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EntropyAQA A-Level Chemistry: Revision notes

Section 1

Why enthalpy change is not enough

Many exothermic changes happen readily, which suggests that energy is released when a change is favourable. But the enthalpy change, ΔH, is not sufficient to explain whether a change is feasible (able to happen without continuing input of energy).

Some endothermic changes occur readily at room temperature: ammonium nitrate dissolving in water, a solid melting, or a liquid evaporating. Some exothermic changes do not proceed at all. A second quantity is needed: entropy.

Key termsfeasible

Section 2

Entropy as disorder

Entropy (S) is a measure of the disorder of a system: the number of ways the particles and their energy can be arranged. The greater the number of arrangements, the greater the entropy.

A positive entropy change (ΔS) means the system becomes more disordered. The units are J K⁻¹ mol⁻¹.

All substances have a positive absolute entropy value. A perfect crystal at 0 K has the lowest possible entropy, so absolute entropy values can be tabulated at 298 K as standard entropies (S°).

Key termsentropydisorder
Common mistake

Entropy is measured in J K⁻¹ mol⁻¹ but enthalpy is in kJ mol⁻¹. Convert one to match the other before combining them.

Section 3

Entropy and physical changes

Entropy depends on the state of a substance. Particles in a solid are held in fixed positions and are the most ordered. A liquid is more disordered, and a gas, whose particles move randomly and are far apart, has by far the highest entropy: S(gas) ≫ S(liquid) > S(solid).

  • Melting and boiling increase entropy, so ΔS is positive.
  • Freezing and condensing decrease entropy, so ΔS is negative.
  • Dissolving a solid usually increases entropy as the ordered lattice breaks into mobile ions.
  • Raising the temperature raises entropy, because the particles have more ways of sharing out the energy.
Key termsstandard entropy

Section 4

Entropy and chemical changes

In a chemical reaction the entropy change is mostly determined by changes in the number of moles of gas.

  • More gas particles in the products than in the reactants: ΔS is positive, e.g. CaCO₃(s) → CaO(s) + CO₂(g).
  • Fewer gas particles in the products: ΔS is negative, e.g. N₂(g) + 3H₂(g) → 2NH₃(g).
  • No change in the number of gas particles: ΔS is small, positive or negative.

A gas produced from solids or solutions gives a large positive ΔS. More particles overall in the same state also gives a positive ΔS.

Exam tip

To predict the sign of ΔS, count moles of gas on each side of the equation first, then consider solids and liquids.

Section 5

Calculating entropy changes

The entropy change of a reaction is found from absolute entropy values:

ΔS = ΣS(products) − ΣS(reactants)

Multiply each value by its coefficient in the balanced equation.

Worked example. C(graphite) + O₂(g) → CO₂(g). S = 5.7, 205 and 214 J K⁻¹ mol⁻¹. ΔS = 214 − (5.7 + 205) = +3.3 J K⁻¹ mol⁻¹

The value is small because the number of moles of gas does not change. Always give a sign and the units J K⁻¹ mol⁻¹.

Common mistake

Forgetting the coefficients, or subtracting products from reactants. Write out ΣS(products) and ΣS(reactants) separately first.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Entropy

  1. A student discusses entropy using the three physical states of water. She considers how the disorder of the particles in ice, liquid water and steam changes as a sample of water is heated from below 0 °C to above 100 °C at constant pressure.
    Explain why the entropy of steam is greater than the entropy of liquid water at the same temperature.2 marks
  2. Ammonia is made industrially in the Haber process: N₂(g) + 3H₂(g) → 2NH₃(g). Standard entropies (J K⁻¹ mol⁻¹): N₂(g) = 192; H₂(g) = 131; NH₃(g) = 193.
    Predict, with a reason, the sign of the entropy change for the reaction Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g).2 marks
  3. Calcium carbonate decomposes on strong heating: CaCO₃(s) → CaO(s) + CO₂(g), ΔH = +178 kJ mol⁻¹. Standard entropies (J K⁻¹ mol⁻¹): CaCO₃(s) = 93; CaO(s) = 40; CO₂(g) = 214. Relative formula mass of CaCO₃ = 100.1.
    Calculate the entropy change for the decomposition of calcium carbonate. Include units.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).