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ThermodynamicsAQA A-Level Chemistry: Topic test

20 questions, 54 marks

AQA A-Level Chemistry

Thermodynamics topic test

Total 54 marks

Name

Class

Date

  1. 1
    A student constructs a Born–Haber cycle for potassium chloride, KCl. The standard enthalpy changes (kJ mol⁻¹) are: enthalpy of formation of KCl(s) = −437; enthalpy of atomisation of potassium = +89; first ionisation energy of potassium = +419; enthalpy of atomisation of chlorine (per mole of Cl atoms) = +122; first electron affinity of chlorine = −349.
    (a)
    Which equation represents the enthalpy of lattice formation of potassium chloride?
    [1 mark]
    • AK⁺(g) + Cl⁻(g) → KCl(s)
    • BKCl(s) → K⁺(g) + Cl⁻(g)
    • CK(s) + ½Cl₂(g) → KCl(s)
    • DK⁺(aq) + Cl⁻(aq) → KCl(s)
    (b)
    What is the enthalpy of lattice formation of KCl calculated from the data?
    [1 mark]
    • A+718 kJ mol⁻¹
    • B−1067 kJ mol⁻¹
    • C−718 kJ mol⁻¹
    • D−1416 kJ mol⁻¹
    (c)
    Define the standard enthalpy of atomisation of chlorine and write an equation, with state symbols, for the process.
    [2 marks]

    Total for question 1: 4 marks

  2. 2
    Potassium bromide dissolves in water with a small endothermic enthalpy change. The enthalpy of lattice dissociation of KBr is +670 kJ mol⁻¹, and the enthalpies of hydration of K⁺ and Br⁻ are −322 kJ mol⁻¹ and −335 kJ mol⁻¹ respectively.
    (a)
    Which equation represents the enthalpy of hydration of the bromide ion?
    [1 mark]
    • ABr⁻(aq) → Br⁻(g) + aq
    • BBr⁻(g) + aq → Br⁻(aq)
    • CKBr(s) + aq → K⁺(aq) + Br⁻(aq)
    • D½Br₂(l) + e⁻ + aq → Br⁻(aq)
    (b)
    What is the enthalpy of solution of KBr?
    [1 mark]
    • A−13 kJ mol⁻¹
    • B+1327 kJ mol⁻¹
    • C−657 kJ mol⁻¹
    • D+13 kJ mol⁻¹
    (c)
    Explain why potassium bromide dissolves in water at room temperature even though the process is endothermic.
    [2 marks]

    Total for question 2: 4 marks

  3. 3
    Hydrogen is made by steam reforming of methane: CH₄(g) + H₂O(g) → CO(g) + 3H₂(g), ΔH = +206 kJ mol⁻¹. The standard entropies (J K⁻¹ mol⁻¹) are: CH₄(g) = 186; H₂O(g) = 189; CO(g) = 198; H₂(g) = 131.
    (a)
    Calculate the entropy change for the reaction, with units, and explain the sign of your answer.
    [3 marks]
    (b)
    Calculate the minimum temperature at which the reaction becomes feasible. Show clearly how you use the units of ΔH and ΔS.
    [4 marks]

    Total for question 3: 7 marks

  4. 4
    Ammonium chloride is used in instant cold packs because it dissolves in water with a small endothermic enthalpy change: NH₄Cl(s) + aq → NH₄⁺(aq) + Cl⁻(aq), ΔH(solution) = +15 kJ mol⁻¹. The enthalpy of lattice dissociation of NH₄Cl is +686 kJ mol⁻¹, and the enthalpy of hydration of Cl⁻ is −364 kJ mol⁻¹. The standard entropies (J K⁻¹ mol⁻¹) are: NH₄Cl(s) = 94.6; NH₄⁺(aq) = 113.4; Cl⁻(aq) = 56.5.
    (a)
    Use an enthalpy cycle to calculate the enthalpy of hydration of the ammonium ion, and explain why the enthalpy of solution is so much smaller than the enthalpy of lattice dissociation.
    [6 marks]
    (b)
    Explain why ammonium chloride dissolves spontaneously at 298 K even though the process is endothermic. Include calculations of the entropy change and of ΔG at 298 K, and find the lowest temperature at which dissolving is feasible.
    [6 marks]

    Total for question 4: 12 marks

  5. 5
    Nitrogen monoxide is oxidised by oxygen in the gas phase: 2NO(g) + O₂(g) → 2NO₂(g). For this reaction ΔH = −114 kJ mol⁻¹ and ΔS = −146 J K⁻¹ mol⁻¹.
    (a)
    Which statement correctly explains why ΔS for this reaction is negative?
    [1 mark]
    • AThe products are more disordered than the reactants
    • BThe reaction is exothermic, so heat is lost to the surroundings
    • CThere are fewer moles of gas in the products, so the products are less disordered
    • DThe temperature of the system decreases during the reaction
    (b)
    Which statement about the feasibility of this reaction is correct?
    [1 mark]
    • AIt is feasible only at temperatures below about 781 K
    • BIt is feasible only at temperatures above about 781 K
    • CIt is feasible at all temperatures
    • DIt is not feasible at any temperature
    (c)
    Calculate ΔG for this reaction at 298 K, and state whether the reaction is feasible at this temperature.
    [2 marks]

    Total for question 5: 4 marks

  6. 6
    A student compares four ionic solids: sodium bromide, NaBr, potassium bromide, KBr, magnesium oxide, MgO, and calcium oxide, CaO.
    (a)
    Which solid has the most exothermic enthalpy of lattice formation?
    [1 mark]
    • ANaBr
    • BKBr
    • CCaO
    • DMgO
    (b)
    Which cation has the most exothermic enthalpy of hydration?
    [1 mark]
    • AK⁺
    • BMg²⁺
    • CNa⁺
    • DCa²⁺
    (c)
    Explain why the enthalpy of lattice formation of NaBr is more exothermic than that of KBr.
    [2 marks]

    Total for question 6: 4 marks

  7. 7
    A Born–Haber cycle is used to find the enthalpy of lattice formation of sodium oxide, Na₂O. The standard enthalpy changes (kJ mol⁻¹) are: enthalpy of formation of Na₂O(s) = −418; enthalpy of atomisation of sodium = +107; enthalpy of atomisation of oxygen (per mole of O atoms) = +249; first ionisation energy of sodium = +496; first electron affinity of oxygen = −141; second electron affinity of oxygen = +798.
    (a)
    Write an equation, with state symbols, for the second electron affinity of oxygen, and explain why this enthalpy change is endothermic although the first is exothermic.
    [3 marks]
    (b)
    Calculate the enthalpy of lattice formation of sodium oxide.
    [4 marks]

    Total for question 7: 7 marks

  8. 8
    Calcium reacts with chlorine to form calcium chloride, CaCl₂: Ca(s) + Cl₂(g) → CaCl₂(s), for which ΔfH = −796 kJ mol⁻¹. A student wonders whether calcium(I) chloride, CaCl, could form instead. For the hypothetical CaCl, the data (kJ mol⁻¹) are: enthalpy of atomisation of calcium = +178; first ionisation energy of calcium = +590; enthalpy of atomisation of chlorine (per mole of Cl atoms) = +122; first electron affinity of chlorine = −349; enthalpy of lattice formation of CaCl = −720. The standard entropies (J K⁻¹ mol⁻¹) are: Ca(s) = 41.6; Cl₂(g) = 223.1; CaCl₂(s) = 104.6.
    (a)
    Calculate the enthalpy of formation of CaCl. By considering the reaction 2CaCl(s) → Ca(s) + CaCl₂(s), explain why CaCl is not formed in practice.
    [6 marks]
    (b)
    Calculate the entropy change and the value of ΔG at 298 K for the formation of CaCl₂ from its elements. Explain the sign of the entropy change, and use your results to explain why the reaction is feasible even though it has a negative entropy 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).