Reversible Reactions and EquilibriumCambridge IGCSE Chemistry: Revision notes
Section 1
What are reversible reactions and how do they differ from irreversible reactions?
A reversible reaction is one that can proceed in both the forward and reverse directions simultaneously. Reversible reactions are represented by the symbol ⇌ (double arrow), whilst irreversible reactions use a single arrow (→).
In a reversible reaction:
- Products can react with each other to reform the original reactants
- The reaction does not go to completion
- Both forward and reverse reactions occur at the same time
Common examples of reversible reactions include:
- Thermal decomposition of hydrated copper(II) sulfate: CuSO₄·5H₂O(s) ⇌ CuSO₄(s) + 5H₂O(g) (blue to white)
- Thermal decomposition of hydrated cobalt(II) chloride: CoCl₂·6H₂O(s) ⇌ CoCl₂(s) + 6H₂O(g) (pink to blue)
- Addition of water to anhydrous copper(II) sulfate reverses the reaction (white to blue)
- Addition of water to anhydrous cobalt(II) chloride reverses the reaction (blue to pink)
A reversible reaction is like a seesaw: both sides can go up and down. An irreversible reaction is like a ball rolling down a hill – it only goes one way.
In exam questions, always use ⇌ not → for reversible reactions. Examiners specifically mark this symbol choice.
Section 2
What is chemical equilibrium and when is it achieved?
Chemical equilibrium occurs in a closed system (one where no reactants or products can enter or leave) when:
- The rate of the forward reaction equals the rate of the reverse reaction
- The concentrations of reactants and products no longer change (remain constant)
- Both forward and reverse reactions continue to occur – the system is in dynamic equilibrium
Key points about equilibrium:
- Equilibrium can only be achieved in a closed system
- Equilibrium is dynamic – reactions continue but no overall change is observed
- At equilibrium, the system is stable and energy is not being released or absorbed overall
- The position of equilibrium describes whether there are more reactants or products present at equilibrium
Time for equilibrium to be reached depends on:
- Temperature (higher temperature = faster)
- Presence of a catalyst (speeds up both forward and reverse reactions equally)
- Nature of the reactants (some reactions are naturally faster)
In the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), equilibrium is reached when the rate at which N₂ and H₂ combine to form NH₃ equals the rate at which NH₃ decomposes back to N₂ and H₂. At this point, the concentrations of all three gases remain constant.
Students often think equilibrium means the reaction has stopped. In fact, at dynamic equilibrium, both forward and reverse reactions continue – just at the same rate, so there is no visible change.
Section 3
How do changes in conditions affect the position of equilibrium?
The position of equilibrium can be shifted by changing the conditions of a system. Understanding these shifts is crucial for exam success.
Effect of Temperature Change:
- Increasing temperature favours the endothermic direction (the reaction that absorbs heat)
- Decreasing temperature favours the exothermic direction (the reaction that releases heat)
- Both forward and reverse reaction rates increase with temperature, but the equilibrium shifts towards the endothermic direction
- Example: CuSO₄·5H₂O(s) ⇌ CuSO₄(s) + 5H₂O(g) – heating shifts right (endothermic), cooling shifts left
Effect of Pressure Change:
- Increasing pressure favours the side with fewer moles of gas
- Decreasing pressure favours the side with more moles of gas
- Pressure changes have no effect on reactions with equal moles of gas on both sides
- Example: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) – 4 moles left, 2 moles right, so increasing pressure shifts right
Effect of Concentration Change:
- Increasing the concentration of reactants shifts equilibrium to the right (towards products)
- Increasing the concentration of products shifts equilibrium to the left (towards reactants)
- Removing reactants or products shifts equilibrium in the opposite direction
Effect of Catalyst:
- A catalyst does not affect the position of equilibrium
- A catalyst speeds up both forward and reverse reactions equally
- The equilibrium position is reached more quickly, but the amounts of reactants and products at equilibrium remain the same
To predict pressure effects, count total moles of gas on each side of the equation. The side with fewer moles is favoured by increased pressure. Examiners expect you to show this counting explicitly.
For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g): Left side has 3 moles of gas (2 + 1), right side has 2 moles. Increasing pressure shifts equilibrium right to produce more SO₃. Increasing temperature shifts left if the forward reaction is exothermic.
Section 4
What is the Haber Process and why are specific conditions used?
The Haber Process is an industrial process for manufacturing ammonia (NH₃) from nitrogen (N₂) and hydrogen (H₂).
Equation: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
Sources of Raw Materials:
- Nitrogen (N₂): Obtained from air (air contains approximately 78% nitrogen)
- Hydrogen (H₂): Obtained from methane (CH₄) via steam reforming (methane is a component of natural gas)
Typical Haber Process Conditions:
- Temperature: 450°C
- Pressure: 20000 kPa (or 200 atm)
- Catalyst: Iron (Fe)
Why These Conditions Are Chosen:
| Condition | Effect on Rate | Effect on Equilibrium Position | Why Chosen |
|---|---|---|---|
| 450°C | Increases forward and reverse reaction rates significantly | Forward reaction is exothermic, so higher temperature shifts equilibrium left (fewer products) | Compromise: high enough for acceptable reaction rate, low enough to avoid excessive shifting left |
| 20000 kPa | Pressure alone does not affect rate of gases | Left side has 4 moles gas (1 + 3), right side has 2 moles; increased pressure shifts right (more NH₃) | Maximises ammonia yield by shifting equilibrium towards products |
| Iron catalyst | Increases both forward and reverse reaction rates equally | Does not affect equilibrium position | Allows equilibrium to be reached more quickly without affecting final yield |
Economic and Safety Considerations:
- High pressure is expensive to maintain but increases NH₃ yield, justifying the cost
- Temperature must be high enough to achieve reasonable reaction rate within production timescales
- The process is continuous: ammonia is liquefied and removed from the system, shifting equilibrium right
- Excess nitrogen and hydrogen are recycled to improve overall yield
Examiners expect you to explain Haber process conditions in terms of both rate and equilibrium. Always state whether the forward reaction is exothermic or endothermic when discussing temperature.
The forward reaction N₂ + 3H₂ ⇌ 2NH₃ is exothermic. Increasing temperature increases reaction rate (both forward and reverse) but shifts equilibrium left, reducing ammonia yield. At 450°C, this balance gives a reasonable rate without sacrificing too much equilibrium yield.
Section 5
What is the Contact Process and why are specific conditions used?
The Contact Process is an industrial process for manufacturing sulfuric acid (H₂SO₄) from sulfur dioxide (SO₂) and oxygen (O₂). This is the key step in sulfuric acid production.
Equation: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
(Sulfur trioxide is then reacted with water to produce sulfuric acid)
Sources of Raw Materials:
- Sulfur dioxide (SO₂): Obtained by:
- Burning sulfur in air: S(s) + O₂(g) → SO₂(g)
- Roasting sulfide ores (heating metal sulfides in air)
- Oxygen (O₂): Obtained from air (air contains approximately 21% oxygen)
Typical Contact Process Conditions:
- Temperature: 450°C
- Pressure: 200 kPa (or 2 atm)
- Catalyst: Vanadium(V) oxide (V₂O₅)
Why These Conditions Are Chosen:
| Condition | Effect on Rate | Effect on Equilibrium Position | Why Chosen |
|---|---|---|---|
| 450°C | Increases forward and reverse reaction rates significantly | Forward reaction is exothermic, so higher temperature shifts equilibrium left (fewer products) | Compromise: high enough for acceptable reaction rate, low enough to avoid excessive shifting left |
| 200 kPa | Pressure alone does not affect rate of gases | Left side has 3 moles gas (2 + 1), right side has 2 moles; increased pressure shifts right (more SO₃) | Lower pressure than Haber because SO₂ yield is already adequate; reduces equipment costs |
| Vanadium(V) oxide catalyst | Increases both forward and reverse reaction rates equally | Does not affect equilibrium position | Allows equilibrium to be reached quickly; more selective than iron catalyst, reducing side reactions |
Economic and Safety Considerations:
- Lower pressure (200 kPa vs 20000 kPa in Haber) means lower equipment costs as less robust apparatus is needed
- The 450°C temperature is the same as in Haber but chosen for the same reason: balancing rate and equilibrium
- Vanadium(V) oxide catalyst is more expensive than iron but more efficient and selective
- The process is continuous: SO₃ is removed and reacted with water, shifting equilibrium right
- Excess SO₂ and O₂ are recycled to improve conversion
Compare Haber and Contact processes: both use 450°C (exothermic forward reactions), but Contact uses lower pressure (200 kPa vs 20000 kPa) because SO₃ yield is adequate at lower pressure, reducing costs.
In the Contact Process: 2SO₂ + O₂ ⇌ 2SO₃ is exothermic. At 450°C, the reaction rate is fast enough for industrial production. Increasing pressure to 200 kPa shifts equilibrium right (3 moles → 2 moles), but the cost saving from lower pressure is justified because SO₃ formation is still efficient.
Must Know
- Reversible reactions are represented by ⇌ and can proceed in both forward and reverse directions simultaneously; irreversible reactions use → and go to completion
- Chemical equilibrium in a closed system occurs when the rate of forward reaction equals the rate of reverse reaction and concentrations of reactants and products no longer change
- Changing temperature shifts equilibrium towards the endothermic direction; changing pressure shifts equilibrium towards the side with fewer moles of gas; changing concentration shifts equilibrium away from the added substance; catalysts do not affect equilibrium position but speed up both directions equally
- Haber Process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) at 450°C, 20000 kPa, with iron catalyst; nitrogen from air, hydrogen from methane; conditions chosen to balance reaction rate with equilibrium yield and justify equipment costs
- Contact Process: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) at 450°C, 200 kPa, with vanadium(V) oxide catalyst; SO₂ from burning sulfur or roasting ores, oxygen from air; lower pressure than Haber reduces equipment costs whilst maintaining adequate yield
- Exam success requires: explaining conditions in both rate and equilibrium terms, counting moles of gas correctly for pressure effects, and understanding why industrial processes use the conditions they do (economic balance, safety, and practicality)
That's the notes covered.
Carry on to the next subtopic.