Lesson 3.1.4.3

3.1.4.3 Applications of Hess’s law Quiz: AQA Chemistry, Unit 1

20 questions

Written by Revision Ninja · Last reviewed 11 October 2026

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Lesson 3.1.4.3, Applications of Hess’s law: 20 multiple choice questions for the AQA Chemistry (7405), Unit 1: Physical chemistry, written with Revision Ninja.

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The 20 questions

  1. According to Hess's law, enthalpy change is independent of which factor?

    • Reaction route
    • Initial state
    • Final state
    • Temperature change

  2. How is enthalpy change calculated from standard enthalpies of combustion?

    • Bonds formed minus broken
    • Reactants minus products
    • Bonds broken minus formed
    • Products minus reactants

  3. How is enthalpy change calculated from standard enthalpies of formation?

    • Reactants minus products
    • Bonds formed minus broken
    • Bonds broken minus formed
    • Products minus reactants

  4. What is the standard enthalpy of formation of an element in its standard state?

    • undefined
    • 0
    • 1
    • -1

  5. What is the main practical application of Hess's law?

    • Calculating unmeasurable enthalpies
    • Determining activation energy
    • Measuring reaction rates
    • Calculating solution pH

  6. What happens to the enthalpy change when a chemical equation is reversed?

    • Sign stays same
    • Magnitude is doubled
    • Sign is reversed
    • Value becomes zero

  7. What happens to the enthalpy change value when an equation is multiplied by two?

    • It stays constant
    • It halves
    • It doubles
    • It squares

  8. C + O2 -> CO2 has delta H = -394 kJ mol-1, and CO + 1/2 O2 -> CO2 has delta H = -283 kJ mol-1. What is delta H for C + 1/2 O2 -> CO?

    • -283 kJ mol-1
    • -677 kJ mol-1
    • +111 kJ mol-1
    • -111 kJ mol-1

  9. Standard enthalpies of formation are: CH4 = -75, CO2 = -394, H2O(l) = -286 kJ mol-1. What is delta H for CH4 + 2O2 -> CO2 + 2H2O(l)?

    • -891 kJ mol-1
    • -75 kJ mol-1
    • -966 kJ mol-1
    • +891 kJ mol-1

  10. Enthalpies of combustion are: C(s) = -394, H2(g) = -286, C2H5OH(l) = -1367 kJ mol-1. What is the standard enthalpy of formation of ethanol from its elements?

    • -1367 kJ mol-1
    • -2549 kJ mol-1
    • +279 kJ mol-1
    • -279 kJ mol-1

  11. Step 1: A -> B has delta H = +40 kJ mol-1. Step 2: B -> C has delta H = -65 kJ mol-1. What is delta H for A -> C?

    • +25 kJ mol-1
    • +105 kJ mol-1
    • -105 kJ mol-1
    • -25 kJ mol-1

  12. A reaction has delta H = -20 kJ mol-1. The whole equation is multiplied by 3. What is the new delta H?

    • -60 kJ mol-1
    • -6.7 kJ mol-1
    • +60 kJ mol-1
    • -20 kJ mol-1

  13. Combustion enthalpies are -394 kJ mol-1 for graphite and -395 kJ mol-1 for diamond. What is delta H for C(graphite) -> C(diamond)?

    • +789 kJ mol-1
    • -1 kJ mol-1
    • -789 kJ mol-1
    • +1 kJ mol-1

  14. For CaCO3 -> CaO + CO2, standard enthalpies of formation are CaCO3 = -1207, CaO = -635 and CO2 = -394 kJ mol-1. What is delta H for the reaction?

    • +2236 kJ mol-1
    • -2236 kJ mol-1
    • -178 kJ mol-1
    • +178 kJ mol-1

  15. A two-step route has delta H1 = -141 kJ mol-1 and an unknown delta H2. The overall delta H is -100 kJ mol-1. What is delta H2?

    • +241 kJ mol-1
    • -41 kJ mol-1
    • +41 kJ mol-1
    • -241 kJ mol-1

  16. Why might a reaction enthalpy change be impossible to measure directly?

    • Side reactions occur
    • Fast reaction speed
    • No temperature change
    • Gas phase reactants

  17. Given delta fH values of CH3OH(l) = -239, CO2 = -394 and H2O(l) = -286 kJ mol-1, what is delta cH of methanol, CH3OH + 3/2 O2 -> CO2 + 2H2O?

    • -727 kJ mol-1
    • -1205 kJ mol-1
    • -239 kJ mol-1
    • +727 kJ mol-1

  18. The overall enthalpy change is -150 kJ mol-1. Two of the steps are -90 kJ mol-1 and -45 kJ mol-1. What is the enthalpy change of the missing step?

    • -195 kJ mol-1
    • +15 kJ mol-1
    • -15 kJ mol-1
    • -135 kJ mol-1

  19. For 2X + Y -> 2Z, delta H = -60 kJ mol-1. What is delta H for Z -> X + 1/2 Y?

    • -30 kJ mol-1
    • +60 kJ mol-1
    • +30 kJ mol-1
    • -60 kJ mol-1

  20. Why can calorimetry measurements for individual steps be combined using Hess's law?

    • Reactions are exothermic
    • Mass is conserved
    • Enthalpy is route-independent
    • Temperature stays constant

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