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A2 · Practice questions · Entropy change

Count the gas.

Six original Cambridge-style questions on entropy and entropy change.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Products minus reactants, in joules.

01
Recall
[2 marks]

State what is meant by entropy, and give its units.

  • Entropy is a measure of the dispersal of energy in a system: the number of ways the particles and their energy can be arranged. ✓
  • Units: J K−1 mol−1. ✓
02
Analysis
[4 marks]

Predict the sign of ΔS for each change and explain: (a) I2(s) → I2(g); (b) 2SO2(g) + O2(g) → 2SO3(g); (c) KNO3(s) dissolving in water; (d) cooling water from 50 °C to 20 °C.

  • (a) Positive: solid to gas, far more freedom of movement. ✓
  • (b) Negative: 3 mol of gas become 2 mol of gas. ✓
  • (c) Positive: ions leave the ordered lattice and spread through the solution. ✓
  • (d) Negative: at lower temperature the energy is spread over fewer energy levels. ✓
03
Calculation
[2 marks]

Calculate ΔS° for N2(g) + 3H2(g) → 2NH3(g). S° / J K−1 mol−1: N2 192, H2 131, NH3 193.

  • ΣS°(products) = 2 × 193 = 386; ΣS°(reactants) = 192 + 3 × 131 = 585. ✓
  • ΔS° = 386 − 585 = −199 J K−1 mol−1. ✓
04
Calculation
[2 marks]

Calculate ΔS° for 2Mg(s) + O2(g) → 2MgO(s). S° / J K−1 mol−1: Mg 33, O2 205, MgO 27.

  • Products: 2 × 27 = 54; reactants: 2 × 33 + 205 = 271. ✓
  • ΔS° = 54 − 271 = −217 J K−1 mol−1. ✓
05
Analysis
[2 marks]

The entropy change for C(graphite) + O2(g) → CO2(g) is only +3.3 J K−1 mol−1. Explain why it is so small.

  • There is 1 mol of gas on each side, so the number of moles of gas does not change. ✓
  • Graphite has a very small entropy (5.7), so the only change is between two similar gases (214 − 205). ✓
06
Calculation
[2 marks]

The standard entropy of H2O(l) is 70 and of H2O(g) is 189 J K−1 mol−1. Calculate ΔS for boiling 1 mol of water and explain the sign.

  • ΔS = 189 − 70 = +119 J K−1 mol−1. ✓
  • Positive: gas particles move freely and randomly in a far larger volume, so energy is much more dispersed. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.