Reversible Reactions and Equilibrium
A reversible reaction can go both ways. When the two directions balance, the system is at equilibrium — and predicting how it responds to change is the whole of this topic.
1. Reversible reactions
A reversible reaction can proceed in both directions, shown by the ⇌ symbol.
Use the reversible arrow ⇌, not a single arrow. Tutors flagged this specifically for gas-phase reactions.
The classic example: hydrated copper(II) sulfate.
CuSO₄·5H₂O ⇌ CuSO₄ + 5H₂O blue (hydrated) ⇌ white (anhydrous)
Heating drives it forward (blue → white); adding water reverses it (white → blue).
2. Dynamic equilibrium
Equilibrium is reached in a CLOSED SYSTEM when the RATE of the FORWARD reaction EQUALS the RATE of the REVERSE reaction, and the CONCENTRATIONS of reactants and products remain CONSTANT.
Three things must be in the definition:
1. A closed system (nothing enters or leaves) 2. The RATES of forward and reverse reactions are EQUAL 3. Concentrations remain CONSTANT
“The rates of the forward and reverse reactions are equal” is the phrase examiners want. Tutors flagged it twice — always mention the rate of the forward reaction, not just that “nothing changes”.
Equilibrium does NOT mean the reactions have stopped. This was a specific recorded misconception. It is dynamic: both reactions continue at the same rate, so there is no net change.
“Constant” is not the same as “equal”. Confusion between the two was recorded. The concentrations stay constant (unchanging) — they are not necessarily equal to each other.
3. Le Chatelier’s principle
If a change is made to a system at equilibrium, the position of equilibrium shifts to OPPOSE that change.
The system “pushes back”. Work through each factor with that idea.
Temperature
INCREASING temperature shifts equilibrium in the ENDOTHERMIC direction (to absorb the extra heat). DECREASING temperature shifts it in the EXOTHERMIC direction (to release heat).
Decreasing temperature favours the EXOTHERMIC direction, not the endothermic one — a recorded error had this backwards. Increasing temperature favours endothermic.
You must know which direction is which. If the forward reaction is exothermic, the reverse is endothermic.
Pressure (gases only)
INCREASING pressure shifts equilibrium to the side with FEWER gas molecules. DECREASING pressure shifts it to the side with MORE gas molecules.
Count the moles of gas on each side from the balanced equation.
Example: N₂ + 3H₂ ⇌ 2NH₃ — 4 molecules on the left, 2 on the right. So increasing pressure shifts it RIGHT, increasing the ammonia yield.
Count the gas molecules carefully. Confusion over which side has more was recorded — and pressure effects were the most-misunderstood factor in the whole topic, appearing wrong five separate times.
If both sides have the SAME number of gas molecules, pressure has NO effect.
Pressure only affects gases — solids and liquids don’t count in the tally.
Concentration
INCREASING the concentration of a reactant shifts equilibrium to the RIGHT (using it up). Removing a product also shifts it RIGHT (replacing it).
Catalyst
A catalyst does NOT change the position of equilibrium. It speeds up both directions equally, so equilibrium is simply reached faster.
Don’t confuse the rate of reaction with the equilibrium position. Recorded as an error — a catalyst changes how fast, not how far.
Describing a shift properly:
Say the equilibrium “shifts to the right” (or left), and state the consequence — e.g. “so the yield of ammonia increases”. A recorded error stated that SO₃ concentration decreases when equilibrium shifts right; shifting right increases the products.
4. The Haber process (ammonia)
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) — the forward reaction is EXOTHERMIC
Raw materials: nitrogen from the air, hydrogen from natural gas (methane).
Conditions:
| Condition | Value | Why |
|---|---|---|
| Temperature | ~450 °C | a compromise |
| Pressure | ~200 atmospheres | high pressure favours the side with fewer molecules (the ammonia side) |
| Catalyst | iron | speeds up equilibrium without shifting it |
The temperature compromise — a standard exam question:
A lower temperature would give a higher yield (favouring the exothermic forward reaction), but the rate would be too slow. A higher temperature gives a faster rate but a lower yield. 450 °C is a compromise between yield and rate.
Learn the name “Haber process”. Forgetting it was recorded, as was calling ammonia production something else entirely.
5. The Contact process (sulfuric acid)
The key equilibrium step:
2SO₂(g) + O₂(g) ⇌ 2SO₃(g) — exothermic
Conditions:
| Condition | Value |
|---|---|
| Temperature | ~450 °C (again a compromise) |
| Pressure | ~2 atmospheres — only slightly above atmospheric |
| Catalyst | vanadium(V) oxide, V₂O₅ |
The Contact process uses a much LOWER pressure than the Haber process. Confusing the pressure conditions of the two was recorded. The Contact process already gives a high yield at low pressure, so high pressure isn’t worth the cost.
The Contact process is not combustion. A recorded confusion — it is a reversible oxidation of sulfur dioxide.
6. Method for equilibrium questions
- Identify the change being made
- Ask: which direction opposes it?
- For temperature — which direction is endothermic?
- For pressure — which side has fewer gas molecules?
- State the shift and the effect on yield
Always say what happens to the yield, not just the direction of the shift.
7. Mistakes that cost marks
Saying reactions stop at equilibrium.
Omitting “rates are equal” from the definition.
Confusing “constant” with “equal”.
Getting the temperature direction backwards.
Miscounting gas molecules for a pressure prediction.
Applying pressure effects to solids and liquids.
Saying a catalyst shifts the equilibrium.
Confusing rate with position of equilibrium.
Swapping the Haber and Contact pressure conditions.
Forgetting the name of the process or its catalyst.
Frequently asked questions
What is a reversible reaction? One that can go both ways, shown by ⇌.
What is dynamic equilibrium? In a closed system, the rates of the forward and reverse reactions are equal and concentrations stay constant.
Do the reactions stop at equilibrium? No — both continue at the same rate.
What is Le Chatelier’s principle? Equilibrium shifts to oppose any change made to the system.
What does increasing temperature do? Shifts equilibrium in the endothermic direction.
What does increasing pressure do? Shifts it to the side with fewer gas molecules.
What if both sides have equal gas molecules? Pressure has no effect.
Does a catalyst change the yield? No — it only makes equilibrium arrive faster.
What are the Haber process conditions? 450 °C, 200 atm, iron catalyst.
Why is 450 °C a compromise? Lower temperature gives a better yield but too slow a rate; higher gives speed but a poorer yield.
Quick revision checklist
- I use the ⇌ symbol
- I can define equilibrium with all three elements
- I know equilibrium is dynamic, not stopped
- I can state Le Chatelier’s principle
- I can predict the effect of temperature
- I can count gas molecules and predict the effect of pressure
- I know pressure has no effect when the sides are equal
- I can predict the effect of concentration
- I know a catalyst changes rate, not position
- I state the shift and the effect on yield
- I know the Haber conditions and the compromise argument
- I know the Contact conditions and its catalyst
- I don’t swap the two processes’ pressures
These notes cover reversible reactions and equilibrium in the Cambridge IGCSE Chemistry (0620) syllabus and are written for Grade 9–11 / Year 10–11 students. They are based on teaching patterns observed across a large set of one-to-one IGCSE Chemistry lessons, with particular attention to the errors students make most often and the wording examiners reward. Always check the current syllabus for your own exam series.
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