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Topic 6 · 6.3–6.4

Reversible reactions, equilibrium and redox

Core and Extended · Papers 1–6

🎯What you need to be able to do

  • State that some reactions are reversible (symbol ⇌) and describe how heating hydrated salts, or adding water to anhydrous ones, reverses the direction.
  • State the two conditions for equilibrium in a closed system EXTENDED.
  • Predict and explain how temperature, pressure, concentration and a catalyst affect the position of equilibrium EXTENDED.
  • State the equations, raw materials and conditions of the Haber and Contact processes, and explain the choice of conditions EXTENDED.
  • Define oxidation and reduction as gain and loss of oxygen, and use Roman numerals for oxidation numbers.
  • Define oxidation and reduction in terms of electrons and oxidation numbers; identify oxidising and reducing agents; use the colour changes of acidified potassium manganate(VII) and potassium iodide EXTENDED.

📚The chemistry

Reversible reactions

A reversible reaction can go in both directions, shown by the symbol ⇌. Changing the conditions changes which direction wins. Heating blue hydrated copper(II) sulfate drives off its water of crystallisation and leaves white anhydrous copper(II) sulfate; adding water turns it blue again and releases heat.

\[ \mathrm{CuSO_4{\cdot}5H_2O(s) \rightleftharpoons CuSO_4(s) + 5H_2O(l)} \]
Blue crystals of hydrated copper(II) sulfate on the left and white anhydrous copper(II) sulfate powder on the right. An arrow labelled heat, endothermic, points from blue to white; an arrow labelled add water, exothermic, points back from white to blue. Beneath: cobalt(II) chloride does the same, pink when hydrated and blue when anhydrous.
The two reversible reactions on the syllabus. Both are used as tests for water: anhydrous copper(II) sulfate turns blue, and blue cobalt(II) chloride paper turns pink.

Equilibrium EXTENDED

In a closed system (nothing gets in or out), a reversible reaction reaches equilibrium when

  • the rate of the forward reaction equals the rate of the reverse reaction, and
  • the concentrations of reactants and products no longer change.

Both reactions are still happening — the equilibrium is dynamic — but they cancel out. The concentrations are constant, not equal.

A graph of rate against time. The forward rate starts high and falls; the reverse rate starts at zero and rises. They meet and stay equal, in a shaded region labelled equilibrium: rates are equal.
As reactants are used up the forward rate falls; as products build up the reverse rate rises, until the two are equal. EXTENDED

Moving the position of equilibrium. If the conditions change, the equilibrium shifts in the direction that opposes the change:

raise the temperature: shifts in the endothermic direction (lower it: the exothermic direction)
raise the pressure (gases): shifts to the side with fewer gas molecules
add a reactant (or remove a product): shifts to the right, making more product
a catalyst: no change to the position — both rates increase equally, so equilibrium is reached sooner

Questions give you the information you need: the \( \Delta H \) sign, the colours, or the number of gas molecules on each side. Count gas molecules from the balanced equation, ignoring solids and liquids.

The Haber process EXTENDED

\[ \mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)} \qquad \Delta H = -92\ \text{kJ/mol} \]

Nitrogen comes from the air; hydrogen from methane. Typical conditions: 450 °C, 20 000 kPa (200 atm) and an iron catalyst. Unreacted gases are recycled.

  • Temperature. The forward reaction is exothermic, so a low temperature gives a higher yield — but the rate would be too slow. 450 °C is a compromise between yield and rate.
  • Pressure. 4 gas molecules become 2, so a high pressure gives a higher yield and also a faster rate. Higher still would cost more to build and run (thicker pipes, more energy for compressors) and is more dangerous, so 200 atm is a compromise between yield, cost and safety.
  • Catalyst. Iron increases the rate; it does not change the yield.
A graph of the percentage of ammonia in the equilibrium mixture against temperature from 300 to 550 degrees Celsius, for 50, 200 and 400 atmospheres. Every curve falls as temperature rises, and higher pressure gives a higher curve. A point marks about 36 percent at 450 degrees and 200 atmospheres.
Lower temperature and higher pressure both raise the equilibrium yield. The industrial choice, 450 °C and 200 atm, gives about 36% at equilibrium. (Calculated for ideal gases; real plants reach less per pass and recycle.)

The Contact process EXTENDED

\[ \mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)} \qquad \Delta H = -197\ \text{kJ/mol} \]

Sulfur dioxide comes from burning sulfur or roasting sulfide ores; oxygen from the air. Typical conditions: 450 °C, 200 kPa (2 atm) and a vanadium(V) oxide catalyst. The temperature is the same compromise as in the Haber process. The pressure is low because the equilibrium already lies far to the right at 2 atm, so a higher pressure would add cost and risk for very little extra yield. The sulfur trioxide is then used to make sulfuric acid.

Redox

A redox reaction is one in which oxidation and reduction happen together. The simplest definitions use oxygen:

  • Oxidation is gain of oxygen; reduction is loss of oxygen. In \( \mathrm{CuO + H_2 \rightarrow Cu + H_2O} \), copper(II) oxide is reduced and hydrogen is oxidised.
  • A Roman numeral gives an element’s oxidation number in a compound: iron(III) oxide contains Fe in the +3 state, Fe2O3.

EXTENDED The wider definitions work for reactions without oxygen too:

Oxidation: loss of electrons; increase in oxidation number
Reduction: gain of electrons; decrease in oxidation number

Rules for oxidation numbers: an uncombined element is 0; a monatomic ion equals its charge (Na+ is +1, O2− is −2); the numbers in a compound add up to zero; those in an ion add up to its charge. In SO42−: \( S + 4(-2) = -2 \), so S is +6.

An oxidising agent oxidises something else and is itself reduced; a reducing agent reduces something else and is itself oxidised. Two colour changes identify them:

A number line of oxidation numbers from minus 2 to plus 7. An arrow along the top labelled oxidation: loses electrons, oxidation number goes up. An arrow along the bottom labelled reduction. A short arrow from minus 1 to 0 shows iodide becoming iodine: potassium iodide goes from colourless to brown and detects an oxidising agent. A long arrow from plus 7 to plus 2 shows manganate(VII) becoming Mn2+: acidified potassium manganate(VII) goes from purple to colourless and detects a reducing agent.
Acidified potassium manganate(VII) is an oxidising agent, so it is decolourised by a reducing agent. Potassium iodide is a reducing agent, so it turns brown with an oxidising agent. EXTENDED

✏️Worked example EXTENDED

(a) In the Contact process, \( \mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)} \) and the forward reaction is exothermic. Predict and explain the effect on the yield of SO3 of (i) raising the temperature, (ii) raising the pressure, (iii) adding more catalyst. [5] (b) In \( \mathrm{Cl_2 + 2KBr \rightarrow 2KCl + Br_2} \), use oxidation numbers to show that the reaction is redox, and name the oxidising agent. [3]

(a)(i) The yield decreases: the equilibrium shifts in the endothermic direction, which is the reverse reaction.

(ii) The yield increases: 3 gas molecules on the left, 2 on the right, so higher pressure shifts it to the side with fewer gas molecules.

(iii) No change: a catalyst speeds up the forward and reverse reactions equally, so equilibrium is reached sooner but its position is the same.

(b) Chlorine: 0 in Cl2 to −1 in KCl, a decrease, so it is reduced. Bromine: −1 in KBr to 0 in Br2, an increase, so it is oxidised. Both happen, so it is redox. Chlorine is the oxidising agent (it oxidises the bromide and is itself reduced). Potassium stays +1: a spectator.

Check it. Electrons lost must equal electrons gained: 2 Br− each lose one electron (2 in total) and one Cl2 gains two ✓. And the answer to (a)(i) matches why industry does not simply use a very high temperature for speed.
“A catalyst increases the yield.” It never does. It increases the rate of both reactions, so the same equilibrium arrives sooner — which is why it still saves money.

📝Practise

In the style of the multiple-choice and theory papers. EXTENDED marks Supplement content.

1. (Theory.) White anhydrous copper(II) sulfate is used to test a liquid. Describe what is seen if the liquid contains water, and state whether this change releases or absorbs energy. [2]
It turns from white to blue (hydrated copper(II) sulfate forms). Energy is released: it is exothermic, the reverse of the endothermic dehydration.
2. (Theory.) In \( \mathrm{2PbO + C \rightarrow 2Pb + CO_2} \), which substance is oxidised and which is reduced? Give a reason for each. [2]
Carbon is oxidised (it gains oxygen). Lead(II) oxide is reduced (it loses oxygen).
3. (Multiple choice.) What is the oxidation number of iron in iron(III) chloride, and what is its formula? A: +2, FeCl2. B: +3, FeCl3. C: +3, Fe3Cl. D: −3, FeCl3.
B. The Roman numeral is the oxidation number, +3; three Cl− balance it.
4. (Theory.) EXTENDED Brown nitrogen dioxide forms colourless dinitrogen tetroxide: \( \mathrm{2NO_2(g) \rightleftharpoons N_2O_4(g)} \); the forward reaction is exothermic. A sealed syringe of the equilibrium mixture is (a) placed in hot water, (b) compressed at constant temperature. Predict the colour change in each case and explain. [4]
(a) Darker brown: raising the temperature shifts the equilibrium in the endothermic (reverse) direction, making more NO2. (b) Paler (after the first darkening from compression): higher pressure shifts it to the side with fewer gas molecules, 1 molecule of N2O4 against 2 of NO2.
5. (Theory.) EXTENDED Use the ammonia graph above. (a) Read the percentage of ammonia at 350 °C and 200 atm. (b) Suggest why 350 °C is not used, even though it gives more ammonia. [3]
(a) About 57%. (b) At the lower temperature the rate is too slow, so equilibrium would take too long to reach and less ammonia would be made per hour; 450 °C is a compromise between rate and yield.
6. (Theory.) EXTENDED Deduce the oxidation number of (a) sulfur in SO3, (b) chromium in Cr2O72−, (c) manganese in MnO4−. [3]
(a) \( S + 3(-2) = 0 \), so +6. (b) \( 2\,Cr + 7(-2) = -2 \), so Cr = +6. (c) \( Mn + 4(-2) = -1 \), so +7.
7. (Theory.) EXTENDED A few drops of an unknown solution Y are added to acidified aqueous potassium manganate(VII), which turns from purple to colourless. What does this show about Y? [2]
Y is a reducing agent: it reduces the manganate(VII) (Mn +7 to +2) and is itself oxidised.
8. (Theory.) EXTENDED In the Contact process, explain why a pressure of only 2 atm is used, and state the catalyst. [3]
The position of equilibrium already lies far to the right at low pressure, giving a high yield; a higher pressure would raise cost (and risk) for little extra yield. Catalyst: vanadium(V) oxide.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • PhET “Reversible Reactions” — watch forward and reverse rates settle to equilibrium
  • Royal Society of Chemistry — the nitrogen dioxide / dinitrogen tetroxide equilibrium demonstration