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AS & A Level Chemistry (9701)

Chemical energetics (A Level: lattice energy, Born–Haber, hydration) Banker

29 questions · about 3% of the marks we indexed · examined in 9 of 9 sessions.

Questions by year

2026

  • 2026 May/June · Paper 41 · Q1 (a, b, c, d(i), d(ii))9 marksLattice energy & Born–Haber cycles
  • 2026 May/June · Paper 43 · Q1 (a, b, c, d(i), d(ii))9 marksLattice energy & Born–Haber cycles
  • 2026 May/June · Paper 44 · Q4 (a(i), a(ii), b)7 marksElectron affinity & ionisation energy (A Level) · Lattice energy & Born–Haber cycles
  • 2026 May/June · Paper 44 · Q5 (b(ii))3 marksEnthalpies of solution & hydration

2025

  • 2025 Oct/Nov · Paper 41 · Q1 (c(i), c(ii), d)5 marksEnthalpies of solution & hydration
  • 2025 Oct/Nov · Paper 42 · Q4 (a, b, c, d)6 marksElectron affinity & ionisation energy (A Level) · Lattice energy & Born–Haber cycles
  • 2025 Oct/Nov · Paper 43 · Q1 (c(i), c(ii), d)5 marksEnthalpies of solution & hydration
  • 2025 Oct/Nov · Paper 44 · Q2 (d(i), d(ii), d(iii), d(iv), e)7 marksElectron affinity & ionisation energy (A Level) · Lattice energy & Born–Haber cycles
  • 2025 May/June · Paper 41 · Q4 (a(i), a(ii), a(iii), a(iv))6 marksEnthalpies of solution & hydration · Lattice energy & Born–Haber cycles
  • 2025 May/June · Paper 43 · Q4 (a(i), a(ii), a(iii), a(iv))6 marksEnthalpies of solution & hydration · Lattice energy & Born–Haber cycles
  • 2025 May/June · Paper 44 · Q8 (a(i), a(ii), b, c, d)9 marksEnthalpies of solution & hydration · Lattice energy & Born–Haber cycles

2024

  • 2024 Oct/Nov · Paper 41 · Q2 (a, b(i), b(ii), b(iii), b(iv), c)10 marksEnthalpies of solution & hydration · Lattice energy & Born–Haber cycles
  • 2024 Oct/Nov · Paper 42 · Q2 (a, b(i), b(ii), b(iii))9 marksEnthalpies of solution & hydration
  • 2024 Oct/Nov · Paper 43 · Q2 (a, b(i), b(ii), b(iii), b(iv), c)10 marksEnthalpies of solution & hydration · Lattice energy & Born–Haber cycles
  • 2024 May/June · Paper 41 · Q4 (c, d(i), d(ii), e)7 marksOfficial paper ↗Enthalpies of solution & hydration · Lattice energy & Born–Haber cycles
  • 2024 May/June · Paper 42 · Q1 (b, c, d(i), d(ii))9 marksLattice energy & Born–Haber cycles
  • 2024 May/June · Paper 43 · Q4 (c, d(i), d(ii), e)7 marksEnthalpies of solution & hydration · Lattice energy & Born–Haber cycles

2023

  • 2023 Oct/Nov · Paper 41 · Q3 (g(i), g(ii), g(iii))4 marksEnthalpies of solution & hydration
  • 2023 Oct/Nov · Paper 43 · Q3 (g(i), g(ii), g(iii))4 marksEnthalpies of solution & hydration
  • 2023 May/June · Paper 41 · Q8 (a, b, c(i), c(ii), c(iii))9 marksEnthalpies of solution & hydration · Lattice energy & Born–Haber cycles
  • 2023 May/June · Paper 42 · Q1 (b(i))2 marksLattice energy & Born–Haber cycles
  • 2023 May/June · Paper 42 · Q3 (a, b, c, d)8 marksLattice energy & Born–Haber cycles
  • 2023 May/June · Paper 43 · Q8 (a, b, c(i), c(ii), c(iii))9 marksEnthalpies of solution & hydration · Lattice energy & Born–Haber cycles

2022

  • 2022 Oct/Nov · Paper 41 · Q1 (a, b, c, d(i), d(ii))8 marksElectron affinity & ionisation energy (A Level) · Enthalpies of solution & hydration · Lattice energy & Born–Haber cycles
  • 2022 Oct/Nov · Paper 42 · Q1 (a, b, c, d(i), d(ii))8 marksElectron affinity & ionisation energy (A Level) · Enthalpies of solution & hydration · Lattice energy & Born–Haber cycles
  • 2022 Oct/Nov · Paper 43 · Q1 (a, b, c, d(i), d(ii))8 marksElectron affinity & ionisation energy (A Level) · Enthalpies of solution & hydration · Lattice energy & Born–Haber cycles
  • 2022 May/June · Paper 41 · Q1 (d(i), d(ii))5 marksLattice energy & Born–Haber cycles
  • 2022 May/June · Paper 42 · Q3 (a(i), a(ii), a(iii), b, c)9 marksElectron affinity & ionisation energy (A Level) · Lattice energy & Born–Haber cycles
  • 2022 May/June · Paper 43 · Q1 (d(i), d(ii))5 marksLattice energy & Born–Haber cycles

References only, not question text. Open Official paper ↗ where Cambridge publishes the paper, or see the official Cambridge past-papers page; for other papers, ask your school. “Paper 42” means Paper 4, variant 2.

Practice questions

Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.

Q1Lattice energy & Born–Haber cycles · Core · Paper 4 style · 4 marks
  1. (a) Define the term lattice energy. [2]
  2. (b) Write an equation, with state symbols, for the lattice energy of magnesium chloride. [1]
  3. (c) State whether lattice energy is exothermic or endothermic and explain why. [1]
Show answer
  1. (a) The enthalpy change when one mole of an ionic compound is formed from its gaseous ions under standard conditions.
    M1 one mole of ionic compound/solid formedM2 from gaseous ions (standard conditions)
  2. (b) Mg²⁺(g) + 2Cl⁻(g) → MgCl₂(s)
    B1
  3. (c) Exothermic: bonds (electrostatic attractions) form between oppositely charged ions, releasing energy.
    B1
Q2Lattice energy & Born–Haber cycles · Core · Paper 4 style · 3 marks

Data (kJ/mol): ΔHf⦵ NaCl(s) −411; ΔHat⦵ Na(s) +107; first ionisation energy of Na +496; ΔHat⦵ of chlorine (½Cl₂ → Cl) +122; first electron affinity of Cl −349.

  1. (a) Use the data to calculate the lattice energy of sodium chloride. [2]
  2. (b) Name the enthalpy change ½Cl₂(g) → Cl(g). [1]
Show answer
  1. (a) −787 kJ/mol
    M1 LE = ΔHf − ΔHat(Na) − IE(Na) − ΔHat(Cl) − EA(Cl) = −411 − 107 − 496 − 122 + 349A1 −787 kJ/mol
  2. (b) Standard enthalpy change of atomisation of chlorine
    B1
Q3Lattice energy & Born–Haber cycles · Core · Paper 4 style · 5 marks
  1. (a) Define the standard enthalpy change of atomisation of an element. [2]
  2. (b) Write equations, with state symbols, for the standard enthalpy change of atomisation of (i) sodium and (ii) chlorine. [2]
  3. (c) Explain why the enthalpy change of atomisation of chlorine is half the Cl–Cl bond energy. [1]
Show answer
  1. (a) The enthalpy change when one mole of gaseous atoms is formed from the element in its standard state under standard conditions.
    M1 one mole of gaseous atoms formedM2 from element in standard state
  2. (b) (i) Na(s) → Na(g) (ii) ½Cl₂(g) → Cl(g)
    B1 each
  3. (c) One mole of Cl atoms comes from half a mole of Cl₂ molecules, so only half a mole of Cl–Cl bonds is broken.
    B1
Q4Lattice energy & Born–Haber cycles · Core · Paper 4 style · 3 marks

Data (kJ/mol): ΔHat⦵ Li +159; 1st IE Li +520; ΔHat⦵ F (½F₂ → F) +79; 1st EA F −328; lattice energy LiF −1046.

  1. (a) Use the data to calculate the standard enthalpy change of formation of lithium fluoride. [2]
  2. (b) Write the equation for the standard enthalpy change of formation of lithium fluoride. [1]
Show answer
  1. (a) −616 kJ/mol
    M1 ΔHf = 159 + 520 + 79 + (−328) + (−1046)A1 −616 kJ/mol
  2. (b) Li(s) + ½F₂(g) → LiF(s)
    B1
Q5Lattice energy & Born–Haber cycles · Core · Paper 4 style · 6 marks
  1. (a) List, in order, the enthalpy changes in a Born–Haber cycle that lead from K(s) and ½Cl₂(g) to K⁺(g) and Cl⁻(g), giving an equation for each. [4]
  2. (b) State which of the steps in (a) are exothermic. [1]
  3. (c) Given ΔHf⦵ KCl = −437, ΔHat K = +89, IE K = +419, ΔHat Cl = +122 and EA Cl = −349 (all kJ/mol), calculate the lattice energy of KCl. [1]
Show answer
  1. (a) atomisation of K: K(s) → K(g); 1st IE of K: K(g) → K⁺(g) + e⁻; atomisation of Cl: ½Cl₂(g) → Cl(g); 1st EA of Cl: Cl(g) + e⁻ → Cl⁻(g) (atomisation steps may come in either order before the ionisation/EA steps)
    B1 each step with correct equation
  2. (b) Only the first electron affinity of chlorine
    B1
  3. (c) −718 kJ/mol
    B1
Q6Lattice energy & Born–Haber cycles · Standard · Paper 4 style · 4 marks

Data (kJ/mol): ΔHf⦵ MgO(s) −602; ΔHat⦵ Mg +148; 1st IE Mg +738; 2nd IE Mg +1451; ΔHat⦵ O (½O₂ → O) +249; 1st EA O −141; 2nd EA O +798.

  1. (a) Calculate the lattice energy of magnesium oxide. [3]
  2. (b) Suggest why magnesium oxide is used to line the inside of furnaces. [1]
Show answer
  1. (a) −3845 kJ/mol
    M1 sum of steps from elements to gaseous ions = 148 + 738 + 1451 + 249 − 141 + 798 = +3243M1 LE = −602 − 3243A1 −3845 kJ/mol
  2. (b) Its very exothermic lattice energy (strong attraction between 2+ and 2− ions) gives it a very high melting point.
    B1
Q7Lattice energy & Born–Haber cycles · Standard · Paper 4 style · 4 marks

Calcium chloride is used to melt ice on roads. Data (kJ/mol): ΔHf⦵ CaCl₂(s) −796; ΔHat⦵ Ca +178; 1st IE Ca +590; 2nd IE Ca +1145; ΔHat⦵ Cl (½Cl₂ → Cl) +122; 1st EA Cl −349.

  1. (a) Calculate the lattice energy of calcium chloride. [3]
  2. (b) State the two ionisation energy equations included in the cycle. [1]
Show answer
  1. (a) −2255 kJ/mol
    M1 atomisation of Cl and EA each ×2: 2 × 122 = 244; 2 × (−349) = −698M1 LE = −796 − 178 − 590 − 1145 − 244 + 698A1 −2255 kJ/mol
  2. (b) Ca(g) → Ca⁺(g) + e⁻; Ca⁺(g) → Ca²⁺(g) + e⁻
    B1 both
Q8Lattice energy & Born–Haber cycles · Standard · Paper 4 style · 4 marks

Data (kJ/mol): ΔHf⦵ KBr(s) −394; ΔHat⦵ K +89; 1st IE K +419; ΔHat⦵ Br (½Br₂(l) → Br(g)) +112; 1st EA Br −325.

  1. (a) Calculate the lattice energy of potassium bromide. [2]
  2. (b) The Br–Br bond energy is 193 kJ/mol. Explain why the enthalpy change of atomisation of bromine (+112 kJ/mol) is greater than ½ × 193. [2]
Show answer
  1. (a) −689 kJ/mol
    M1 LE = −394 − 89 − 419 − 112 + 325A1 −689 kJ/mol
  2. (b) Bromine is a liquid in its standard state, so atomisation also includes vaporisation (overcoming London dispersion forces between Br₂ molecules) as well as breaking half a mole of Br–Br bonds.
    M1 standard state is liquidM2 energy needed to vaporise / overcome intermolecular forces
Q9Lattice energy & Born–Haber cycles · Standard · Paper 4 style · 4 marks

Lattice energies (kJ/mol): NaF −918; NaCl −787; NaBr −747.

  1. (a) Explain the trend in lattice energy from NaF to NaBr. [2]
  2. (b) Explain why the lattice energy of MgO (−3845 kJ/mol) is much more exothermic than that of NaF. [2]
Show answer
  1. (a) The anion radius increases from F⁻ to Br⁻, so the distance between ion centres increases and the electrostatic attraction becomes weaker; the lattice energy becomes less exothermic.
    M1 anion size increases (same charge)M2 weaker attraction, less exothermic
  2. (b) Mg²⁺ and O²⁻ have double the charges of Na⁺ and F⁻ (and Mg²⁺ is smaller), so the attraction between the ions is much stronger.
    M1 higher charges on both ionsM2 stronger electrostatic attraction (Mg²⁺ smaller)
Q10Lattice energy & Born–Haber cycles · Standard · Paper 4 style · 4 marks

Data (kJ/mol): ΔHf⦵ SrO(s) −592; 1st IE Sr +550; 2nd IE Sr +1064; ΔHat⦵ O +249; 1st EA O −141; 2nd EA O +798; lattice energy SrO −3276.

  1. (a) Use the data to calculate the enthalpy change of atomisation of strontium. [3]
  2. (b) Suggest why the lattice energy of SrO is less exothermic than that of CaO (−3454 kJ/mol). [1]
Show answer
  1. (a) +164 kJ/mol
    M1 ΔHf = ΔHat(Sr) + IE1 + IE2 + ΔHat(O) + EA1 + EA2 + LEM1 ΔHat(Sr) = −592 − 550 − 1064 − 249 + 141 − 798 + 3276A1 +164 kJ/mol
  2. (b) Sr²⁺ is larger than Ca²⁺, so the ions are further apart and attract less strongly.
    B1
Q11Lattice energy & Born–Haber cycles · Standard · Paper 4 style · 4 marks

Data (kJ/mol): ΔHf⦵ MgF₂(s) −1124; ΔHat⦵ Mg +148; 1st IE Mg +738; 2nd IE Mg +1451; ΔHat⦵ F +79; 1st EA F −328.

  1. (a) Calculate the lattice energy of magnesium fluoride. [2]
  2. (b) The lattice energy of MgO is −3845 kJ/mol. Predict which of MgO and MgF₂ has the higher melting point and explain your answer. [2]
Show answer
  1. (a) −2963 kJ/mol
    M1 −1124 − 148 − 738 − 1451 − 2(79) − 2(−328)A1 −2963 kJ/mol
  2. (b) MgO: O²⁻ has twice the charge of F⁻, so stronger attraction between ions; more energy needed to separate them.
    B1 MgOB1 O²⁻ double charge → stronger ionic attraction / more exothermic lattice energy
Q12Lattice energy & Born–Haber cycles · Challenge · Paper 4 style · 5 marks

Data (kJ/mol): ΔHf⦵ CaO(s) −635; ΔHat⦵ Ca +178; 1st IE Ca +590; 2nd IE Ca +1145; ΔHat⦵ O +249; 1st EA O −141; lattice energy CaO −3454.

  1. (a) Use the data to calculate the second electron affinity of oxygen. [3]
  2. (b) Explain why the second electron affinity of oxygen is endothermic. [2]
Show answer
  1. (a) +798 kJ/mol
    M1 ΔHf = ΔHat(Ca) + IE1 + IE2 + ΔHat(O) + EA1 + EA2 + LEM1 EA2 = −635 − 178 − 590 − 1145 − 249 + 141 + 3454A1 +798 kJ/mol
  2. (b) The electron is being added to a negative ion, O⁻, so energy is needed to overcome the repulsion between the O⁻ ion and the incoming electron.
    M1 adding electron to negative ionM2 repulsion must be overcome
Q13Lattice energy & Born–Haber cycles · Challenge · Paper 4 style · 7 marks

Data (kJ/mol): ΔHat⦵ Mg +148; 1st IE Mg +738; 2nd IE Mg +1451; ΔHat⦵ Cl +122; 1st EA Cl −349.

  1. (a) Calculate ΔHf⦵ for the hypothetical compound MgCl(s), taking its lattice energy as −753 kJ/mol. [2]
  2. (b) Calculate ΔHf⦵ for MgCl₂(s), whose lattice energy is −2526 kJ/mol. [2]
  3. (c) Use your answers to explain why magnesium forms MgCl₂ and not MgCl, even though forming Mg²⁺ needs much more energy than forming Mg⁺. [2]
  4. (d) Suggest why MgCl₃ does not form. [1]
Show answer
  1. (a) −94 kJ/mol
    M1 148 + 738 + 122 − 349 − 753A1 −94 kJ/mol
  2. (b) −643 kJ/mol
    M1 148 + 738 + 1451 + 2(122) + 2(−349) − 2526A1 −643 kJ/mol
  3. (c) The much more exothermic lattice energy of MgCl₂ (2+ ion, smaller and more highly charged) more than compensates for the second ionisation energy, so MgCl₂ is far more energetically stable (ΔHf much more negative).
    M1 LE of MgCl₂ far more exothermic, outweighs IE2M2 ΔHf(MgCl₂) much more negative, so MgCl₂ is more stable
  4. (d) The third electron would be removed from an inner (2p) shell, so the third IE (about 7700 kJ/mol) is too large to be recovered by the lattice energy.
    B1
Q14Lattice energy & Born–Haber cycles · Challenge · Paper 4 style · 4 marks

Aluminium oxide is mined as bauxite. Data (kJ/mol): ΔHf⦵ Al₂O₃(s) −1676; ΔHat⦵ Al +326; IE of Al: 1st +578, 2nd +1817, 3rd +2745; ΔHat⦵ O +249; 1st EA O −141; 2nd EA O +798.

  1. (a) Write the equation for the lattice energy of aluminium oxide. [1]
  2. (b) Calculate the lattice energy of aluminium oxide. [3]
Show answer
  1. (a) 2Al³⁺(g) + 3O²⁻(g) → Al₂O₃(s)
    B1
  2. (b) −15 326 kJ/mol
    M1 2 × (326 + 578 + 1817 + 2745) = 10 932M1 3 × (249 − 141 + 798) = 2718A1 LE = −1676 − 10 932 − 2718 = −15 326 kJ/mol
Q15Lattice energy & Born–Haber cycles · Challenge · Paper 4 style · 5 marks

Data (kJ/mol): ΔHf⦵ CsCl(s) −443; ΔHat⦵ Cs +76; 1st IE Cs +376; ΔHat⦵ Cl +122; lattice energy CsCl −668.

  1. (a) Use the data to calculate the first electron affinity of chlorine. [3]
  2. (b) Suggest why the first ionisation energy of caesium is low, and explain why the lattice energy of CsCl (−668 kJ/mol) is less exothermic than that of NaCl (−787 kJ/mol). [2]
Show answer
  1. (a) −349 kJ/mol
    M1 cycle: ΔHf = ΔHat(Cs) + IE(Cs) + ΔHat(Cl) + EA(Cl) + LEM1 EA = −443 − 76 − 376 − 122 + 668A1 −349 kJ/mol
  2. (b) Cs has its outer electron in the 6th shell, far from the nucleus with a lot of shielding, so little energy is needed to remove it. Cs⁺ is much larger than Na⁺ (same charge), so the distance between ion centres is larger and the attraction between Cs⁺ and Cl⁻ is weaker.
    M1 outer electron far from nucleus / much shielding, so low IEM2 Cs⁺ larger than Na⁺, ions further apart, weaker attraction → less exothermic lattice energy
Q16Enthalpies of solution & hydration · Core · Paper 4 style · 5 marks
  1. (a) Define the standard enthalpy change of hydration of an ion. [2]
  2. (b) Define the standard enthalpy change of solution. [1]
  3. (c) Write equations, with state symbols, for (i) the enthalpy change of hydration of K⁺ and (ii) the enthalpy change of solution of NaCl. [2]
Show answer
  1. (a) The enthalpy change when one mole of gaseous ions dissolves in water to form an infinitely dilute solution (one mole of aqueous ions), under standard conditions.
    M1 one mole of gaseous ionsM2 dissolved in (excess) water to form aqueous ions
  2. (b) The enthalpy change when one mole of a solute dissolves in enough solvent to form an infinitely dilute solution, under standard conditions.
    B1
  3. (c) (i) K⁺(g) + aq → K⁺(aq) (ii) NaCl(s) + aq → Na⁺(aq) + Cl⁻(aq)
    B1 each
Q17Enthalpies of solution & hydration · Core · Paper 4 style · 3 marks

Data (kJ/mol): lattice energy NaCl −787; ΔHhyd⦵ Na⁺ −419; ΔHhyd⦵ Cl⁻ −364.

  1. (a) Describe, in words, an energy cycle linking the lattice energy, the hydration enthalpies and the enthalpy change of solution of NaCl. [1]
  2. (b) Calculate the enthalpy change of solution of NaCl. [2]
Show answer
  1. (a) NaCl(s) → Na⁺(g) + Cl⁻(g) (reverse of lattice energy, +787) → Na⁺(aq) + Cl⁻(aq) (sum of hydration enthalpies); this equals NaCl(s) → Na⁺(aq) + Cl⁻(aq) (ΔHsol).
    B1
  2. (b) +4 kJ/mol
    M1 ΔHsol = −LE + ΔHhyd(Na⁺) + ΔHhyd(Cl⁻) = 787 − 419 − 364A1 +4 kJ/mol
Q18Enthalpies of solution & hydration · Core · Paper 4 style · 3 marks
  1. (a) Explain why the hydration enthalpy of F⁻ (−506 kJ/mol) is more exothermic than that of Cl⁻ (−364 kJ/mol). [2]
  2. (b) State the type of interaction between an ion and the water molecules around it. [1]
Show answer
  1. (a) F⁻ is smaller than Cl⁻ (same charge), so it has a higher charge density and attracts the δ+ hydrogen atoms of water molecules more strongly.
    M1 F⁻ smaller / higher charge densityM2 stronger attraction to δ+ H of water
  2. (b) Ion–dipole attraction
    B1
Q19Enthalpies of solution & hydration · Standard · Paper 4 style · 5 marks

Hydration enthalpies (kJ/mol): Li⁺ −519; Na⁺ −419; K⁺ −322; Mg²⁺ −1920.

  1. (a) Explain the trend in hydration enthalpy from Li⁺ to K⁺. [2]
  2. (b) Explain why the hydration enthalpy of Mg²⁺ is much more exothermic than that of Na⁺. [2]
  3. (c) Describe how water molecules are arranged around a cation and around an anion. [1]
Show answer
  1. (a) Ionic radius increases from Li⁺ to K⁺ (same charge), so charge density falls and the attraction between the ion and the δ− oxygen of water molecules becomes weaker; hydration is less exothermic.
    M1 ionic radius increases / charge density decreasesM2 weaker attraction to (O of) water molecules
  2. (b) Mg²⁺ has twice the charge and a smaller radius, so a much higher charge density and stronger attraction for water molecules.
    M1 higher charge and smallerM2 stronger ion–dipole attraction
  3. (c) Around a cation the δ− oxygen atoms point towards the ion; around an anion the δ+ hydrogen atoms point towards the ion.
    B1
Q20Enthalpies of solution & hydration · Standard · Paper 4 style · 4 marks

Data (kJ/mol): enthalpy change of solution of KCl +17.2; lattice energy KCl −703; ΔHhyd⦵ K⁺ −322.

  1. (a) Calculate the hydration enthalpy of the chloride ion. Give your answer to 3 significant figures. [3]
  2. (b) Predict whether the temperature rises or falls when KCl dissolves in water. [1]
Show answer
  1. (a) −364 kJ/mol
    M1 ΔHsol = −LE + ΔHhyd(K⁺) + ΔHhyd(Cl⁻)M1 ΔHhyd(Cl⁻) = 17.2 + (−703) − (−322)A1 −364 kJ/mol
  2. (b) Falls (ΔHsol is positive / endothermic).
    B1
Q21Enthalpies of solution & hydration · Standard · Paper 5 style · 6 marks

A student investigates instant cold packs used by sports teams. 8.00 g of ammonium nitrate is added to 50.0 g of water in a polystyrene cup. The temperature falls from 21.5 °C to 10.3 °C.

  1. (a) Calculate the heat energy absorbed from the water. [c = 4.18 J g⁻¹ K⁻¹] [1]
  2. (b) Calculate the enthalpy change of solution of ammonium nitrate, in kJ/mol, including its sign. [Mr: NH₄NO₃, 80.0] [2]
  3. (c) State two assumptions made in your calculation. [2]
  4. (d) Suggest one improvement to the method that would make the result more accurate. [1]
Show answer
  1. (a) 2340 J (2.34 kJ)
    B1 50.0 × 4.18 × 11.2 = 2341 J
  2. (b) +23.4 kJ/mol
    M1 n = 8.00/80.0 = 0.100 molA1 +2.34/0.100 = +23.4 kJ/mol (positive sign)
  3. (c) The solution has the same specific heat capacity (and density) as water; mass of solid is ignored (mass of solution = 50.0 g); no heat is gained from the surroundings / cup has negligible heat capacity.
    B1 each for any two
  4. (d) Insulate the cup (lid, second cup) or take readings over time and extrapolate the cooling curve back to the time of mixing.
    B1
Q22Enthalpies of solution & hydration · Standard · Paper 4 style · 3 marks

Data (kJ/mol): lattice energy MgCl₂ −2526; ΔHhyd⦵ Mg²⁺ −1920; ΔHhyd⦵ Cl⁻ −364.

  1. (a) Calculate the enthalpy change of solution of magnesium chloride. [2]
  2. (b) Explain why the chloride hydration enthalpy must be multiplied by 2 in this calculation. [1]
Show answer
  1. (a) −122 kJ/mol
    M1 2526 + (−1920) + 2(−364)A1 −122 kJ/mol
  2. (b) One mole of MgCl₂ produces two moles of Cl⁻ ions.
    B1
Q23Enthalpies of solution & hydration · Standard · Paper 4 style · 4 marks

Data (kJ/mol): lattice energy CaCl₂ −2255; ΔHhyd⦵ Ca²⁺ −1650; ΔHhyd⦵ Cl⁻ −364.

  1. (a) Calculate the enthalpy change of solution of calcium chloride. [2]
  2. (b) 5.55 g of anhydrous CaCl₂ is dissolved in 100 g of water. Using your answer to (a), calculate the expected temperature rise. [Mr: CaCl₂, 111.1; c = 4.18 J g⁻¹ K⁻¹; assume the solution has mass 100 g] [2]
Show answer
  1. (a) −123 kJ/mol
    M1 2255 + (−1650) + 2(−364)A1 −123 kJ/mol
  2. (b) 14.7 °C
    M1 n = 0.0500 mol; heat released = 0.0500 × 123 000 = 6.14 × 10³ JA1 ΔT = 6145/(100 × 4.18) = 14.7 °C
Q24Enthalpies of solution & hydration · Challenge · Paper 4 style · 5 marks

Data (kJ/mol): enthalpy change of solution of LiF +4.9; ΔHhyd⦵ Li⁺ −519; ΔHhyd⦵ F⁻ −506.

  1. (a) Calculate the lattice energy of lithium fluoride from these data. Give your answer to 3 significant figures. [3]
  2. (b) Lithium fluoride is only slightly soluble in water, but lithium chloride is very soluble. Suggest an explanation in terms of the enthalpy changes involved. [2]
Show answer
  1. (a) −1030 kJ/mol
    M1 ΔHsol = −LE + ΔHhyd(Li⁺) + ΔHhyd(F⁻)M1 LE = −519 + (−506) − 4.9A1 −1030 kJ/mol
  2. (b) F⁻ is very small, so the lattice energy of LiF is very exothermic; the large hydration enthalpies of Li⁺ and F⁻ only just balance it (ΔHsol slightly positive). For LiCl the lattice energy falls more (larger Cl⁻) than the hydration enthalpy, so ΔHsol is negative.
    M1 LiF lattice energy very large (small ions)M2 for LiCl lattice energy decreases more than hydration enthalpy, so dissolving is more favourable
Q25Enthalpies of solution & hydration · Challenge · Paper 4 style · 5 marks

Data (kJ/mol): lattice energy MgSO₄ −2833, BaSO₄ −2469; ΔHhyd⦵ Mg²⁺ −1920, Ba²⁺ −1360, SO₄²⁻ −1059.

  1. (a) Calculate the enthalpy change of solution of MgSO₄ and of BaSO₄. [2]
  2. (b) Use the data to explain why the solubility of the Group 2 sulfates decreases down the group. [3]
Show answer
  1. (a) MgSO₄ −146 kJ/mol; BaSO₄ +50 kJ/mol
    B1 2833 − 1920 − 1059 = −146B1 2469 − 1360 − 1059 = +50
  2. (b) The sulfate ion is large, so the lattice energy changes only a little as the cation gets bigger (−2833 to −2469, a change of 364). The hydration enthalpy of the cation falls much more (−1920 to −1360, a change of 560). So ΔHsol becomes less exothermic / more endothermic down the group and the sulfates become less soluble.
    M1 lattice energy decreases only slightly (large SO₄²⁻ dominates the inter-ionic distance)M1 cation hydration enthalpy decreases much moreM1 ΔHsol becomes more endothermic → less soluble
Q26Enthalpies of solution & hydration · Challenge · Paper 5 style · 7 marks

A student dissolves different masses of potassium nitrate, KNO₃, in 50.0 g portions of water and records the temperature fall.mass of KNO₃ / g: 2.02, 4.04, 6.06, 8.08temperature fall / °C: 3.2, 6.5, 8.1, 13.0

  1. (a) Identify the anomalous result and suggest one reason for it. [2]
  2. (b) Use the result for 8.08 g to calculate the enthalpy change of solution of potassium nitrate. [Mr: KNO₃, 101.1; c = 4.18 J g⁻¹ K⁻¹] [3]
  3. (c) The data-book value is +34.9 kJ/mol. Suggest why the experimental value is lower and how this error could be reduced. [2]
Show answer
  1. (a) 6.06 g (ΔT 8.1 °C), which is well below the trend (about 9.8 °C expected); e.g. not all the solid dissolved, or the minimum temperature was missed.
    B1 6.06 g resultB1 sensible reason
  2. (b) +34.0 kJ/mol
    M1 q = 50.0 × 4.18 × 13.0 = 2717 JM1 n = 8.08/101.1 = 0.0799 molA1 +34.0 kJ/mol (positive)
  3. (c) Heat is gained from the surroundings while the solution is below room temperature, so the measured ΔT is too small. Use a lid/insulation, or plot temperature against time and extrapolate to find the true minimum.
    M1 heat gained from surroundingsM2 insulation / extrapolation of cooling curve
Q27Enthalpies of solution & hydration · Challenge · Paper 4 style · 4 marks

Data (kJ/mol): ΔHf⦵ NaCl(s) −411; ΔHat⦵ Na +107; 1st IE Na +496; ΔHat⦵ Cl +122; 1st EA Cl −349; enthalpy change of solution of NaCl +3.9; ΔHhyd⦵ Cl⁻ −364.

  1. (a) Calculate the lattice energy of NaCl. [1]
  2. (b) Hence calculate the hydration enthalpy of Na⁺. Give your answer to 3 significant figures. [2]
  3. (c) Calculate the enthalpy change for Na(s) + ½Cl₂(g) → Na⁺(aq) + Cl⁻(aq). [1]
Show answer
  1. (a) −787 kJ/mol
    B1 −411 − 107 − 496 − 122 + 349
  2. (b) −419 kJ/mol
    M1 ΔHhyd(Na⁺) = ΔHsol + LE − ΔHhyd(Cl⁻) = 3.9 + (−787) − (−364)A1 −419 kJ/mol
  3. (c) −407 kJ/mol
    B1 −411 + 3.9 = −407.1
Q28Electron affinity & ionisation energy (A Level) · Core · Paper 4 style · 4 marks
  1. (a) Define the first electron affinity of an element. [2]
  2. (b) Write the equation for the first electron affinity of oxygen. [1]
  3. (c) Explain why the first electron affinity of oxygen is exothermic. [1]
Show answer
  1. (a) The enthalpy change when one electron is added to each atom in one mole of gaseous atoms to form one mole of gaseous 1− ions.
    M1 one mole of gaseous atoms each gaining one electronM2 forming gaseous 1− ions
  2. (b) O(g) + e⁻ → O⁻(g)
    B1
  3. (c) The incoming electron is attracted by the nucleus (effective nuclear charge), so energy is released.
    B1
Q29Electron affinity & ionisation energy (A Level) · Standard · Paper 4 style · 4 marks

First electron affinities (kJ/mol): F −328; Cl −349; Br −325; I −295.

  1. (a) Explain the general trend from chlorine to iodine. [2]
  2. (b) Suggest why fluorine's first electron affinity is less exothermic than chlorine's. [1]
  3. (c) The second electron affinity of sulfur is +640 kJ/mol. Calculate the enthalpy change for S(g) + 2e⁻ → S²⁻(g) given that the first electron affinity of sulfur is −200 kJ/mol. [1]
Show answer
  1. (a) The atoms get larger with more shielding, so the incoming electron is further from the nucleus and attracted less strongly; less energy is released.
    M1 larger atom / more shieldingM2 weaker attraction for added electron
  2. (b) F is very small, so the added electron is repelled strongly by the electrons already in the small, crowded 2p sub-shell.
    B1
  3. (c) +440 kJ/mol
    B1
Q30Electron affinity & ionisation energy (A Level) · Challenge · Paper 4 style · 4 marks

Data (kJ/mol): 1st IE Mg +738; 2nd IE Mg +1451; 1st EA O −141; 2nd EA O +798.

  1. (a) Calculate the enthalpy change for O(g) + 2e⁻ → O²⁻(g). [1]
  2. (b) Calculate the total enthalpy change for Mg(g) + O(g) → Mg²⁺(g) + O²⁻(g). [1]
  3. (c) Forming the gaseous ions is very endothermic, yet magnesium burns vigorously in oxygen to form MgO. Explain why. [2]
Show answer
  1. (a) +657 kJ/mol
    B1 −141 + 798
  2. (b) +2846 kJ/mol
    B1 738 + 1451 + 657
  3. (c) The lattice energy of MgO is very exothermic (small, doubly charged ions attract strongly), which more than compensates for the energy needed to form the ions, so ΔHf of MgO is very negative.
    M1 very exothermic lattice energyM2 outweighs the endothermic steps so overall ΔHf negative

Similar questions

Questions that test the same subtopic in different years, grouped so you can work on one skill at a time.

References only: find each question in its paper (official link where available, otherwise through your school).

Lattice energy & Born–Haber cycles Banker23 questions · 8/9 sessions
Enthalpies of solution & hydration Banker18 questions · 8/9 sessions
Electron affinity & ionisation energy (A Level) 7 questions · 4/9 sessions

Subtopics

Our analysis of how often each subtopic was examined and how many marks it carried, 2022 to May/June 2026.

SubtopicMarks indexedSessions
Lattice energy & Born–Haber cycles Banker1038/9
Enthalpies of solution & hydration Banker828/9
Electron affinity & ionisation energy (A Level) 184/9

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