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Energetics IEdexcel A-Level Chemistry: Topic test

20 questions, 54 marks

Edexcel A-Level Chemistry

Energetics I topic test

Total 54 marks

Name

Class

Date

  1. 1
    Magnesium burns in oxygen with a bright white flame: 2Mg(s) + O₂(g) → 2MgO(s). Under standard conditions the enthalpy change for the equation as written is −1204 kJ.
    (a)
    What is the standard enthalpy change of formation of magnesium oxide, in kJ mol⁻¹?
    [1 mark]
    • A−1204
    • B−602
    • C+602
    • D−301
    (b)
    Which set of conditions is used for a standard enthalpy change?
    [1 mark]
    • AA pressure of 100 kPa and a stated temperature, usually 298 K
    • BA pressure of 100 kPa and a temperature of exactly 273 K
    • CA concentration of 1 mol dm⁻³ for every substance, at 298 K
    • DAny pressure, provided that the temperature is 298 K
    (c)
    Define the term standard enthalpy change of formation.
    [2 marks]

    Total for question 1: 4 marks

  2. 2
    4.00 g of solid ammonium nitrate, NH₄NO₃ (molar mass 80.0 g mol⁻¹), is dissolved in 50.0 cm³ of water in an insulated polystyrene cup. The temperature falls from 21.0 °C to 14.6 °C. Treat the mass of solution as 50.0 g, and assume the specific heat capacity of the solution is 4.18 J g⁻¹ K⁻¹.
    (a)
    What is the enthalpy change for the dissolving of ammonium nitrate, in kJ per mole of ammonium nitrate?
    [1 mark]
    • A−26.8
    • B+1.34
    • C+26.8
    • D+26 800
    (b)
    Which statement describes the enthalpy level diagram for this change?
    [1 mark]
    • AThe products are at a lower enthalpy than the reactants, and ΔH is negative
    • BThe products are at the same enthalpy as the reactants, and ΔH is zero
    • CThe reactants are at a higher enthalpy than the products, because the temperature has fallen
    • DThe products are at a higher enthalpy than the reactants, and ΔH is positive
    (c)
    The true enthalpy change is more endothermic than the value found. Suggest an experimental reason, and explain its effect.
    [2 marks]

    Total for question 2: 4 marks

  3. 3
    Standard enthalpy changes of combustion at 298 K: carbon (graphite) −394 kJ mol⁻¹, hydrogen −286 kJ mol⁻¹, propene, C₃H₆(g), −2058 kJ mol⁻¹, propane, C₃H₈(g), −2219 kJ mol⁻¹.
    (a)
    Use Hess's law to calculate the standard enthalpy change for the hydrogenation of propene, C₃H₆(g) + H₂(g) → C₃H₈(g).
    [3 marks]
    (b)
    Calculate the standard enthalpy change of formation of propene, 3C(s) + 3H₂(g) → C₃H₆(g).
    [4 marks]

    Total for question 3: 7 marks

  4. 4
    Anhydrous copper(II) sulfate reacts with water to form the hydrate: CuSO₄(s) + 5H₂O(l) → CuSO₄·5H₂O(s). This enthalpy change cannot be measured directly, so a student uses Hess's law. In experiment 1, 2.00 g of anhydrous CuSO₄ (molar mass 159.6 g mol⁻¹) is added to 25.0 cm³ of water and the temperature rises by 8.0 K. In experiment 2, 3.13 g of CuSO₄·5H₂O (molar mass 249.6 g mol⁻¹) is added to 25.0 cm³ of water and the temperature falls by 1.4 K. In both experiments treat the mass of solution as 25.0 g with a specific heat capacity of 4.18 J g⁻¹ K⁻¹. The thermometer reads to ±0.1 °C.
    (a)
    Calculate the enthalpy change of solution for each solid, in kJ mol⁻¹, and use them to calculate the enthalpy change for CuSO₄(s) + 5H₂O(l) → CuSO₄·5H₂O(s).
    [6 marks]
    (b)
    Evaluate the method: explain why Hess's law is needed, calculate the percentage uncertainty in the temperature change in each experiment, and suggest one improvement to the procedure.
    [6 marks]

    Total for question 4: 12 marks

  5. 5
    Hydrazine, N₂H₄, is used as a rocket fuel: N₂H₄(g) + O₂(g) → N₂(g) + 2H₂O(g). Mean bond enthalpies in kJ mol⁻¹: N–N 158, N–H 391, O=O 498, N≡N 945, O–H 464. The enthalpy change for this reaction calculated from standard enthalpies of formation, with all species gaseous, is −543 kJ mol⁻¹.
    (a)
    What is the enthalpy change for the reaction calculated from the mean bond enthalpies?
    [1 mark]
    • A−581 kJ mol⁻¹
    • B+581 kJ mol⁻¹
    • C−2145 kJ mol⁻¹
    • D−2801 kJ mol⁻¹
    (b)
    Why does the answer from mean bond enthalpies differ from −543 kJ mol⁻¹?
    [1 mark]
    • ABond breaking is exothermic, so the signs in the bond enthalpy calculation are unreliable.
    • BHess's law does not apply to reactions that involve gases.
    • CMean bond enthalpies are measured at 0 K, but formation data are measured at 298 K.
    • DMean bond enthalpies are averages taken over many compounds, so they are not exact for the bonds in these molecules.
    (c)
    State what is meant by the term mean bond enthalpy.
    [2 marks]

    Total for question 5: 4 marks

  6. 6
    A student burns 1.48 g of butan-1-ol (molar mass 74.0 g mol⁻¹) in a spirit burner under a metal can containing 250 g of water. The temperature of the water rises by 19.0 K. The specific heat capacity of water is 4.18 J g⁻¹ K⁻¹. The data book value for the standard enthalpy change of combustion of butan-1-ol is −2676 kJ mol⁻¹.
    (a)
    What is the experimental enthalpy change of combustion of butan-1-ol, in kJ mol⁻¹?
    [1 mark]
    • A+993
    • B−19.9
    • C−993
    • D−993 000
    (b)
    Which is the main reason the experimental value is much less exothermic than the data book value?
    [1 mark]
    • AThe specific heat capacity of water increases as it is heated.
    • BMuch of the energy is lost to the surroundings and the can, and combustion may be incomplete.
    • CThe data book value applies only to solutions, not to liquids.
    • DThe mass of water used is too large for the sample of alcohol.
    (c)
    Suggest two changes to the apparatus that would give a value closer to the data book value.
    [2 marks]

    Total for question 6: 4 marks

  7. 7
    Hydrogen peroxide decomposes in the presence of a catalyst: 2H₂O₂(l) → 2H₂O(l) + O₂(g). Standard enthalpy changes of formation: H₂O₂(l) −188 kJ mol⁻¹, H₂O(l) −286 kJ mol⁻¹.
    (a)
    Calculate the standard enthalpy change for the reaction as written.
    [3 marks]
    (b)
    1.00 mol of hydrogen peroxide decomposes in 500 g of aqueous solution. Calculate the maximum temperature rise, assuming the specific heat capacity of the solution is 4.18 J g⁻¹ K⁻¹, and state one reason why the actual rise is smaller.
    [4 marks]

    Total for question 7: 7 marks

  8. 8
    Ethyne is hydrogenated to ethane in the presence of a nickel catalyst: C₂H₂(g) + 2H₂(g) → C₂H₆(g). Mean bond enthalpies in kJ mol⁻¹: C≡C 839, C–C 347, C–H 413, H–H 436. Standard enthalpies of formation in kJ mol⁻¹: C₂H₂(g) +227, C₂H₆(g) −84.
    (a)
    Calculate the enthalpy change for the reaction using the mean bond enthalpies, and again using the enthalpies of formation. Comment on the difference between the two values.
    [6 marks]
    (b)
    Explain, in terms of bonds, why the reaction is exothermic. Describe the enthalpy level diagram for the reaction and state the standard conditions for the enthalpy change.
    [6 marks]

    Total for question 8: 12 marks

End of questions

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).