Lesson 13B.2

13B.2 Calculating entropy changes Quiz: Pearson Edexcel Chemistry, Unit 13

20 questions

Written by Revision Ninja · Last reviewed 11 October 2026

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Lesson 13B.2, Calculating entropy changes: 20 multiple choice questions for the Pearson Edexcel Chemistry (9CH0), Unit 13: Energetics II, written with Revision Ninja.

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The 20 questions

  1. Which expression gives the entropy change of the system from standard entropies?

    • dS = sum of S(products) minus sum of S(reactants)
    • dS = S(products) divided by S(reactants)
    • dS = sum of S(products) plus sum of S(reactants)
    • dS = sum of S(reactants) minus sum of S(products)

  2. Which factor affects the standard entropy of a substance most strongly?

    • Its physical state, with gases having higher entropy than liquids and solids of the same substance
    • Its density only, because denser substances pack their particles more closely and so have more order
    • Its colour, because coloured substances absorb more visible light and so have more ordered electronic structures
    • Its melting point only, because higher melting points give a wider range of arrangements and more disorder

  3. Why are standard entropies positive for pure substances at 298 K?

    • Standard entropies are positive only for elements, not compounds
    • The third law: a perfect crystal has zero entropy at 0 K
    • Entropy values are defined as the enthalpy at 298 K
    • Entropy is measured relative to the enthalpy of formation, so it is always positive

  4. Why does a gas have a higher entropy than the liquid of the same substance?

    • Liquids have stronger bonds, so they store more energy as entropy
    • Gas particles have more freedom of movement and occupy many more microstates
    • Gas particles have fewer possible positions than liquid particles
    • Gas particles are heavier, so they store more entropy

  5. What does a negative entropy change of the system indicate?

    • The temperature of the system has risen
    • The reaction is endothermic
    • The system has become more ordered
    • The system has become more disordered

  6. Which of CO2(g), H2O(l) and diamond has the highest standard entropy at 298 K?

    • CO2(g), because gases have much higher standard entropies than liquids or solids
    • H2O(l), because liquids have the highest standard entropies of all the states at 298 K
    • All three have the same standard entropy, because each contains only a fixed number of atoms
    • Diamond, because it has the most atoms in its giant covalent structure and so the most arrangements

  7. Which statement about the standard entropy of a pure element at 298 K is correct?

    • It is zero, because elements have no disorder
    • It is equal to its standard enthalpy of formation
    • It is positive, because the element has entropy above zero at 0 K
    • It is negative, because elements are more stable than compounds

  8. For N2(g) + 3H2(g) gives 2NH3(g), with S(N2) = 191.5, S(H2) = 130.7 and S(NH3) = 192.8 J K-1 mol-1, what is dS for the system?

    • -198.0 J K-1 mol-1
    • +775.2 J K-1 mol-1
    • -391.2 J K-1 mol-1
    • +198.0 J K-1 mol-1

  9. For CaCO3(s) gives CaO(s) + CO2(g), with S(CaCO3) = 92.9, S(CaO) = 39.8 and S(CO2) = 213.8 J K-1 mol-1, what is dS for the system?

    • +266.9 J K-1 mol-1
    • -160.7 J K-1 mol-1
    • +160.7 J K-1 mol-1
    • +346.5 J K-1 mol-1

  10. For 2Mg(s) + O2(g) gives 2MgO(s), with S(Mg) = 32.7, S(O2) = 205.2 and S(MgO) = 26.9 J K-1 mol-1, what is dS for the system?

    • -216.8 J K-1 mol-1
    • +270.6 J K-1 mol-1
    • -54.5 J K-1 mol-1
    • +216.8 J K-1 mol-1

  11. For H2O(l) gives H2O(g), with S(l) = 70.0 and S(g) = 188.8 J K-1 mol-1, what is dS for the system?

    • -118.8 J K-1 mol-1
    • +258.8 J K-1 mol-1
    • +118.8 J K-1 mol-1
    • +70.0 J K-1 mol-1

  12. For N2O4(g) gives 2NO2(g), with S(N2O4) = 304.3 and S(NO2) = 240.0 J K-1 mol-1, what is dS for the system?

    • +544.3 J K-1 mol-1
    • +175.7 J K-1 mol-1
    • +64.3 J K-1 mol-1
    • -175.7 J K-1 mol-1

  13. For CH4(g) + 2O2(g) gives CO2(g) + 2H2O(l), with S(CH4) = 186.3, S(O2) = 205.2, S(CO2) = 213.8 and S(H2O, l) = 69.9 J K-1 mol-1, what is dS for the system?

    • +596.7 J K-1 mol-1
    • -43.1 J K-1 mol-1
    • +243.1 J K-1 mol-1
    • -243.1 J K-1 mol-1

  14. For a combustion reaction with dH = -890 kJ mol-1 at 298 K, what is the entropy change of the surroundings?

    • +890 J K-1 mol-1
    • +2.99 J K-1 mol-1
    • +2987 J K-1 mol-1
    • -2987 J K-1 mol-1

  15. Convert dS = -243.1 J K-1 mol-1 into kJ K-1 mol-1.

    • -2.431 kJ K-1 mol-1
    • -24.31 kJ K-1 mol-1
    • -0.2431 kJ K-1 mol-1
    • -0.02431 kJ K-1 mol-1

  16. For a reaction with dS(system) = -243.1 J K-1 mol-1, dH = -890 kJ mol-1 and T = 298 K, what is dStotal?

    • +2743.5 J K-1 mol-1
    • -3229.7 J K-1 mol-1
    • +3229.7 J K-1 mol-1
    • -2743.5 J K-1 mol-1

  17. A student calculates the entropy of a reaction but ignores the surroundings. Which statement best describes the error?

    • Feasibility depends on dStotal
    • The error is using kelvin, because entropy is measured in degrees Celsius
    • The error is in units, because entropy should be given in kJ
    • There is no error, because dSsystem alone always gives dStotal

  18. A reaction has dSsystem = -240 J K-1 mol-1 and dH = -120 kJ mol-1. At what temperature is dStotal equal to zero?

    • 500 K
    • 250 K
    • 50 K
    • 2 K

  19. A reaction has dSsystem = +100 J K-1 mol-1 and dH = +30 kJ mol-1 at 298 K. What is dStotal and what does it indicate?

    • About +100.7 J K-1 mol-1, so the reaction is feasible at 298 K
    • About +0.1 J K-1 mol-1, so the reaction is feasible at 298 K
    • About -0.7 J K-1 mol-1
    • Exactly zero, so the reaction is at equilibrium at 298 K

  20. Why must entropy changes be converted to the same units as dH/T before they are added?

    • dH must be converted to kelvin, since the entropy units are J K-1 and the temperature must match
    • The units are unimportant because all entropy values are dimensionless and so can be added directly
    • dSsystem must be multiplied by 1000 only at room temperature, because temperature changes the units
    • dH is in kJ mol-1, so dH/T must be converted to J K-1 mol-1 before adding it to dSsystem in J K-1 mol-1

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