6.3 Reversible reactions and equilibrium
- Syllabus
- 0620–2026–2027
- Topic
- 6.3
- Level
- —
Learning objectives
6.3.1Some chemical reactions are reversible• State: some chemical reactions are reversible as shown by the symbol ⇌6.3.2Changing the conditions can change the• Describe how changing the conditions can change the direction of a reversible reaction for: (a) the effect of heat on hydrated compounds (b) the addition of water to anhydrous compounds limited to copper(II) sulfate and cobalt(II) chloride6.3.3Reversible reaction in a closed system• State: a reversible reaction in a closed system is at equilibrium when: (a) the rate of the forward reaction is equal to the rate of the reverse reaction (b) the concentrations of reactants and products are no longer changing6.3.4Predict and explain, for a reversible• Predict and explain, for a reversible reaction, how the position of equilibrium is affected by: (a) changing temperature (b) changing pressure (c) changing concentration (d) using a catalyst using information provided6.3.5Symbol equation for the production of• State the symbol equation for the production of ammonia in the Haber process, N2(g) + 3H2(g) ⇌ 2NH3(g)6.3.6Sources of the hydrogen (methane) and• State the sources of the hydrogen (methane) and nitrogen (air) in the Haber process6.3.7Typical conditions in the Haber• State the typical conditions in the Haber process as 450 °C, 20 000 kPa/200 atm and an iron catalyst6.3.8Symbol equation for the conversion of• State the symbol equation for the conversion of sulfur dioxide to sulfur trioxide in the Contact process, 2SO2(g) + O2(g) ⇌ 2SO3(g)6.3.9Sources of the sulfur dioxide (burning• State the sources of the sulfur dioxide (burning sulfur or roasting sulfide ores) and oxygen (air) in the Contact process6.3.10Typical conditions for the conversion• State the typical conditions for the conversion of sulfur dioxide to sulfur trioxide in the Contact process as 450 °C, 200 kPa/2 atm and a vanadium(V) oxide catalyst6.3.11Typical Haber and Contact process• Explain why typical Haber and Contact process conditions are used, considering reaction rate, equilibrium position, safety and economics
Recognise a reversible reaction
In a reversible reaction, products can react to form the original reactants. The forward and reverse reactions are represented together by the symbol ⇌.
For A + B ⇌ C + D, the forward reaction forms C and D from A and B, while the reverse reaction forms A and B from C and D.
Changing conditions can favour one direction, so the mixture may contain different proportions of reactants and products without changing the balanced equation.
The double arrow does not mean the reaction repeatedly switches on and off. Both directions can occur, and their relative rates depend on conditions.
Reverse hydration by heating or adding water
Heating a hydrated salt removes water and forms the anhydrous salt. Adding water to the anhydrous salt reverses the change and reforms the hydrated salt.
| Compound | Hydrated form | After heating | Reverse change |
|---|---|---|---|
| copper(II) sulfate | blue | white anhydrous copper(II) sulfate | add water: white → blue |
| cobalt(II) chloride | pink | blue anhydrous cobalt(II) chloride | add water: blue → pink |
Use the condition to choose direction: heat drives dehydration; water drives hydration. The observed colour identifies which form is present.
Do not swap the colour pairs: hydrated copper(II) sulfate is blue, while hydrated cobalt(II) chloride is pink.
Define dynamic equilibrium
A reversible reaction is at equilibrium in a closed system when the forward and reverse reactions occur at equal rates and the concentrations of reactants and products are no longer changing.
| Microscopic view | Macroscopic view |
|---|---|
| both forward and reverse reactions continue | concentrations remain constant |
| their rates are equal | observable properties no longer change |
A closed system prevents substances entering or leaving, allowing the opposing reactions to establish and maintain equilibrium.
Equal rates do not mean equal concentrations, and equilibrium does not mean both reactions have stopped.
Predict shifts in equilibrium
When a condition changes, the equilibrium position shifts in the direction that opposes that change, using the equation and energy information provided.
| Change | Direction favoured |
|---|---|
| increase temperature | endothermic direction |
| decrease temperature | exothermic direction |
| increase gas pressure | side with fewer moles of gas |
| decrease gas pressure | side with more moles of gas |
| add a reactant or product | direction that uses the added substance |
| remove a reactant or product | direction that replaces the removed substance |
| add a catalyst | no change in equilibrium position |
For pressure, count gaseous coefficients only. For temperature, label the forward direction exothermic or endothermic. For concentration, identify which side consumes the changed species. Then state the shift and the resulting yield change.
A catalyst speeds up forward and reverse reactions equally, so equilibrium is reached sooner but its position and equilibrium composition do not change.
Write the Haber process equation
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
One mole of nitrogen reacts reversibly with three moles of hydrogen to form two moles of ammonia. All three substances are gases under the reaction conditions.
The equation is balanced: two nitrogen atoms and six hydrogen atoms appear on each side. The reversible arrow is essential because ammonia can decompose back to nitrogen and hydrogen.
Do not use an ordinary one-way arrow or write NH₄. The Haber product is ammonia, NH₃.
State the Haber process feedstock sources
| Feed gas | Main source |
|---|---|
| nitrogen, N₂ | air |
| hydrogen, H₂ | methane, usually from natural gas |
Nitrogen is separated from air. Hydrogen is produced industrially from methane, commonly by reaction with steam before purification.
The purified gases are supplied in the stoichiometric ratio shown by the equation: one volume of nitrogen to three volumes of hydrogen.
Air is the source of nitrogen, not hydrogen. Methane is the named syllabus source of hydrogen.
Recall the typical Haber process conditions
| Variable | Typical Haber condition |
|---|---|
| temperature | 450 °C |
| pressure | 20 000 kPa, equivalent to 200 atm |
| catalyst | iron |
Keep the values attached to the correct process: Haber uses a very high pressure of 200 atm and an iron catalyst.
Iron increases the rate by lowering activation energy; it does not change the equilibrium yield.
Do not substitute vanadium(V) oxide: that is the catalyst for the Contact process.
Write the Contact process equilibrium equation
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
Two moles of sulfur dioxide react reversibly with one mole of oxygen to form two moles of sulfur trioxide. Each substance is gaseous under the reaction conditions.
The equation balances two sulfur atoms and six oxygen atoms on each side, and gas moles decrease from three to two in the forward direction.
This is the catalysed equilibrium step. Directly adding water to sulfur trioxide is not this equation and is not the industrial absorption route.
State the Contact process reactant sources
| Reactant | Source |
|---|---|
| sulfur dioxide, SO₂ | burn sulfur in air, or roast sulfide ores in air |
| oxygen, O₂ | air |
Burning sulfur gives S + O₂ → SO₂. Roasting a metal sulfide ore in air also produces sulfur dioxide as the sulfur is oxidised.
The sulfur dioxide is purified before it enters the catalytic converter with oxygen from air.
The syllabus asks for sources of sulfur dioxide and oxygen, not merely a source of elemental sulfur.
Recall the typical Contact process conditions
| Variable | Typical Contact condition |
|---|---|
| temperature | 450 °C |
| pressure | 200 kPa, equivalent to 2 atm |
| catalyst | vanadium(V) oxide, V₂O₅ |
Contact and Haber both use 450 °C, but Contact uses only 2 atm and the catalyst V₂O₅.
Vanadium(V) oxide increases the rates of both directions and helps equilibrium to be reached faster without changing its position.
Do not use iron or 200 atm; those are Haber process conditions.
Explain industrial equilibrium compromises
| Haber choice | Rate and equilibrium benefit | Safety/economic limit |
|---|---|---|
| 450 °C | fast enough; a lower temperature would favour exothermic ammonia formation | lower temperature is too slow; higher temperature lowers equilibrium yield |
| 200 atm | increases rate and shifts equilibrium to fewer gas moles, increasing NH₃ yield | still higher pressure needs stronger, costlier equipment and increases hazard |
| iron catalyst | increases rate so a moderate temperature can be used | does not increase equilibrium yield |
| Contact choice | Rate and equilibrium benefit | Safety/economic limit |
|---|---|---|
| 450 °C | fast enough; lower temperature would favour exothermic SO₃ formation | lower temperature is too slow; higher temperature lowers yield |
| 2 atm | pressure favours the side with fewer gas moles | much higher pressure gives insufficient extra benefit for its equipment, energy and safety costs |
| V₂O₅ catalyst | increases rate at the compromise temperature | does not change equilibrium yield |
Industrial conditions maximise neither rate nor single-pass equilibrium yield alone. They balance production speed, equilibrium composition, plant cost, energy use, equipment strength, and safety; unreacted gases can be recycled.
A high temperature always speeds both reactions, but for these exothermic forward reactions it lowers the equilibrium product yield. A catalyst improves rate only, not the equilibrium position.