction where the reactants form products, which can then react together to form the original reactants. This means the reaction can proceed in both forward and backward directions. Reversible reactions are a key concept in the Cambridge IGCSE Chemistry curriculum as they help explain how chemical equilibrium is established.

How does dynamic equilibrium relate to reversible reactions?

Dynamic equilibrium occurs in a reversible reaction when the rate of the forward reaction equals the rate of the backward reaction, resulting in no net change in the concentrations of reactants and products. At this point, the system is said to be in equilibrium. This concept is crucial for understanding how chemical reactions behave in closed systems, as taught in the Cambridge IGCSE Chemistry syllabus.

What factors affect the position of equilibrium in a chemical reaction?

The position of equilibrium in a chemical reaction can be affected by changes in concentration, temperature, and pressure. According to Le Chatelier's Principle, if a system at equilibrium is subjected to a change, the system will adjust itself to counteract that change and restore a new equilibrium. This principle is an important part of the Cambridge IGCSE Chemistry curriculum.

Why is the concept of equilibrium important in industrial chemical processes?

The concept of equilibrium is crucial in industrial chemical processes because it helps in optimizing the yield of products. By manipulating conditions such as temperature and pressure, industries can shift the position of equilibrium to favor the production of desired products. This understanding is essential for students studying Cambridge IGCSE Chemistry, as it applies to real-world chemical manufacturing.

Can you provide an example of a reversible reaction and its equilibrium in everyday life?

A common example of a reversible reaction is the synthesis of ammonia in the Haber process, where nitrogen and hydrogen gases react to form ammonia. This reaction is reversible, and the equilibrium can be shifted by changing temperature and pressure. Understanding such examples is part of the Cambridge IGCSE Chemistry curriculum, highlighting the practical applications of reversible reactions and equilibrium in everyday life.

Why is "Saying the reaction stops at equilibrium" a common mistake in Reversible Reactions and Equilibrium?

Why it happens: Concentrations are constant. How to avoid it: Equilibrium is DYNAMIC — forward and reverse continue at equal rates. Source: 0620/42 — recurring

Why is "Thinking exothermic always favours high T" a common mistake in Reversible Reactions and Equilibrium?

Why it happens: Confusing rate with yield. How to avoid it: Higher T favours ENDOTHERMIC direction. So for exothermic forward reaction → LOWER T → more product.

Why is "Saying pressure changes always shift the equilibrium" a common mistake in Reversible Reactions and Equilibrium?

Why it happens: Generalising. How to avoid it: Only if the number of gas moles differs between sides. Equal moles → no shift.

Why is "Saying a catalyst shifts equilibrium" a common mistake in Reversible Reactions and Equilibrium?

Why it happens: Confusing 'speed up' with 'shift'. How to avoid it: Catalyst speeds BOTH directions equally → no change in position.

Chemical ReactionsCambridge IGCSE Chemistry (0620)

Reversible Reactions and Equilibrium

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Reversible Reactions and Equilibrium — Cambridge IGCSE 0620 Chemistry Extended (2026)

Reactions that go both ways. Dynamic equilibrium, Le Chatelier's principle, and the conditions chosen for the Haber and Contact processes.

What you’ll learn

Mapped to the Cambridge IGCSE 0620 syllabus (2026-2028).

8.1 — Describe a reversible reaction.
8.1 — Describe dynamic equilibrium.
8.1 — Apply Le Chatelier's principle qualitatively.
8.1 — Describe conditions in the Haber and Contact processes.

Reversible reactions and dynamic equilibrium

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 $\rightleftharpoons$ instead of $\to$ .

Worked example. $\text{N}_{2}(g) + 3\text{H}_{2}(g) \rightleftharpoons 2\text{NH}_{3}(g)$ . Both directions occur at the same time.

Dynamic equilibrium. When a reversible reaction is in a CLOSED SYSTEM, eventually:

Forward rate = reverse rate.
Concentrations of all species stay constant (don't change with time).
The reactions are STILL HAPPENING — just balancing each other.

That's why it's called DYNAMIC, not static. The molecules don't stop reacting; they just keep up.

Conditions for equilibrium.

Reaction must be reversible.
System must be CLOSED (no substance enters or leaves).
Conditions (T, P) must be constant.

Worked qualitative. A reaction $A + B \rightleftharpoons C + D$ 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.

Reversible: both directions possible.
Equilibrium: forward rate = reverse rate.
Concentrations CONSTANT, not zero.
Reactions still occurring (dynamic).
Closed system needed.

Le Chatelier's principle

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).

E.g. for $A + B \rightleftharpoons C$ : adding more $A$ → forward shifted → more $C$ formed.

Disturbance 2: Pressure (gas-phase reactions only). Increasing pressure → equilibrium shifts to the side with FEWER MOLES of gas (to reduce pressure).

E.g. $\text{N}_{2} + 3\text{H}_{2} \rightleftharpoons 2\text{NH}_{3}$ . Forward direction: 4 mol → 2 mol. Higher pressure favours forward (NH₃).
E.g. $\text{H}_{2} + \text{I}_{2} \rightleftharpoons 2\text{HI}$ . Both sides: 2 mol. Pressure has NO effect.

Disturbance 3: Temperature. Increasing temperature → shifts in the ENDOTHERMIC direction (which absorbs heat to oppose the change).

E.g. forward exothermic ( $\Delta H < 0$ ): higher T → reverse favoured → less product.
E.g. forward endothermic ( $\Delta H > 0$ ): higher T → forward favoured → more product.

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.

Worked. Haber process: $\text{N}_{2} + 3\text{H}_{2} \rightleftharpoons 2\text{NH}_{3}$ , exothermic.

High pressure: favours NH₃ (forward, fewer moles).
High temp: favours reverse (away from exothermic NH₃ formation) — REDUCES yield.
BUT low temp = slow reaction. So compromise: $\sim 450\degree\text{C}$ — fast enough to be commercial, hot enough to keep yield reasonable.

Concentration: shifts AWAY from added species.
Pressure: shifts to side with FEWER moles of gas.
Temperature: shifts in ENDOTHERMIC direction.
Catalyst: equilibrium faster, position unchanged.
Often a compromise needed for industrial processes.

Haber and Contact processes — applying Le Chatelier

Industry chooses conditions that balance YIELD against RATE and COST.

Haber process: making ammonia. $\text{N}_{2}(g) + 3\text{H}_{2}(g) \rightleftharpoons 2\text{NH}_{3}(g) \quad (\Delta H < 0)$

Condition	Choice	Reason
Pressure	$200\,\text{atm}$	High pressure favours forward (4 mol → 2 mol). Higher pressures used to be used but cost more.
Temperature	$\sim 450\degree\text{C}$	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: $\sim 15\%$ per pass. Unreacted N₂ and H₂ recycled.

Contact process: making sulfuric acid. $2\text{SO}_{2}(g) + \text{O}_{2}(g) \rightleftharpoons 2\text{SO}_{3}(g) \quad (\Delta H < 0)$

Condition	Choice	Reason
Pressure	$\sim 2\,\text{atm}$	Forward direction favoured at higher P (3 mol → 2 mol), but yield is already $> 95\%$ at $2\,\text{atm}$ — no need to spend on higher pressure.
Temperature	$\sim 450\degree\text{C}$	Compromise as in Haber.
Catalyst	Vanadium(V) oxide ( $\text{V}_{2}\text{O}_{5}$ )	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 $1000\degree\text{C}$ with iron catalyst? Higher T speeds up the reaction but DECREASES yield (forward is exothermic). Iron catalyst loses effectiveness above $\sim 500\degree\text{C}$ . Energy costs would be huge. The chosen $450\degree\text{C}$ is the engineering sweet spot.

Haber: $\text{N}_{2} + 3\text{H}_{2} \rightleftharpoons 2\text{NH}_{3}$ , $200\,\text{atm}$ , $450\degree\text{C}$ , Fe.
Contact: $2\text{SO}_{2} + \text{O}_{2} \rightleftharpoons 2\text{SO}_{3}$ , $2\,\text{atm}$ , $450\degree\text{C}$ , $\text{V}_{2}\text{O}_{5}$ .
Both compromise temperature for rate vs yield.
Catalysts speed equilibrium without changing position.

How it’s examined

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.

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Step-by-step worked examples — Reversible Reactions and Equilibrium

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.

1Reversible hydration
Core• reversible
Question
Describe what happens when blue copper(II) sulfate is heated and then cooled with water added.
Step-by-step solution
1. Step 1
  Heating: $\text{CuSO}_{4}\cdot 5\text{H}_{2}\text{O}\,(\text{blue}) \to \text{CuSO}_{4}\,(\text{white}) + 5\text{H}_{2}\text{O}$ .
2. Step 2
  Adding water: white turns back to blue.
Answer
Forward: blue → white + water (endothermic). Reverse: white + water → blue (exothermic). Reversible.
2Le Chatelier — pressure on the Haber process
Extended• Adapted from 0620/42 May/Jun 2024 Q16• Haber
Question
$\text{N}_{2}(\text{g}) + 3\text{H}_{2}(\text{g}) \rightleftharpoons 2\text{NH}_{3}(\text{g})$ . Explain the effect of increasing pressure on the equilibrium yield of ammonia.
Step-by-step solution
1. Step 1
  Count gas moles: 4 on left, 2 on right.
2. Step 2
  Increasing pressure shifts the position to the side with FEWER gas moles → toward $\text{NH}_{3}$ .
Answer
Higher pressure → more $\text{NH}_{3}$ at equilibrium.
3Contact process — temperature compromise
Extended• Contact
Question
$2\text{SO}_{2} + \text{O}_{2} \rightleftharpoons 2\text{SO}_{3}$ ( $\Delta H < 0$ ). Why is a temperature of $\sim 450\degree\text{C}$ used instead of a much lower temperature?
Step-by-step solution
1. Step 1
  Forward exothermic → LOWER temperature would increase yield (Le Chatelier).
2. Step 2
  But too low → reaction is too SLOW.
3. Step 3
  $450\degree\text{C}$ is a COMPROMISE between yield and rate.
Answer
Compromise between high yield (favoured by low T) and acceptable rate (favoured by high T).
4Effect of changing concentration
Extended• concentration
Question
$\text{H}_{2}(\text{g}) + \text{I}_{2}(\text{g}) \rightleftharpoons 2\text{HI}(\text{g})$ . Predict the shift if more $\text{H}_{2}$ is added.
Step-by-step solution
1. Step 1
  System opposes the change → consumes the added $\text{H}_{2}$ .
2. Step 2
  Equilibrium shifts FORWARDS → more HI produced.
Answer
Equilibrium shifts to the right; more HI forms.

Key Formulae — Reversible Reactions and Equilibrium

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.

Le Chatelier's principle
$\text{System opposes any change imposed on it.}$
When to use
Predicting how T, P or concentration changes affect equilibrium position.

Key Definitions and Keywords — Reversible Reactions and 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.

Reversible reaction
Examiner keyword
A reaction in which products can react back to form the original reactants under suitable conditions. Shown by $\rightleftharpoons$ .
Dynamic equilibrium
Examiner keyword
Forward and reverse reactions occur at EQUAL RATES; concentrations of reactants and products remain constant.
Le Chatelier's principle
Examiner keyword
If a system at equilibrium is disturbed, the system shifts to oppose the disturbance.

Common Mistakes and Misconceptions — Reversible Reactions and Equilibrium

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.

✕Saying the reaction stops at equilibrium
0620/42 — recurring
Why it happens
Concentrations are constant.
How to avoid it
Equilibrium is DYNAMIC — forward and reverse continue at equal rates.
✕Thinking exothermic always favours high T
Why it happens
Confusing rate with yield.
How to avoid it
Higher T favours ENDOTHERMIC direction. So for exothermic forward reaction → LOWER T → more product.
✕Saying pressure changes always shift the equilibrium
Why it happens
Generalising.
How to avoid it
Only if the number of gas moles differs between sides. Equal moles → no shift.
✕Saying a catalyst shifts equilibrium
Why it happens
Confusing 'speed up' with 'shift'.
How to avoid it
Catalyst speeds BOTH directions equally → no change in position.

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Reversible Reactions and Equilibrium — frequently asked questions

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Reversible Reactions and Equilibrium

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Reversible hydration

2Le Chatelier — pressure on the Haber process

3Contact process — temperature compromise

4Effect of changing concentration

Le Chatelier's principle

Reversible reaction

Dynamic equilibrium

Le Chatelier's principle

✕Saying the reaction stops at equilibrium

✕Thinking exothermic always favours high T

✕Saying pressure changes always shift the equilibrium

✕Saying a catalyst shifts equilibrium

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Reversible Reactions and Equilibrium — frequently asked questions